CODEWITHSUNDEEP
cbasharguments
Bruce Chidester
Bruce Chidester

Reputation: 621

Bash and Double-Quotes passing to argv

I have re-purposed this example to keep it simple, but what I am trying to do is get a nested double-quote string as a single argv value when the bash shell executes it.

Here is the script example:

set -x
command1="key1=value1 \"key2=value2 key3=value3\""
command2="keyA=valueA keyB=valueB keyC=valueC"
echo $command1
echo $command2

the output is:

++ command1='key1=value1 "key2=value2 key3=value3"'
++ command2='keyA=valueA keyB=valueB keyC=valueC'
++ echo key1=value1 '"key2=value2' 'key3=value3"'
key1=value1 "key2=value2 key3=value3"
++ echo keyA=valueA keyB=valueB keyC=valueC
keyA=valueA keyB=valueB keyC=valueC

I did test as well, that when you do everything on the command line, the nested quote message IS set as a single argv value. i.e.

prog.exe argument1 "argument2 argument3"

argv[0] = prog.exe
argv[1] = argument1
argv[2] = argument2 argument3

Using the above example:

command1="key1=value1 \"key2=value2 key3=value3\""

The error is, my argv is comming back like:

arg[1] = echo 
arg[2] = key1=value1 
arg[3] = "key2=value2
arg[4] = key3=value3"

where I really want my argv[3] value to be "key2=value2 key3=value3"

I noticed that debug (set -x) shows a single-quote at the points where my arguments get broken which kinda indicates that it is thinking about the arguments at these break point...just not sure.

Any idea what is really going on here? How can I change the script?

Thanks in advance.

Upvotes: 4

Views: 1466

Answers (2)

Mindaugas Kubilius
Mindaugas Kubilius

Reputation: 341

A kind of crazy "answer" is to set IFS to double quote like this (save/restore original IFS):

SAVED_IFS=$IFS
IFS=$'\"'
prog.exe $command1

IFS=$SAVED_IFS

It kind of illustrates word splitting which occurs on unquoted arguments but does not affect variables or text inside ".." quotes. Text inside double quotes (after various expansions) is passed to the program as a single argument. However a bare variable $command1 (unquoted) undergoes word splitting which does not care about " inside the variable (taking it literal). A stupid IFS hack forces word splitting to be made at ". Also beware of a trailing whitespace at the end of argv[1] which appears because of word splitting at the " boundary.

jordanm's answer is much better for production use than mine :) The array is quoted, i.e. each array element is expanded as individual string and no word splitting occurs afterwards. This is essential. If it is unquoted like ${args[@]} it would be word split into three arguments instead of two.

Upvotes: 1

jordanm
jordanm

Reputation: 34934

What is happening is that your nested quotes are literal and not parsed into separate arguments by the shell. The best way to handle this using bash is to use an array instead of a string:

args=('key1=value1', 'key2=value2 key3=value3')
prog.exe "${args[@]}"

The Bash FAQ50 has some more examples and use cases for dynamic commands.

Upvotes: 5

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