Counter increment in Bash loop not working

Question

I have the following simple script where I am running a loop and want to maintain a COUNTER  I am unable to figure out why the counter is not updating  Is it due to subshell thats getting created  How can I potentially fix this      bin bash  WFY PATH  var log nginx WFY FILE error log COUNTER 0 grep  GET  log    WFY PATH  WFY FILE   grep  upstream timed out    awk -F        print  2  4  0     awk   print  http   domain com  5  amp ip   2  amp date   7  amp time   8  amp end 1      awk -F   amp end 1    print  1  amp end 1        while read WFY URL do     echo  WFY URL  Some more action     COUNTER    COUNTER 1   done    echo  COUNTER   output   0

User · Answer

This is all you need to do      COUNTER       Here s an excerpt from Learning the bash Shell  3rd Edition  pp  147  148      bash arithmetic expressions are equivalent to their counterparts in   the Java and C languages  9  Precedence and associativity are the same   as in C  Table 6-2 shows the arithmetic operators that are supported    Although some of these are  or contain  special characters  there is   no need to backslash-escape them  because they are within the            syntax                                   The    and - operators are useful when you want to increment or   decrement a value by one  11  They work the same as in Java and C    e g   value   increments value by 1  This is called post-increment    there is also a pre-increment    value  The difference becomes evident   with an example      i 0   echo  i 0   echo    i     0   echo  i 1   echo      i   2   echo  i 2   See http   www safaribooksonline com a learning-the-bash 7572399

User · Answer

This is a simple example   COUNTER 1 for i in  1  5  do       echo  COUNTER       echo  Welcome  i times       COUNTER          done

User · Answer

Instead of using a temporary file  you can avoid creating a subshell around the while loop by using process substitution   while     do        done  lt   lt  grep        By the way  you should be able to transform all that grep  grep  awk  awk  awk into a single awk   Starting with Bash 4 2  there is a lastpipe option that     runs the last command of a       pipeline in the current shell context   The lastpipe option has no       effect if job control is enabled    bash -c  echo foo   while read -r s  do c 3  done  echo   c    bash -c  shopt -s lastpipe  echo foo   while read -r s  do c 3  done  echo   c   3

User · Answer

There were two conditions that caused the expression   var     to fail for me    If I set bash to strict mode  set -euo pipefail  and if I start my increment at zero  0   Starting at one  1  is fine but zero causes the increment to return  1  when evaluating      which is a non-zero return code failure in strict mode    I can either use   var  1   or var    var 1   to escape this behavior

User · Answer

I think this single awk call is equivalent to your grep grep awk awk pipeline  please test it  Your last awk command appears to change nothing at all   The problem with COUNTER is that the while loop is running in a subshell  so any changes to the variable vanish when the subshell exits  You need to access the value of COUNTER in that same subshell  Or take  DennisWilliamson s advice  use a process substitution  and avoid the subshell altogether   awk      GET   log    amp  amp   upstream timed out        split  0  a            split a 2  FS a 4  FS  0  b      print  http   example com  b 5    amp ip   b 2    amp date   b 7    amp time   b 8    amp end 1                while read WFY URL     do         echo  WFY URL  Some more action            COUNTER          done     echo  COUNTER

User · Answer

count 0    base 1    count    base

User · Answer

It seems that you didn t update the counter is the script  use counter

User · Answer

COUNTER    COUNTER 1      is quite a clumsy construct in modern programming      COUNTER        looks more  modern   You can also use   let COUNTER     if you think that improves readability  Sometimes  Bash gives too many ways of doing things - Perl philosophy I suppose - when perhaps the Python  there is only one right way to do it  might be more appropriate  That s a debatable statement if ever there was one  Anyway  I would suggest the aim  in this case  is not just to increment a variable but  general rule  to also write code that someone else can understand and support  Conformity goes a long way to achieving that   HTH

User · Answer

Source script has some problem with subshell  First example  you probably do not need subshell  But We don t know what is hidden under  Some more action   The most popular answer has hidden bug  that will increase I O  and won t work with subshell  because it restores couter inside loop   Do not fortot add     sign  it will inform bash interpreter about line continuation  I hope it will help you or anybody  But in my opinion this script should be fully converted to AWK script  or else rewritten to python using regexp  or perl  but perl popularity over years is degraded  Better do it with python   Corrected Version without subshell      bin bash WFY PATH  var log nginx WFY FILE error log COUNTER 0 grep  GET  log    WFY PATH  WFY FILE   grep  upstream timed out     awk -F        print  2  4  0      awk   print  http   example com  5  amp ip   2  amp date   7  amp time   8  amp end 1       awk -F   amp end 1    print  1  amp end 1            unneeded bracket while read WFY URL do     echo  WFY URL  Some more action     COUNTER    COUNTER 1   done     unneeded bracket  echo  COUNTER   output   0   Version with subshell if it is really needed     bin bash  TEMPFILE  tmp    tmp   I ve got it from the most popular answer WFY PATH  var log nginx WFY FILE error log COUNTER 0 grep  GET  log    WFY PATH  WFY FILE   grep  upstream timed out     awk -F        print  2  4  0      awk   print  http   example com  5  amp ip   2  amp date   7  amp time   8  amp end 1       awk -F   amp end 1    print  1  amp end 1         while read WFY URL do     echo  WFY URL  Some more action     COUNTER    COUNTER 1   done echo  COUNTER  gt   TEMPFILE   store counter only once  do it after loop  you will save I O    COUNTER   cat  TEMPFILE    restore counter unlink  TEMPFILE echo  COUNTER   output   0

User · Answer

Try to use   COUNTER    COUNTER 1     instead of  COUNTER    COUNTER

User · Answer

First  you are not increasing the counter  Changing COUNTER    COUNTER   into COUNTER    COUNTER   1   or COUNTER   COUNTER   1  will increase it   Second  it s trickier to back-propagate subshell variables to the callee as you surmise  Variables in a subshell are not available outside the subshell  These are variables local to the child process   One way to solve it is using a temp file for storing the intermediate value   TEMPFILE  tmp    tmp echo 0  gt   TEMPFILE    Loop goes here     Fetch the value and increase it   COUNTER     cat  TEMPFILE    1       Store the new value   echo  COUNTER  gt   TEMPFILE    Loop done  script done  delete the file unlink  TEMPFILE

User · Answer

COUNTER 1 while   Your     done    do      echo    COUNTER        COUNTER    COUNTER  1  done   TESTED BASH  Centos  SuSE  RH

User · Answer

minimalist  counter 0   counter     echo  counter

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