Reputation: 29872

How to detect if a script is being sourced

I have a script where I do not want it to call exit if it's being sourced.

I thought of checking if $0 == bash but this has problems if the script is sourced from another script, or if the user sources it from a different shell like ksh.

Is there a reliable way of detecting if a script is being sourced?

Upvotes: 361

Answers (28)

Jim

Reputation: 111

This ancient question still has no complete, robust, reliable, answer. (Though kudos to the late, highly detailed 2022 answer - but even it misses at least one common edge case - i.e. an sh script sourcing another script.)

The fundamental problem that all of these answers struggle at great length and complexity with - by trying to solve programmatically - is to compare the name or ID of the calling script or process, with that of the "this" script.

But the various shell environments make it all but impossible to do so cleanly, consistently, and/or reliably. Especially when the script may be sourced from another script.

But there's IS a dead-simple workaround for most if not all shells:

Just don't do it programmatically!

You just have to accept hard-coding the name of your script, within the script itself. (Which I know, feels like feeding your dog to itself. And it will make you pause whenever renaming any script in the future.)

Then just compare that hard-coded value, to what the environment thinks it is. If it's the same, then the script was invoked directly. If it's different, then you know it was sourced - exactly how (via shell or another script), well for this question and for most purposes, it isn't really the point.

It's "inelegant" because as programmers (and/or scripters), we don't like hard-coding values that SHOULD be obtainable via reflection of some sort. But let's face it - all this gnashing of teeth is because that's just not in the cards in a reliable way across shells and circumstances.

Here's an example:

# The ugly bit we don't want to do but avoids so much of this pain
meName="my-script-name.sh"


# The test. Works are sh, bash, dash - possibly others? And on all platforms.
isSourced=0
[ "${0##*/}" != "${meName}" ] && isSourced=1


# In use
if [ $isSourced = 1 ]; then
    echo "I'm sourced."
echo
    echo "I'm NOT sourced."
fi

Upvotes: 1

Wechar Yu

Reputation: 337

Even though this question was raised some time ago, I'd like to share a solution that is compatible with almost all shells.

Edit: support macOS

#!/bin/sh

if [ -n "$BASH_VERSION" ]; then
    # Replace `$0` with `BASH_SOURCE[0]` in Bash because it returns the shell when using `source`.
    SCRIPT_PATH=${BASH_SOURCE[0]}
else
    SCRIPT_PATH="$0"
fi

OS=$(uname -s)
if [[ $OS == "Linux" ]]; then
  CMD=$(cat /proc/$$/cmdline)
elif [[ $OS == "Darwin" ]]; then
  CMD=$(ps -p $$ -o command=)
else
  echo "Unsupport kernel: $OS"
  exit 1
fi

# source command does not start a new process, so the cmdline would not contains current script name.
if [[ ${CMD} == *"$SCRIPT_PATH"* ]]; then
    echo "This script should only be sourced" >&2
    exit 1
fi

The key point is that the source command is executed in current shell, so the command line is not recorded in the /proc/{process_id}/cmdline file. In contrast, executing a script spawns a new shell, which will record the command line.

Upvotes: 0

Kim Taylor

Reputation: 356

I upvoted a couple of very good answers and ended up with these 2 lines for bash:

function do_not_source { [[ "${FUNCNAME[-1]}" != "source" ]]; }
do_not_source || return 2

Upvotes: 0

Gabriel Staples

Reputation: 52569

The most beautiful way to detect if a Bash script is being executed or sourced (imported)

I really think this is the most beautiful way to do it:

From my if__name__==__main___check_if_sourced_or_executed_best.sh file in my eRCaGuy_hello_world repo:

#!/usr/bin/env bash

main() {
    echo "Running main."
    # Add your main function code here
}

if [ "${BASH_SOURCE[0]}" = "$0" ]; then
    # This script is being run.
    __name__="__main__"
else
    # This script is being sourced.
    __name__="__source__"
fi

# Only run `main` if this script is being **run**, NOT sourced (imported)
if [ "$__name__" = "__main__" ]; then
    echo "This script is being run."
    main "$@"
else
    echo "This script is being sourced."
fi

References:

See also my other answer here for additional details on the above technique, including showing the run output: What is the bash equivalent to Python's if __name__ == '__main__'?
This answer where I first learned about "${BASH_SOURCE[0]}" = "$0"

You can also explore the following alternatives if you like, but I prefer to use the code chunk above.

Important: Using the "${FUNCNAME[-1]}" technique does not properly handle nested scripts, where one script calls or sources another, whereas the if [ "${BASH_SOURCE[0]}" = "$0" ] technique does. That's another huge reason to use if [ "${BASH_SOURCE[0]}" = "$0" ] instead.

4 ways to determine whether a bash script is being sourced or executed

I have read a bunch of answers all over the place on this and a few other questions, and have come up with 4 ways I'd like to summarize and put in one place.

if __name__ == "__main__":

See: What does if __name__ == "__main__": do? for what that does in Python.

