Bash - bad assignment value to variable

Question

I want to do a very simple script: just want to find the newest version of a program, say svn, on my computer. I want to load the result into a variable, say mysvn

So I make this script:


#!/bin/sh

mysvn="foobar"
best_ver=0
which -a svn | while read p
do
    version=$("$p" --version | grep 'version ' | grep -oE '[0-9.]+' | head -1)
    if [[ "$version" > "$best_ver" ]]
    then
        best_ver=$version
        mysvn="$p"
    fi
    echo $mysvn
done

echo $mysvn

Very simple in fact ... but it does not work under rxvt (my pseudo-linux terminal), version 2.7.10, running under XP: the final output string is foobar.

Does anybody know why I have this problem?

I have been writing some scripts for the past few months, it is the first time I encounter such a behaviour.

Note: I know how to make it work, with a few changes (just put the main lines into $() )

try doing temp=$( which -a svn ) and then for p in $temp I think this pipe is your problem but don't have a way to test it right now. — Ivaylo Strandjev
– Ivaylo Strandjev, Commented Jul 6, 2012 at 14:38
@izomorphius: write this up as a solution. The problem is indeed that the pipe causes the while loop to be executed in a subshell, so any assignments to mysvn are local to that shell. — chepner
– chepner, Commented Jul 6, 2012 at 14:53

Community · Accepted Answer · 2017-05-23 11:48:12Z

2

The reason this occurs is that the while loop is part of a pipeline, and (at least in bash) any shell commands in a pipeline always get executed in a subshell. When you set mysvn inside the loop, it gets set in the subshell; when the loop ends, the subshell exits and this value is lost. The parent shell's value of mysvn never gets changed. See BashFAQ #024 and this previous question.

The standard solution in bash is to use process substitution rather than a pipeline:

while
...
done < <(which -a svn)

But note that this is a bash-only feature, and you must use #!/bin/bash as your shebang for this to work.

edited May 23, 2017 at 11:48

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answered Jul 6, 2012 at 15:26

Gordon Davisson

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3 Comments

Dennis Williamson Over a year ago

That's referred to a "process substitution". "Command substitution" looks like $(). Zsh supports process substitution, but it (and ksh93) don't create a subshell when something is piped into while. Ksh93 doesn't seem to support process substitution redirected into done or other builtins. Bash 4.2 has shopt -s lastpipe which prevents a subshell under these circumstances.

Bentoy13 Over a year ago

Thanks, your answers are both valuable. I see now why my script cannot work properly.

Bentoy13 Over a year ago

By the way, concerning svn, my script was a little too complicated ... just calling svn --version --quiet is nicer than those awkward greps and if-construct :) So I finally came into this, running with /bin/sh:

mysvn=$(which -a svn 2> /dev/null | while read p; do echo $("$p" --version --quiet) "$p"; done | sort -r | head -1 | grep -Eo '".*$')

Miguel · Accepted Answer · 2012-07-06 14:54:07Z

0

Here on Ubuntu:

:~$ which -a svn | while read p
> do
>   version=$("$p" --version | grep 'version ' | grep -oE '[0-9.]+' | head -1)
>   echo $version
> done
.

so, your version is ., not very nice.

I tried this, and I think it's what you're looking for:

:~$ which -a svn | while read p
> do
>   version=$("$p" --version | grep -oE '[0-9.]+' | head -1)
>   echo $version
> done
1.7.5

answered Jul 6, 2012 at 14:54

Miguel

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