williampli williampli - 6 months ago 16
Bash Question

Why doesn't my variable seem to increment in my bash while loop?

I am fairly new to bash scripting. I can't seem to get the correct value of my counting variables to display at the end of of a

while
loop in my bash script.

Background: I have a fairly simple task: I would like to pass a text file containing a list of file paths to a bash script, have it check for the existence of those files, and count the number of existing/missing files. I got most of the script to work, except for the counting part.

N=0
correct=0
incorrect=0
cat $1 | while read filename ; do
N=$((N+1))
echo "$N"

if ! [ -f $filename ]; then

incorrect=$((incorrect+1))
else
correct=$((correct+1))

fi

done

echo "# of Correct Paths: $correct"
echo "# of Incorrect Paths: $incorrect"
echo "Total # of Files: $N"


If I have a list of 5 files, 4 of which exist, I expect to get the following output (note the
echo
command within the
while
loop):

1
2
3
4
5
# of Correct Paths: 4
# of Incorrect Paths: 1
Total # of Files: 5


Instead, I get:

1
2
3
4
5
# of Correct Paths: 0
# of Incorrect Paths: 0
Total # of Files: 0


What happened to the values of these variables? Google had many suggestions of questionable quality and I think I could get it to work with a little more searching, but a brief explanation of what I'm doing wrong would be very helpful.

Answer

This is because you are using the useless cat command with a pipe, causing a subshell to be created. Try it without the cat:

while read filename ; do
    N=$((N+1))
    ....
done < file
Comments