You can see a full demonstration of all 4 techniques below in my check_if_sourced_or_executed.sh script in my eRCaGuy_hello_world repo.
You can see one of the techniques in-use in my advanced bash program with help menu, argument parsing, main function, automatic execute vs source detection (akin to if __name__ == "__main__": in Python), etc, see my demo/template program in this list here. It is currently called argument_parsing__3_advanced__gen_prog_template.sh, but if that name changes in the future I'll update it in the list at the link just above

Anyway, here are the 4 Bash techniques:

Technique 1 (can be placed anywhere; handles nested scripts): See: https://unix.stackexchange.com/questions/424492/how-to-define-a-shell-script-to-be-sourced-not-run/424495#424495
```
if [ "${BASH_SOURCE[0]}" -ef "$0" ]; then
    echo "  This script is being EXECUTED."
    run="true"
else
    echo "  This script is being SOURCED."
fi
```
Technique 2 [My favorite technique] (can be placed anywhere; handles nestes scripts): See this type of technique in-use in my most-advanced bash demo script yet, here: argument_parsing__3_advanced__gen_prog_template.sh, near the bottom.

Modified from: What is the bash equivalent to Python's `if __name__ == '__main__'`?
```
if [ "${BASH_SOURCE[0]}" == "$0" ]; then
    echo "  This script is being EXECUTED."
    run="true"
else
    echo "  This script is being SOURCED."
fi
```

Technique 3 (requires another line which MUST be outside all functions): Modified from: How to detect if a script is being sourced

# A. Place this line OUTSIDE all functions:
(return 0 2>/dev/null) && script_is_being_executed="false" || script_is_being_executed="true"

# B. Place these lines anywhere
if [ "$script_is_being_executed" == "true" ]; then
    echo "  This script is being EXECUTED."
    run="true"
else
    echo "  This script is being SOURCED."
fi

Technique 4 [Limitation: does not handle nested scripts!] (MUST be inside a function): Modified from: How to detect if a script is being sourced
and Unix & Linux: How to define a shell script to be sourced not run.
```
if [ "${FUNCNAME[-1]}" == "main" ]; then
    echo "  This script is being EXECUTED."
    run="true"
elif [ "${FUNCNAME[-1]}" == "source" ]; then
    echo "  This script is being SOURCED."
else
    echo "  ERROR: THIS TECHNIQUE IS BROKEN"
fi
```
1. This is where I first learned about the ${FUNCNAME[-1]} trick: @mr.spuratic: How to detect if a script is being sourced - he learned it from Dennis Williamson apparently.

Explanations

bash

(return 0 2>/dev/null) && sourced=1 || sourced=0

^{Note: The technique was adapted from user5754163's answer, as it turned out to be more robust than the original solution, [[ $0 != "$BASH_SOURCE" ]] && sourced=1 || sourced=0[1]}

Bash allows return statements only from functions and, in a script's top-level scope, only if the script is sourced.
- If return is used in the top-level scope of a non-sourced script, an error message is emitted, and the exit code is set to 1.
(return 0 2>/dev/null) executes return in a subshell and suppresses the error message; afterwards the exit code indicates whether the script was sourced (0) or not (1), which is used with the && and || operators to set the sourced variable accordingly.
- Use of a subshell is necessary, because executing return in the top-level scope of a sourced script would exit the script.
- ^{Tip of the hat to @Haozhun, who made the command more robust by explicitly using 0 as the return operand; he notes: per bash help of return [N]: "If N is omitted, the return status is that of the last command." As a result, the earlier version [which used just return, without an operand]
  produces incorrect result if the last command on the user's shell has a non-zero return value.}

ksh

[[ "$(cd -- "$(dirname -- "$0")" && pwd -P)/$(basename -- "$0")" != "$(cd -- "$(dirname -- "${.sh.file}")" && pwd -P)/$(basename -- "${.sh.file}")" ]] && sourced=1 || sourced=0

Special variable ${.sh.file} is somewhat analogous to $BASH_SOURCE; note that ${.sh.file} causes a syntax error in bash, zsh, and dash, so be sure to execute it conditionally in multi-shell scripts.

Unlike in bash, $0 and ${.sh.file} are NOT guaranteed to be the same - at different times either may be a relative path or mere file name, while the other may be a full one; therefore, both $0 and ${.sh.file} must be resolved to full paths before comparing. If the full paths differ, sourcing is implied.

zsh

[[ $ZSH_EVAL_CONTEXT =~ :file$) ]] && sourced=1 || sourced=0

$ZSH_EVAL_CONTEXT contains information about the evaluation context: substring file, separated with :, is only present if the script is being sourced.

In a sourced script's top-level scope, $ZSH_EVAL_CONTEXT ends with :file, and that's what this test is limited to. Inside a function, :shfunc is appended to :file; inside a command substitution, :cmdsubst, is appended.

