Problem displaying the value of a counter

dgiles79 · 10-14-2008, 09:17 PM

I am brand new to Linux and Shell Scripting, so this may be a very basic problem, but I am having this one issue with my script (using the BASH shell).

I am trying to display the value of my count, but it always reverts back to the default value of 0

This is my code

Code:

#!/bin/bash
COUNTER=0

cd /home/dave
ls-1 > directory.txt

cat directory.txt |while read line; do
if [ -d ${line} ]; then
let COUNTER+=1
echo $COUNTER
fi
done

echo The total is: $COUNTER

The point is to count the number of directory folders, and the echo within my if statement displays correctly, but when I get to my final statement it displays "The total is: 0". Even though the previous value displayed was 5.

I am totally confused why it is doing this, any help would be much appreciated.

jlinkels · 10-14-2008, 09:31 PM

The problem is that the COUNTER is in a loop and the loop is in a piped subproces. In other words, while...done is a different subprocess and hence the scope of COUNTER is limited to the loop.

Suggestion:

Code:

while read line; do
if [ -d ${line} ]; then
let COUNTER+=1
echo $COUNTER
fi
done < directory.txt

Not tested, but it should be somethinh like that with the '< file' at the end of the loop.

Test your script with 'sh -x yourscript' to see the trace of the program.

jlinkels

dgiles79 · 10-14-2008, 09:55 PM

Thanks, that was exactly what I needed. The quick response is greatly appreciated.

chrism01 · 10-14-2008, 10:13 PM

Or

Code:

for line in `cat directory.txt`
do
...

instead of the piped while