It is not a bug in bash. it is just how it works. the while loop
creates a subshell and changes to the variables are not visable
outside of the subshell. if you put the while loop first, then it
will not create the subshell.
do this:
result=""
while read line; do
extracteddata=`echo "$line" | sed -e 's/X/Y/'`
result="$result $extracteddata"
done < <(/usr/bin/output_generator)
/usr/bin/another_tool "$result"
the <() is syntax for a named pipes. it makes a command look like a file.
Be aware that this may leave files in your /tmp directory.
BTW: I would use $() syntax instead of the backtic syntax; just easier to see.
--
potter
On 30 Jan 2008 11:21:34 GMT, Stefan Palme <[EMAIL PROTECTED]> wrote:
> Hi,
> don't know if this is the right newsgroup, but it's the only one
> I can find with "bash" in its name :-)
>
> I want to do something like this:
>
> result=""
> /usr/bin/output_generator | while read line; do
> extracteddata=`echo "$line" | sed -e 's/X/Y/'`
> result="$result $extracteddata"
> done
> /usr/bin/another_tool "$result"
>
> In the last line is "result" as empty as at the start of the
> whole thing - I guess because the inner "while" loop is executed
> in a subshell, so that changing the value of "result" in this
> loop does not affect the "outer result".
>
> How can I solve this? I have some very ugly solutions, but
> I guess there must be something "nice" :-)
>
> (using bash-3.2.17(1)-release)
>
> Thanks and regards
> -stefan-
>
>
>