Using POSIX features only

If you're willing to make certain assumptions, you can make a reasonable, but not fool-proof guess as to whether your script is being sourced, based on knowing the binary filenames of the shells that may be executing your script.
Notably, this means that this approach doesn't detect the case when your script is being sourced by another script.

^{The section "How to handle sourced invocations" in this answer discusses the edge cases that cannot be handled with POSIX features only in detail.}

Examining the binary filename relies on the standard behavior of $0, which zsh, for instance, does not exhibit.

Thus, the safest approach is to combine the robust, shell-specific methods above - which do not rely on $0 - with a $0-based fallback solution for all remaining shells.

In short: The following solution:

in the shells that are covered with shell-specific tests: works robustly.
in all other shells: works only as expected when the script is being sourced directly from such a shell, as opposed to from another script.

^{Tip of the hat to Stéphane Desneux and his answer for the inspiration (transforming my cross-shell statement expression into a sh-compatible if statement and adding a handler for other shells).}

sourced=0
if [ -n "$ZSH_VERSION" ]; then 
  case $ZSH_EVAL_CONTEXT in *:file) sourced=1;; esac
elif [ -n "$KSH_VERSION" ]; then
  [ "$(cd -- "$(dirname -- "$0")" && pwd -P)/$(basename -- "$0")" != "$(cd -- "$(dirname -- "${.sh.file}")" && pwd -P)/$(basename -- "${.sh.file}")" ] && sourced=1
elif [ -n "$BASH_VERSION" ]; then
  (return 0 2>/dev/null) && sourced=1 
else # All other shells: examine $0 for known shell binary filenames.
  # Detects `sh` and `dash`; add additional shell filenames as needed.
  case ${0##*/} in sh|-sh|dash|-dash) sourced=1;; esac
fi

Note that, for robustness, each shell binary filename (e.g. sh) is represented twice - once as-is and a second time, prefixed with -. This is to account for environments, such as macOS, where interactive shells are launched as login shells with a custom $0 value that is the (path-less) shell filename prefixed with -.^{Thanks, t7e.} (While sh and dash are perhaps unlikely to be used as interactive shells, others that you may need to add to the list may.)

^{[1] user1902689 discovered that [[ $0 != "$BASH_SOURCE" ]] yields a false positive when you execute a script located in the $PATH by passing its mere filename to the bash binary; e.g., bash my-script, because $0 is then just my-script, whereas $BASH_SOURCE is the full path. While you normally wouldn't use this technique to invoke scripts in the $PATH - you'd just invoke them directly (my-script) - it is helpful when combined with -x for debugging.}

Upvotes: 318

t7e

Reputation: 509

A small addition to the @mklement0 answer. This is the custom function I used in my script to determine whether it is sourced or not:

replace_shell(){
   if [ -n "$ZSH_EVAL_CONTEXT" ]; then 
      case $ZSH_EVAL_CONTEXT in *:file*) echo "Zsh is sourced";; esac
   else
      case ${0##*/} in sh|dash|bash) echo "Bash is sourced";; esac
   fi
}

In a function, the output of "$ZSH_EVAL_CONTEXT" for zsh is toplevel:file:shfunc and not just toplevel:file during sorcing; thus, *:file* should fix this issue.

Upvotes: 1

usretc

Reputation: 824

Not exactly what the OP wanted, but I often find myself needing to source a script just to load its functions (i.e. as a library). For example, for benchmarking or testing purposes.

Here's a design that works in all shells (including POSIX):

Wrap all your top-level actions in a run_main() function.
Have your sourced script check for an initial --no-run argument which doesn't perform any actions; without --no-run, it can call run_main.
source the script using:

set -- --no-run "$@"
. script.sh
shift

The problem with . or source is that it's impossible to pass arguments to the script portably. POSIX shells ignore arguments to . and pass the caller's "$@" no matter what.

Upvotes: 2

Kaz

Reputation: 58568

The fix for this issue is not to write code that needs to know such a thing in order to behave correctly. And the way to do that is to put the code into a function, and not into the mainline of a script that needs to be sourced.

Code inside a function can just return 0 or return 1. This terminates just the function, so that control returns to whatever invoked the function.

This works whether the function is called from the mainline of a sourced script, from the mainline of a top-level script, or from another function.

Use sourcing to bring in "library" scripts that only define functions and perhaps variables, but don't actually execute any other top-level commands:

. path/to/lib.sh # defines libfunction
libfunction arg

or else:

path/to/script.sh arg # call script as a child process

and not:

. path/to/script.sh arg  # shell programming anti-pattern

Upvotes: 1

BuvinJ

Reputation: 11046

This is a spin off from some other answers, regarding "universal" cross shell support. This is admittedly very similar to https://stackoverflow.com/a/2942183/3220983 in particular, though slightly different. The weakness with this, is that a client script must respect how to use it (i.e. by exporting a variable first). The strength is that this is simple and should work "anywhere". Here's a template for your cut & paste pleasure:

# NOTE: This script may be used as a standalone executable, or callable library.
# To source this script, add the following *prior* to including it:
# export ENTRY_POINT="$0"

main()
{
    echo "Running in direct executable context!"
}

if [ -z "${ENTRY_POINT}" ]; then main "$@"; fi

Note: I use export just be sure this mechanism can be extended into sub processes.

