unable to access a variable not local to a while loop

codeman007 · #1 (**permalink**) 12-23-2008

I have a while loop like this

cat ${filename} | while read fileline

do
...

done

I need to access a variable value $newfile inside this while loop
How will i do that??

geekosaur · #2 (**permalink**) 12-24-2008

You should be able to read any variable that has been set at that point. Assigning to it won't work because the pipe forces the while loop to run in a subshell; assignments will be visible within the while loop but will be lost when the loop is exited. I generally use temporary files to work around this, but sometimes run it in a "eval $(expression)" and have it echo the variable assignment(s) to stdout. (The latter avoids the various issues with temporary files but is rather more confusing to most people.)

codeman007 · #3 (**permalink**) 12-24-2008

i didnt get this really. I understood since i am using the pipe ; variables declared outside the while loop are not accessible inside. Is there any means to pass the varible inside the while loop...

Franklin52 · #4 (**permalink**) 12-24-2008

You can avoid the use of a pipe and cat is redundant in such loops:

Code:

while read fileline
do
  ...
done < ${filename}

Regards

codeman007 · #5 (**permalink**) 12-24-2008

Franklin I tried this before as i saw it in some other post of yours but my requirement is different

cat ${filename} | while read fileline

do
size=`echo $fileline | cut -d "," -f2`
value=`echo $fileline | cut -d "," -f3`
info=`echo $fileline | cut -d "," -f4`

variable=$passedvariable

done
I need to read each line from the ${filename} and at the same time get the value of $passedvariable which is declared outside while loop inside the while

Christoph Spohr · #6 (**permalink**) 12-24-2008

Hi,

Code:

variable=1 && while IFS="${IFS}," read size value info 
do 
  echo $size $value $info
  echo "$variable"
done < ${filename}

$variable is declared outside and can be accessed from inside
the loop.

HTH Chris

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