Incrementing a variable inside a Bash loop - bash

Question

I'm trying to write a small script that will count entries in a log file, and I'm incrementing a variable (USCOUNTER) which I'm trying to use after the loop is done.
But at that moment USCOUNTER looks to be 0 instead of the actual value. Any idea what I'm doing wrong? Thanks!
FILE=$1
tail -n10 mylog > $FILE
USCOUNTER=0
cat $FILE | while read line; do
country=$(echo "$line" | cut -d' ' -f1)
if [ "US" = "$country" ]; then
USCOUNTER=`expr $USCOUNTER + 1`
echo "US counter $USCOUNTER"
fi
done
echo "final $USCOUNTER"
It outputs:
US counter 1
US counter 2
US counter 3
..
final 0

Answer 1

You are using USCOUNTER in a subshell, that's why the variable is not showing in the main shell.
Instead of cat FILE | while ..., do just a while ... done < $FILE. This way, you avoid the common problem of I set variables in a loop that's in a pipeline. Why do they disappear after the loop terminates? Or, why can't I pipe data to read?:
while read country _; do
if [ "US" = "$country" ]; then
USCOUNTER=$(expr $USCOUNTER + 1)
echo "US counter $USCOUNTER"
fi
done < "$FILE"
Note I also replaced the `` expression with a $().
I also replaced while read line; do country=$(echo "$line" | cut -d' ' -f1) with while read country _. This allows you to say while read var1 var2 ... varN where var1 contains the first word in the line, $var2 and so on, until $varN containing the remaining content.

Answer 2

Always use -r with read.
There is no need to use cut, you can stick with pure bash solutions.
In this case passing read a 2nd var (_) to catch the additional "fields"
Prefer [[ ]] over [ ].
Use arithmetic expressions.
Do not forget to quote variables! Link includes other pitfalls as well
while read -r country _; do
if [[ $country = 'US' ]]; then
((USCOUNTER++))
echo "US counter $USCOUNTER"
fi
done < "$FILE"

Answer 3

minimalist
counter=0
((counter++))
echo $counter

Answer 4

You're getting final 0 because your while loop is being executed in a sub (shell) process and any changes made there are not reflected in the current (parent) shell.
Correct script:
while read -r country _; do
if [ "US" = "$country" ]; then
((USCOUNTER++))
echo "US counter $USCOUNTER"
fi
done < "$FILE"

Answer 5

I had the same $count variable in a while loop getting lost issue.
#fedorqui's answer (and a few others) are accurate answers to the actual question: the sub-shell is indeed the problem.
But it lead me to another issue: I wasn't piping a file content... but the output of a series of pipes & greps...
my erroring sample code:
count=0
cat /etc/hosts | head | while read line; do
((count++))
echo $count $line
done
echo $count
and my fix thanks to the help of this thread and the process substitution:
count=0
while IFS= read -r line; do
((count++))
echo "$count $line"
done < <(cat /etc/hosts | head)
echo "$count"

Answer 6

USCOUNTER=$(grep -c "^US " "$FILE")

Answer 7

Incrementing a variable can be done like that:
_my_counter=$[$_my_counter + 1]
Counting the number of occurrence of a pattern in a column can be done with grep
grep -cE "^([^ ]* ){2}US"
-c count
([^ ]* ) To detect a colonne
{2} the colonne number
US your pattern

Answer 8

Using the following 1 line command for changing many files name in linux using phrase specificity:
find -type f -name '*.jpg' | rename 's/holiday/honeymoon/'
For all files with the extension ".jpg", if they contain the string "holiday", replace it with "honeymoon". For instance, this command would rename the file "ourholiday001.jpg" to "ourhoneymoon001.jpg".
This example also illustrates how to use the find command to send a list of files (-type f) with the extension .jpg (-name '*.jpg') to rename via a pipe (|). rename then reads its file list from standard input.