Upvotes: 0

Tino

Reputation: 10459

FWIW, after reading all of the other answers, I came up with following solution for me:

Update: Actually, somebody spotted a since-corrected error in another answer which affected mine, too. I think the update here also is an improvement (see edits if you are curious).

This works for all scripts, which start with #!/bin/bash but might be sourced by different shells as well to learn some information (like settings) which is are kept outside the main function.

According to the comments below, this answer here apparently does not work for all bash variants. Also not for systems, where /bin/sh is based on bash. I. E. it fails for bash v3.x on MacOS. (Currenty I do not know how to solve this.)

#!/bin/bash

# Function definitions (API) and shell variables (constants) go here
# (This is what might be interesting for other shells, too.)

# this main() function is only meant to be meaningful for bash
main()
{
# The script's execution part goes here
}

BASH_SOURCE=".$0" # cannot be changed in bash
test ".$0" != ".$BASH_SOURCE" || main "$@"

Instead of the last 2 lines you can use following (in my opinion less readable) code to not set BASH_SOURCE in other shells and allow set -e to work in main:

if ( BASH_SOURCE=".$0" && exec test ".$0" != ".$BASH_SOURCE" ); then :; else main "$@"; fi

This script-recipe has following properties:

If executed by bash the normal way, main is called. Please note that this does not include a call like bash -x script (where script does not contain a path), see below.
If sourced by bash, main is only called, if the calling script happens to have the same name. (For example, if it sources itself or via bash -c 'someotherscript "$@"' main-script args.. where main-script must be, what test sees as $BASH_SOURCE).
If sourced/executed/read/evaled by a shell other than bash, main is not called (BASH_SOURCE is always different to $0).
main is not called if bash reads the script from stdin, unless you set $0 to be the empty string like so: ( exec -a '' /bin/bash ) <script
If evaluated by bash with eval (eval "`cat script`" all quotes are important!) from within some other script, this calls main. If eval is run from commandline directly, this is similar to previous case, where the script is read from stdin. (BASH_SOURCE is blank, while $0 usually is /bin/bash if not forced to something completely different.)
If main is not called, it does return true ($?=0).
This does not rely on unexpected behavior (previously I wrote undocumented, but I found no documentation that you cannot unset nor alter BASH_SOURCE either):
- BASH_SOURCE is a bash reserved array. But allowing BASH_SOURCE=".$0" to change it would open a very dangerous can of worms, so my expectation is, that this must have no effect (except, perhaps, some ugly warning shows up in some future version of bash).
- There is no documentation that BASH_SOURCE works outside functions. However the opposite (that it only works in functions) is neither documented. The observation is, that it works (tested with bash v4.3 and v4.4, unfortunately I have no bash v3.x anymore) and that quite too many scripts would break, if $BASH_SOURCE stops working as observed. Hence my expectation is, that BASH_SOURCE stays as is for future versions of bash, too.
- In contrast (nice find, BTW!) consider ( return 0 ), which gives 0 if sourced and 1 if not sourced. This comes a bit unexpected not only for me , and (according to the readings there) POSIX says, that return from subshell is undefined behavior (and the return here is clearly from a subshell). Perhaps this feature eventually gets enough widespread use such that it can no more be changed, but AFAICS there is a much higher chance that some future bash version accidental changes the return behavior in that case.
Unfortunately bash -x script 1 2 3 does not run main. (Compare script 1 2 3 where script has no path). Following can be used as workaround:
- bash -x "`which script`" 1 2 3
- bash -xc '. script' "`which script`" 1 2 3
- That bash script 1 2 3 does not run main can be considered a feature.
Note that ( exec -a none script ) calls main (bash does not pass it's $0 to the script, for this you need to use -c as shown in the last point).

Thus, except for some some corner cases, main is only called, when the script is executed the usual way. Normally this is, what you want, especially because it lacks complex hard to understand code.

Note that it is very similar to the Python code:
if __name__ == '__main__': main()
Which also prevents calling of main, except for some corner cases, as you can import/load the script and enforce that __name__='__main__'

Why I think this is a good general way to solve the challenge

If you have something, which can be sourced by multiple shells, it must be compatible. However (read the other answers), as there is no (easy to implement) portable way to detect the sourceing, you should change the rules.

By enforcing that the script must be executed by /bin/bash, you exactly do this.

This solves all cases but following in which case the script cannot run directly:

/bin/bash is not installed or disfunctional (i. E. in a boot environment)
If you pipe it to a shell like in curl https://example.com/script | $SHELL
(Note: This is only true if your bash is recent enough. This recipe is reported to fail for certain variants. So be sure to check it works for your case.)

