Read values into a shell variable from a pipe

Question

I am trying to get bash to process data from stdin that gets piped into, but no luck. What I mean is none of the following work:

echo "hello world" | test=($(< /dev/stdin)); echo test=$test
test=

echo "hello world" | read test; echo test=$test
test=

echo "hello world" | test=`cat`; echo test=$test
test=

where I want the output to be test=hello world. I've tried putting "" quotes around "$test" that doesn't work either.

Answer 1

The questions is how to catch output from a command to save in variable(s) for use later in a script. I might repeat some earlier answers but I try to line up all the answers I can think up to compare and comment, so bear with me.

The intuitive construct

echo test | read x
echo x=$x

is valid in Korn shell because ksh have implemented that the last command in a piped series is part of the current shell ie. the previous pipe commands are subshells. In contrast other shells define all piped commands as subshells including the last. This is the exact reason I prefer ksh. But having to copy with other shells, bash f.ex., another construct must be used.

To catch 1 value this construct is viable:

x=$(echo test)
echo x=$x

But that only caters for 1 value to be collected for later use.

To catch more values this construct is useful and works in bash and ksh:

read x y <<< $(echo test again)
echo x=$x y=$y

There is a variant which I have noticed work in bash but not in ksh:

read x y < <(echo test again)
echo x=$x y=$y

The <<< $(...) is a here-document variant which gives all the meta handling of a standard command line. < <(...) is an input redirection of a file-substitution operator.

I use "<<< $(" in all my scripts now because it seems the most portable construct between shell variants. I have a tools set I carry around on jobs in any Unix flavor.

Of course there is the universally viable but crude solution:

command-1 | {command-2; echo "x=test; y=again" > file.tmp; chmod 700 file.tmp}
. ./file.tmp
rm file.tmp
echo x=$x y=$y

Answer 2

A smart script that can both read data from PIPE and command line arguments:

#!/bin/bash
if [[ -p /dev/stdin ]]
    then
    PIPE=$(cat -)
    echo "PIPE=$PIPE"
fi
echo "ARGS=$@"

Output:

$ bash test arg1 arg2
ARGS=arg1 arg2

$ echo pipe_data1 | bash test arg1 arg2
PIPE=pipe_data1
ARGS=arg1 arg2

Explanation: When a script receives any data via pipe, then the /dev/stdin (or /proc/self/fd/0) will be a symlink to a pipe.

/proc/self/fd/0 -> pipe:[155938]

If not, it will point to the current terminal:

/proc/self/fd/0 -> /dev/pts/5

The bash [[ -p option can check it it is a pipe or not.

cat - reads the from stdin.

If we use cat - when there is no stdin, it will wait forever, that is why we put it inside the if condition.

Answer 3

I wanted something similar - a function that parses a string that can be passed as a parameter or piped.

I came up with a solution as below (works as #!/bin/sh and as #!/bin/bash)

#!/bin/sh

set -eu

my_func() {
    local content=""
    
    # if the first param is an empty string or is not set
    if [ -z ${1+x} ]; then
 
        # read content from a pipe if passed or from a user input if not passed
        while read line; do content="${content}$line"; done < /dev/stdin

    # first param was set (it may be an empty string)
    else
        content="$1"
    fi

    echo "Content: '$content'"; 
}


printf "0. $(my_func "")\n"
printf "1. $(my_func "one")\n"
printf "2. $(echo "two" | my_func)\n"
printf "3. $(my_func)\n"
printf "End\n"

Outputs:

0. Content: ''
1. Content: 'one'
2. Content: 'two'
typed text
3. Content: 'typed text'
End

For the last case (3.) you need to type, hit enter and CTRL+D to end the input.

Answer 4

read won't read from a pipe (or possibly the result is lost because the pipe creates a subshell). You can, however, use a here string in Bash:

$ read a b c <<< $(echo 1 2 3)
$ echo $a $b $c
1 2 3

But see @chepner's answer for information about lastpipe.

Answer 5

Because I fall for it, I would like to drop a note. I found this thread, because I have to rewrite an old sh script to be POSIX compatible. This basically means to circumvent the pipe/subshell problem introduced by POSIX by rewriting code like this:

some_command | read a b c

into:

read a b c << EOF
$(some_command)
EOF

And code like this:

some_command |
while read a b c; do
    # something
done

into:

while read a b c; do
    # something
done << EOF
$(some_command)
EOF

But the latter does not behave the same on empty input. With the old notation the while loop is not entered on empty input, but in POSIX notation it is! I think it's due to the newline before EOF, which cannot be ommitted. The POSIX code which behaves more like the old notation looks like this:

while read a b c; do
    case $a in ("") break; esac
    # something
done << EOF
$(some_command)
EOF

In most cases this should be good enough. But unfortunately this still behaves not exactly like the old notation if some_command prints an empty line. In the old notation the while body is executed and in POSIX notation we break in front of the body.

