Incrementing a variable inside a Bash loop [duplicate]

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

fedorqui · Accepted Answer · 2016-07-25 11:21:31Z

71

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.

edited Jul 25, 2016 at 11:21

answered Dec 19, 2013 at 11:59

fedorqui

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2 Comments

geert3 Over a year ago

Note you can use $(($USCOUNTER+1)) to evaluate the expression

fedorqui Over a year ago

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.

codeasaurus · Accepted Answer · 2024-05-04 13:41:26Z

33

minimalist

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

This method uses arithmetic expansion as defined here: Bash Arithmetic Expansion

Valid expressions are defined here: Shell Arithmetic

edited May 4, 2024 at 13:41

codeasaurus

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answered Nov 5, 2015 at 5:01

geekzspot

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3 Comments

tu4n Over a year ago

Some explanations would be appreciated

tdaget Over a year ago

Where is the wanted while loop in your example ?

zzapper Over a year ago

echo $((++counter))

guettli · Accepted Answer · 2024-07-24 08:51:20Z

31

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 Jul 24, 2024 at 8:51

guettli

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answered Dec 20, 2013 at 2:26

Aleks-Daniel Jakimenko-A.

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2 Comments

Krzysztof Jabłoński Over a year ago

Why not echo "US counter $((USCOUNTER++))" within the if?

Aleks-Daniel Jakimenko-A. Over a year ago

@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!

fedorqui · Accepted Answer · 2016-09-02 11:36:42Z

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 2, 2016 at 11:36

fedorqui

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answered Dec 19, 2013 at 11:59

anubhava

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Comments

fedorqui · Accepted Answer · 2017-03-08 14:49:09Z

10

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"

edited Mar 8, 2017 at 14:49

fedorqui

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answered Oct 27, 2016 at 22:52

jobwat

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1 Comment

fedorqui Over a year ago

Nice to see my answer was helpful :) Note also that cat file | head can be simplified to head file.

Walter A · Accepted Answer · 2014-02-26 16:11:42Z

1

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

answered Feb 26, 2014 at 16:11

Walter A

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Comments

trax · Accepted Answer · 2016-11-02 11:03:58Z

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 2, 2016 at 11:03

answered Apr 21, 2016 at 8:34

trax

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3 Comments

Toby Speight Over a year ago

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.

trax Over a year ago

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)

mklement0 Over a year ago

The $[ ... ] syntax for arithmetic expansion is deprecated. 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.

Draken · Accepted Answer · 2017-01-04 12:22:12Z

0

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.

edited Jan 4, 2017 at 12:22

Draken

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answered Jan 4, 2017 at 12:18

thanh

1

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