However I cannot think about any real reason where you need that and also the ability to source the exactly same script in parallel! Usually you can wrap it to execute the main by hand. Like that:

$SHELL -c '. script && main'
{ curl https://example.com/script && echo && echo main; } | $SHELL
$SHELL -c 'eval "`curl https://example.com/script`" && main'
echo 'eval "`curl https://example.com/script`" && main' | $SHELL

Notes

This answer would not have been possible without the help of all the other answers! Even the wrong ones - which initially made me posting this.
Update: Edited due to the new discoveries found in https://stackoverflow.com/a/28776166/490291

Upvotes: 21

Dennis Williamson

Reputation: 360035

This seems to be portable between Bash and Korn:

[[ $_ != $0 ]] && echo "Script is being sourced" || echo "Script is a subshell"

A line similar to this or an assignment like pathname="$_" (with a later test and action) must be on the first line of the script or on the line after the shebang (which, if used, should be for ksh in order for it to work under the most circumstances).

Upvotes: 89

user5754163

Reputation: 287

^{Editor's note: This answer's solution works robustly, but is bash-only. It can be streamlined to

(return 2>/dev/null).}

TL;DR

Try to execute a return statement. If the script isn't sourced, that will raise an error. You can catch that error and proceed as you need.

Put this in a file and call it, say, test.sh:

#!/usr/bin/env sh

# Try to execute a `return` statement,
# but do it in a sub-shell and catch the results.
# If this script isn't sourced, that will raise an error.
$(return >/dev/null 2>&1)

# What exit code did that give?
if [ "$?" -eq "0" ]
then
    echo "This script is sourced."
else
    echo "This script is not sourced."
fi

Execute it directly:

shell-prompt> sh test.sh
output: This script is not sourced.

Source it:

shell-prompt> source test.sh
output: This script is sourced.

For me, this works in zsh and bash.

Explanation

The return statement will raise an error if you try to execute it outside of a function or if the script is not sourced. Try this from a shell prompt:

shell-prompt> return
output: ...can only `return` from a function or sourced script

You don't need to see that error message, so you can redirect the output to dev/null:

shell-prompt> return >/dev/null 2>&1

Now check the exit code. 0 means OK (no errors occurred), 1 means an error occurred:

shell-prompt> echo $?
output: 1

You also want to execute the return statement inside of a sub-shell. When the return statement runs it . . . well . . . returns. If you execute it in a sub-shell, it will return out of that sub-shell, rather than returning out of your script. To execute in the sub-shell, wrap it in $(...):

shell-prompt> $(return >/dev/null 2>$1)

Now, you can see the exit code of the sub-shell, which should be 1, because an error was raised inside the sub-shell:

shell-prompt> echo $?
output: 1

Upvotes: 27

osexp2000

Reputation: 3165

I ended up with checking [[ $_ == "$(type -p "$0")" ]]

if [[ $_ == "$(type -p "$0")" ]]; then
    echo I am invoked from a sub shell
else
    echo I am invoked from a source command
fi

When use curl ... | bash -s -- ARGS to run remote script on-the-fly, the $0 will be just bash instead of normal /bin/bash when run actual script file, so I use type -p "$0" to show full path of bash.

test:

curl -sSL https://github.com/jjqq2013/bash-scripts/raw/master/common/relpath | bash -s -- /a/b/c/d/e /a/b/CC/DD/EE

source <(curl -sSL https://github.com/jjqq2013/bash-scripts/raw/master/common/relpath)
relpath /a/b/c/d/e /a/b/CC/DD/EE

wget https://github.com/jjqq2013/bash-scripts/raw/master/common/relpath
chmod +x relpath
./relpath /a/b/c/d/e /a/b/CC/DD/EE

Upvotes: 0

ivanleoncz

Reputation: 10005

Straight to the point: you must evaluate if the variable "$0" is equal to the name of your Shell.

Like this:

#!/bin/bash

echo "First Parameter: $0"
echo
if [[ "$0" == "bash" ]] ; then
    echo "The script was sourced."
else
    echo "The script WAS NOT sourced."
fi

Via SHELL:

$ bash check_source.sh 
First Parameter: check_source.sh

The script WAS NOT sourced.

Via SOURCE:

$ source check_source.sh
First Parameter: bash

The script was sourced.

It's pretty hard to have a 100% portable way of detecting if a script was sourced or not.

Regarding my experience (7 years with Shellscripting), the only safe way (not relying on environment variables with PIDs and so on, which is not safe due to the fact that it is something VARIABLE), you should:

extend the possibilities from your if
using switch/case, if you want to.

Both options cannot be auto scaled, but it is the safer way.

For example:

when you source a script via an SSH session, the value returned by the variable "$0" (when using source), is -bash.

