Inifinite loop in bash - bash

Question

I have written the following command to loop over a set of strings in the second column of my file and then do sorting for each string on column 11, then take the second and eleventh column and count the number of unique occurrences. Very simple but it seems that it enters an infinite loop and I can't see why. I would appreciate your help very much.
for item in $(cat file.txt | cut -f2 -d " "| uniq)
do
sort -k11,11 file.txt | cut -f2,11 -d " " | uniq -c | sort -k2,2 > output
done

Answer 1

There's no infinite loop here, but it is a very silly loop (that takes a long time to run, while not accomplishing the script's stated purpose). Let's look at how one might accomplish that purpose more sanely:
Using a temporary file for counts.txt to avoid needing to rerun the sort, cut and uniq steps on each iteration:
sort -k11,11 file.txt | cut -f2,11 -d " " | uniq -c >counts.txt
while read -r item; do
fgrep -e " ${item}" counts.txt
done < <(cut -f2 -d' ' <file.txt | uniq)
Even better, using bash 4 associative arrays and no temporary file:
# reads counts into an array
declare -A counts=( )
while read -r count item; do
counts[$item]=count
done < <(sort -k11,11 file.txt | cut -f2,11 -d " " | sort | uniq -c)
# reads counts back out
while read -r item; do
echo "$item ${counts[$item]}"
done < <(cat file.txt | cut -f2 -d " "| sort | uniq)
...that said, that's only if you want to use sort for ordering on pulling data back out. If you don't need to do that, the latter part could be replaced as such:
# read counts back out
for item in "${!counts[#]}"; do
echo "$item ${counts[$item]}"
done