An approach to fix this might look like this:

while read a b c; do
    case $a in ("something_guaranteed_not_to_be_printed_by_some_command") break; esac
    # something
done << EOF
$(some_command)
echo "something_guaranteed_not_to_be_printed_by_some_command"
EOF

Answer 6

In my eyes the best way to read from stdin in bash is the following one, which also lets you work on the lines before the input ends:

while read LINE; do
    echo $LINE
done < /dev/stdin

Answer 7

How about this:

echo "hello world" | echo test=$(cat)

Answer 8

The following code:

echo "hello world" | ( test=($(< /dev/stdin)); echo test=$test )

will work too, but it will open another new sub-shell after the pipe, where

echo "hello world" | { test=($(< /dev/stdin)); echo test=$test; }

won't.

---

I had to disable job control to make use of chepnars' method (I was running this command from terminal):

set +m;shopt -s lastpipe
echo "hello world" | read test; echo test=$test
echo "hello world" | test="$(</dev/stdin)"; echo test=$test

Bash Manual says:

lastpipe

If set, and job control is not active, the shell runs the last command of a pipeline not executed in the background in the current shell environment.

Note: job control is turned off by default in a non-interactive shell and thus you don't need the set +m inside a script.

Answer 9

This is another option

$ read test < <(echo hello world)

$ echo $test
hello world

Answer 10

Use

IFS= read var << EOF
$(foo)
EOF

You can trick read into accepting from a pipe like this:

echo "hello world" | { read test; echo test=$test; }

or even write a function like this:

read_from_pipe() { read "$@" <&0; }

But there's no point - your variable assignments may not last! A pipeline may spawn a subshell, where the environment is inherited by value, not by reference. This is why read doesn't bother with input from a pipe - it's undefined.

FYI, http://www.etalabs.net/sh_tricks.html is a nifty collection of the cruft necessary to fight the oddities and incompatibilities of bourne shells, sh.

Answer 11

bash 4.2 introduces the lastpipe option, which allows your code to work as written, by executing the last command in a pipeline in the current shell, rather than a subshell.

shopt -s lastpipe
echo "hello world" | read test; echo test=$test

Answer 12

The first attempt was pretty close. This variation should work:

echo "hello world" | { test=$(< /dev/stdin); echo "test=$test"; };

and the output is:

test=hello world

You need braces after the pipe to enclose the assignment to test and the echo.

Without the braces, the assignment to test (after the pipe) is in one shell, and the echo "test=$test" is in a separate shell which doesn't know about that assignment. That's why you were getting "test=" in the output instead of "test=hello world".

Answer 13

I'm no expert in Bash, but I wonder why this hasn't been proposed:

stdin=$(cat)

echo "$stdin"

One-liner proof that it works for me:

$ fortune | eval 'stdin=$(cat); echo "$stdin"'

Answer 14

I think you were trying to write a shell script which could take input from stdin. but while you are trying it to do it inline, you got lost trying to create that test= variable. I think it does not make much sense to do it inline, and that's why it does not work the way you expect.

I was trying to reduce

$( ... | head -n $X | tail -n 1 )

to get a specific line from various input. so I could type...

cat program_file.c | line 34

so I need a small shell program able to read from stdin. like you do.

22:14 ~ $ cat ~/bin/line 
#!/bin/sh

if [ $# -ne 1 ]; then echo enter a line number to display; exit; fi
cat | head -n $1 | tail -n 1
22:16 ~ $ 

there you go.

Answer 15

if you want to read in lots of data and work on each line separately you could use something like this:

cat myFile | while read x ; do echo $x ; done

if you want to split the lines up into multiple words you can use multiple variables in place of x like this:

cat myFile | while read x y ; do echo $y $x ; done

alternatively:

while read x y ; do echo $y $x ; done < myFile

But as soon as you start to want to do anything really clever with this sort of thing you're better going for some scripting language like perl where you could try something like this:

perl -ane 'print "$F[0]\n"' < myFile

There's a fairly steep learning curve with perl (or I guess any of these languages) but you'll find it a lot easier in the long run if you want to do anything but the simplest of scripts. I'd recommend the Perl Cookbook and, of course, The Perl Programming Language by Larry Wall et al.

Answer 16

The syntax for an implicit pipe from a shell command into a bash variable is

var=$(command)

or

var=`command`

In your examples, you are piping data to an assignment statement, which does not expect any input.

Answer 17

Piping something into an expression involving an assignment doesn't behave like that.

Instead, try:

test=$(echo "hello world"); echo test=$test