#!/bin/bash

echo "First Parameter: $0"
echo
if [[ "$0" == "bash" || "$0" == "-bash" ]] ; then
    echo "The script was sourced."
else
    echo "The script WAS NOT sourced."
fi

#!/bin/bash

echo "First Parameter: $0"
echo
if [[ "$0" == "bash" ]] ; then
    echo "The script was sourced."
elif [[ "$0" == "-bash" ]] ; then
    echo "The script was sourced via SSH session."
else
    echo "The script WAS NOT sourced."
fi

Upvotes: 0

mr.spuratic

Reputation: 10107

The BASH_SOURCE[] answer (bash-3.0 and later) seems simplest, though BASH_SOURCE[] is not documented to work outside a function body (it currently happens to work, in disagreement with the man page).

The most robust way, as suggested by Wirawan Purwanto, is to check FUNCNAME[1] within a function:

function mycheck() { declare -p FUNCNAME; }
mycheck

Then:

$ bash sourcetest.sh
declare -a FUNCNAME='([0]="mycheck" [1]="main")'
$ . sourcetest.sh
declare -a FUNCNAME='([0]="mycheck" [1]="source")'

This is the equivalent to checking the output of caller, the values main and source distinguish the caller's context. Using FUNCNAME[] saves you capturing and parsing caller output. You need to know or calculate your local call depth to be correct though. Cases like a script being sourced from within another function or script will cause the array (stack) to be deeper. (FUNCNAME is a special bash array variable, it should have contiguous indexes corresponding to the call stack, as long as it is never unset.)

function issourced() {
    [[ ${FUNCNAME[@]: -1} == "source" ]]
}

(In bash-4.2 and later you can use the simpler form ${FUNCNAME[-1]} instead for the last item in the array. Improved and simplified thanks to Dennis Williamson's comment below.)

However, your problem as stated is "I have a script where I do not want it to call 'exit' if it's being sourced". The common bash idiom for this situation is:

return 2>/dev/null || exit

If the script is being sourced then return will terminate the sourced script and return to the caller.

If the script is being executed, then return will return an error (redirected), and exit will terminate the script as normal. Both return and exit can take an exit code, if required.

Sadly, this doesn't work in ksh (at least not in the AT&T derived version I have here), it treats return as equivalent to exit if invoked outside a function or dot-sourced script.

Updated: What you can do in contemporary versions of ksh is to check the special variable .sh.level which is set to the function call depth. For an invoked script this will initially be unset, for a dot-sourced script it will be set to 1.

function issourced {
    [[ ${.sh.level} -eq 2 ]]
}

issourced && echo this script is sourced

This is not quite as robust as the bash version, you must invoke issourced() in the file you are testing from at the top level or at a known function depth.

(You may also be interested in this code on github which uses a ksh discipline function and some debug trap trickery to emulate the bash FUNCNAME array.)

The canonical answer here: http://mywiki.wooledge.org/BashFAQ/109 also offers $- as another indicator (though imperfect) of the shell state.

Notes:

it is possible to create bash functions named "main" and "source" (overriding the builtin), these names may appear in FUNCNAME[] but as long as only the last item in that array is tested there is no ambiguity.
I don't have a good answer for pdksh. The closest thing I can find applies only to pdksh, where each sourcing of a script opens a new file descriptor (starting with 10 for the original script). Almost certainly not something you want to rely on...

Upvotes: 34

F. Hauri - Give Up GitHub

Reputation: 70772

After reading @DennisWilliamson's answer, there are some issues, see below:

As this question stand for ksh and bash, there is another part in this answer concerning ksh... see below.

Simple bash way

[ "$0" = "$BASH_SOURCE" ]

Let's try (on the fly because that bash could ;-):

source <(echo $'#!/bin/bash
           [ "$0" = "$BASH_SOURCE" ] && v=own || v=sourced;
           echo "process $$ is $v ($0, $BASH_SOURCE)" ')
process 29301 is sourced (bash, /dev/fd/63)

bash <(echo $'#!/bin/bash
           [ "$0" = "$BASH_SOURCE" ] && v=own || v=sourced;
           echo "process $$ is $v ($0, $BASH_SOURCE)" ')
process 16229 is own (/dev/fd/63, /dev/fd/63)

I use source instead off . for readability (as . is an alias to source):

. <(echo $'#!/bin/bash
           [ "$0" = "$BASH_SOURCE" ] && v=own || v=sourced;
           echo "process $$ is $v ($0, $BASH_SOURCE)" ')
process 29301 is sourced (bash, /dev/fd/63)

Note that process number don't change while process stay sourced:

echo $$
29301

Why not to use `$_ == $0` comparison

For ensuring many case, I begin to write a true script:

#!/bin/bash

# As $_ could be used only once, uncomment one of two following lines

#printf '_="%s", 0="%s" and BASH_SOURCE="%s"\n' "$_" "$0" "$BASH_SOURCE"
[[ "$_" != "$0" ]] && DW_PURPOSE=sourced || DW_PURPOSE=subshell

[ "$0" = "$BASH_SOURCE" ] && BASH_KIND_ENV=own || BASH_KIND_ENV=sourced;
echo "proc: $$[ppid:$PPID] is $BASH_KIND_ENV (DW purpose: $DW_PURPOSE)"

Copy this to a file called testscript:

cat >testscript   
chmod +x testscript

Now we could test:

./testscript 
proc: 25758[ppid:24890] is own (DW purpose: subshell)

That's ok.

