Incrementing a variable inside a Bash loop

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

Possible duplicate of [A variable modified inside a while loop is not remembered](https://stackoverflow.com/questions/16854280/a-variable-modified-inside-a-while-loop-is-not-remembered) — codeforester, May 02 '18 at 02:31

fedorqui · Accepted Answer · 2016-07-25T11:21:31.200

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.

Note you can use $(($USCOUNTER+1)) to evaluate the expression — geert3, Dec 19 '13 at 14:27
Thanks @geert3, good one. Also `((USCOUNTER++))` as anubhava's solution is using. I am not adding it to my answer in case it is incompatible with OPs shell. — fedorqui, Dec 19 '13 at 14:34

score 28 · Answer 2 · edited Jan 20 '19 at 21:08

28

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"

edited Jan 20 '19 at 21:08

Bruno Bronosky

66,273
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answered Dec 20 '13 at 02:26

Aleks-Daniel Jakimenko-A.

10,335
3
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1

Why not `echo "US counter $((USCOUNTER++))"` within the if? – Krzysztof Jabłoński May 25 '16 at 15:12
@KrzysztofJabłoński if you want it to be equivalent to the code above, then you should use ``"US counter $((++USCOUNTER))"``. That's a very good point, you are right! – Aleks-Daniel Jakimenko-A. May 25 '16 at 19:26

score 26 · Answer 3 · answered Nov 05 '15 at 05:01

26

minimalist

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

answered Nov 05 '15 at 05:01

geekzspot

787
7
7

2

Some explanations would be appreciated – tu4n Jul 22 '18 at 15:35
3

Where is the wanted while loop in your example ? – tdaget Mar 20 '19 at 14:16
echo $((++counter)) – zzapper Jan 27 '22 at 12:44

score 16 · Answer 4 · edited Sep 02 '16 at 11:36

16

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"

edited Sep 02 '16 at 11:36

fedorqui

275,237
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548
598

answered Dec 19 '13 at 11:59

anubhava

761,203
64
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643

score 8 · Answer 5 · edited Mar 08 '17 at 14:49

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"

Nice to see my answer was helpful :) Note also that `cat file | head` can be simplified to `head file`. — fedorqui, Mar 08 '17 at 14:49

score 1 · Answer 6 · answered Feb 26 '14 at 16:11

1

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

answered Feb 26 '14 at 16:11

Walter A

19,067
2
23
43

trax · Answer 7 · 2016-11-02T11:03:58.253

0

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

edited Nov 02 '16 at 11:03

answered Apr 21 '16 at 08:34

trax

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Although this code may answer the question, providing additional context regarding _why_ and/or _how_ it answers the question would significantly improve its long-term value. Please [edit] your answer to add some explanation. – Toby Speight Apr 21 '16 at 15:23
1

The title is: "Bash incrementing variable in loop" My line does: "incrementing a variable in bash" So I think it does answer a part of the question (at least as much as the example from geekzspot) – trax Apr 23 '16 at 07:37
1

The `$[ ... ]` syntax for arithmetic expansion is [deprecated](http://wiki.bash-hackers.org/syntax/expansion/arith). Please use `$(( ... ))` instead. Both your answer and geekzspot's address an _incidental_ aspect of the OP's question that doesn't solve the problem (the variable-incrementing part of the OP's code was working (however inefficiently)). If you want to suggest an _incidental_ improvement without answering the question, please clearly state so. – mklement0 Oct 27 '16 at 23:03

score 0 · Answer 8 · edited Jan 04 '17 at 12:22

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.

Incrementing a variable inside a Bash loop

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