Bash executing bash command

Question

I have problem running bash command in bash.

Simple:

C="ls"
$C

works

angel@php56:/tmp$ C="ls"
angel@php56:/tmp$ $C
testFile1

Complicated

C="bash -c \"ls\""
$C

works

angel@php56:/tmp$ C="bash -c \"ls\""
angel@php56:/tmp$ $C
testFile1

More complicated

C="bash -c \"bash -c 'ls'\""
$C

doesn't work

angel@php56:/tmp$ C="bash -c \"bash -c 'ls'\""
angel@php56:/tmp$ $C
-c: line 0: unexpected EOF while looking for matching `"'
-c: line 1: syntax error: unexpected end of file

same for C="bash -c \"bash -c \\\"ls\\\"\""

angel@php56:/tmp$ C="bash -c \"bash -c \\\"ls\\\"\""                                                                                                                                                                                               
angel@php56:/tmp$ $C                                                                                                                                                                                                                               
-c: line 0: unexpected EOF while looking for matching `"'                                                                                                                                                                                          
-c: line 1: syntax error: unexpected end of file

Why I can run bash in bash, but cant bash in bash in bash :) May be something with quotes ? Online code: click

I'm trying to put a command in a variable, but the complex cases always fail! — chepner
– chepner, Commented May 26, 2016 at 13:05

Mark Reed · Accepted Answer · 2016-05-26 13:06:03Z

3

The problem is the order of operations with word splitting vs parameter expansion.
It works with eval:

eval "$C"

This gets very tricky very quickly. The best way to deal with dynamic commands that don't include pipes or other redirection is to use arrays:

C=(bash -c "bash -c 'ls'")
"${C[@]}"

answered May 26, 2016 at 13:06

Mark Reed

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