. ./testscript 
proc: 24890[ppid:24885] is sourced (DW purpose: sourced)

source ./testscript 
proc: 24890[ppid:24885] is sourced (DW purpose: sourced)

That's ok.

But,for testing a script before adding -x flag:

bash ./testscript 
proc: 25776[ppid:24890] is own (DW purpose: sourced)

Or to use pre-defined variables:

env PATH=/tmp/bintemp:$PATH ./testscript 
proc: 25948[ppid:24890] is own (DW purpose: sourced)

env SOMETHING=PREDEFINED ./testscript 
proc: 25972[ppid:24890] is own (DW purpose: sourced)

This won't work anymore.

Moving comment from 5th line to 6th would give more readable answer:

./testscript 
_="./testscript", 0="./testscript" and BASH_SOURCE="./testscript"
proc: 26256[ppid:24890] is own

. testscript 
_="_filedir", 0="bash" and BASH_SOURCE="testscript"
proc: 24890[ppid:24885] is sourced

source testscript 
_="_filedir", 0="bash" and BASH_SOURCE="testscript"
proc: 24890[ppid:24885] is sourced

bash testscript 
_="/bin/bash", 0="testscript" and BASH_SOURCE="testscript"
proc: 26317[ppid:24890] is own

env FILE=/dev/null ./testscript 
_="/usr/bin/env", 0="./testscript" and BASH_SOURCE="./testscript"
proc: 26336[ppid:24890] is own

Harder: ksh now...

As I don't use ksh a lot, after some read on the man page, there is my tries:

#!/bin/ksh

set >/tmp/ksh-$$.log

Copy this in a testfile.ksh:

cat >testfile.ksh
chmod +x testfile.ksh

Than run it two time:

./testfile.ksh
. ./testfile.ksh

ls -l /tmp/ksh-*.log
-rw-r--r-- 1 user user   2183 avr 11 13:48 /tmp/ksh-9725.log
-rw-r--r-- 1 user user   2140 avr 11 13:48 /tmp/ksh-9781.log

echo $$
9725

and see:

diff /tmp/ksh-{9725,9781}.log | grep ^\> # OWN SUBSHELL:
> HISTCMD=0
> PPID=9725
> RANDOM=1626
> SECONDS=0.001
>   lineno=0
> SHLVL=3

diff /tmp/ksh-{9725,9781}.log | grep ^\< # SOURCED:
< COLUMNS=152
< HISTCMD=117
< LINES=47
< PPID=9163
< PS1='$ '
< RANDOM=29667
< SECONDS=23.652
<   level=1
<   lineno=1
< SHLVL=2

There is some variable herited in a sourced run, but nothing really related...

You could even check that $SECONDS is close to 0.000, but that's ensure only manualy sourced cases...

You even could try to check for what's parent is:

Place this into your testfile.ksh:

ps $PPID

Than:

./testfile.ksh
  PID TTY      STAT   TIME COMMAND
32320 pts/4    Ss     0:00 -ksh

. ./testfile.ksh
  PID TTY      STAT   TIME COMMAND
32319 ?        S      0:00 sshd: user@pts/4

or ps ho cmd $PPID, but this work only for one level of subsessions...

Sorry, I couldn't find a reliable way of doing that, under ksh.

Upvotes: 84

Stéphane Desneux

Reputation: 11

I followed mklement0 compact expression.

That's neat, but I noticed that it can fail in the case of ksh when invoked as this:

/bin/ksh -c ./myscript.sh

(it thinks it's sourced and it's not because it executes a subshell) But the expression will work to detect this:

/bin/ksh ./myscript.sh

Also, even if the expression is compact, the syntax is not compatible with all shells.

So I ended with the following code, which works for bash,zsh,dash and ksh

SOURCED=0
if [ -n "$ZSH_EVAL_CONTEXT" ]; then 
    [[ $ZSH_EVAL_CONTEXT =~ :file$ ]] && SOURCED=1
elif [ -n "$KSH_VERSION" ]; then
    [[ "$(cd $(dirname -- $0) && pwd -P)/$(basename -- $0)" != "$(cd $(dirname -- ${.sh.file}) && pwd -P)/$(basename -- ${.sh.file})" ]] && SOURCED=1
elif [ -n "$BASH_VERSION" ]; then
    [[ $0 != "$BASH_SOURCE" ]] && SOURCED=1
elif grep -q dash /proc/$$/cmdline; then
    case $0 in *dash*) SOURCED=1 ;; esac
fi

Feel free to add exotic shells support :)

Upvotes: 1

user354193

Reputation: 129

I would like to suggest a small correction to Dennis' very helpful answer, to make it slightly more portable, I hope:

[ "$_" != "$0" ] && echo "Script is being sourced" || echo "Script is a subshell"

because [[ isn't recognized by the (somewhat anal retentive IMHO) Debian POSIX compatible shell, dash. Also, one may need the quotes to protect against filenames containing spaces, again in said shell.

Upvotes: 3

Karsten

Reputation: 501

I needed a one-liner that works on [mac, linux] with bash.version >= 3 and none of these answers fit the bill.

[[ ${BASH_SOURCE[0]} = $0 ]] && main "$@"

Upvotes: 0

Wirawan Purwanto

Reputation: 4043

I will give a BASH-specific answer. Korn shell, sorry. Suppose your script name is include2.sh ; then make a function inside the include2.sh called am_I_sourced. Here's my demo version of include2.sh:

am_I_sourced()
{
  if [ "${FUNCNAME[1]}" = source ]; then
    if [ "$1" = -v ]; then
      echo "I am being sourced, this filename is ${BASH_SOURCE[0]} and my caller script/shell name was $0"
    fi
    return 0
  else
    if [ "$1" = -v ]; then
      echo "I am not being sourced, my script/shell name was $0"
    fi
    return 1
  fi
}

if am_I_sourced -v; then
  echo "Do something with sourced script"
else
  echo "Do something with executed script"
fi

Now try to execute it in many ways:

~/toys/bash $ chmod a+x include2.sh

~/toys/bash $ ./include2.sh 
I am not being sourced, my script/shell name was ./include2.sh
Do something with executed script

~/toys/bash $ bash ./include2.sh 
I am not being sourced, my script/shell name was ./include2.sh
Do something with executed script

~/toys/bash $ . include2.sh
I am being sourced, this filename is include2.sh and my caller script/shell name was bash
Do something with sourced script

So this works without exception, and it is not using the brittle $_ stuff. This trick uses BASH's introspection facility, i.e. built-in variables FUNCNAME and BASH_SOURCE; see their documentation in bash manual page.

Only two caveat:

1) the call to am_I_called must take place in the sourced script, but not within any function, lest ${FUNCNAME[1]} returns something else. Yeah...you could have checked ${FUNCNAME[2]} -- but you just make your life harder.

2) function am_I_called must reside in the sourced script if you want to find out what the name of the file being included.

Upvotes: 7

Mikel

Reputation: 25606

$_ is quite brittle. You have to check it as the first thing you do in the script. And even then, it is not guaranteed to contain the name of your shell (if sourced) or the name of the script (if executed).

For example, if the user has set BASH_ENV, then at the top of a script, $_ contains the name of the last command executed in the BASH_ENV script.

The best way I have found is to use $0 like this:

name="myscript.sh"

main()
{
    echo "Script was executed, running main..."
}

case "$0" in *$name)
    main "$@"
    ;;
esac

Unfortunately, this way doesn't work out of the box in zsh due to the functionargzero option doing more than its name suggests, and being on by default.

To work around this, I put unsetopt functionargzero in my .zshenv.

Upvotes: 2

jim mcnamara

Reputation: 16379

This works later on in the script and does'nt depend on the _ variable:

## Check to make sure it is not sourced:
Prog=myscript.sh
if [ $(basename $0) = $Prog ]; then
   exit 1  # not sourced
fi

[ $(basename $0) = $Prog ] && exit

Upvotes: 13

barroyo

Reputation: 2335

If your Bash version knows about the BASH_SOURCE array variable, try something like:

# man bash | less -p BASH_SOURCE
#[[ ${BASH_VERSINFO[0]} -le 2 ]] && echo 'No BASH_SOURCE array variable' && exit 1

[[ "${BASH_SOURCE[0]}" != "${0}" ]] && echo "script ${BASH_SOURCE[0]} is being sourced ..."

Upvotes: 231

Michal Čihař

Reputation: 10091

I don't think there is any portable way to do this in both ksh and bash. In bash you could detect it using caller output, but I don't think there exists equivalent in ksh.

Upvotes: 1

How to detect if a script is being sourced

Answers (28)

The most beautiful way to detect if a Bash script is being executed or sourced (imported)

4 ways to determine whether a bash script is being sourced or executed

See also:

Explanations

bash

ksh

zsh

Using POSIX features only

Why I think this is a good general way to solve the challenge

Notes

Simple bash way

Why not to use `$_ == $0` comparison

Harder: ksh now...

Related Questions

How to detect if a script is being sourced

Answers (28)

The most beautiful way to detect if a Bash script is being executed or sourced (imported)

4 ways to determine whether a bash script is being sourced or executed

See also:

Explanations

bash

ksh

zsh

Using POSIX features only

Why I think this is a good general way to solve the challenge

Notes

Simple bash way

Why not to use $_ == $0 comparison

Harder: ksh now...

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Why not to use `$_ == $0` comparison