CODEWITHSUNDEEP
bashunixloopsscope
Learner
Learner

Reputation: 1634

How to use export in bash shell?

I am making one script to delete all printer jobs older than one day and no of jobs older than one day has to be print also the jobid which has been canceled

Below are few conditions which I have to follow

I do not have access to cups directory , I can not create any temporary file . I have to use Bash shell.

I tried to use a variable, declared outside while loop but the variable value remains unchanged ~came to know it is because of child process Tried to use export variable but that also does not work in Bash shell , same is working in ksh shell.

I have tried below logic:

count=0
currDate=`date +%Y%m%d`
lpstat -o|while read line
do
  jobid=`echo $line|awk '{print $1}'|cut -d"-" -f2` 
  jobDate=`echo $line|awk -F ' ' 'BEGIN{OFS="-";} {print $5,$6,$7;}'`
  formattedDate=`date -d"${jobDate}" +%Y%m%d`

  if [ `expr $currDate - $formattedDate` -gt 1 ]
  then 
  count=`expr $count + 1`
      echo " cancelling printer job with jobid $jobid "
  cancel $jobid
  fi;
done
[ $count -gt 0 ] ; echo "NO of Printer jobs pending more than 1 days are $count";

Not getting how to handle the count variable as the above mentioned way it will not work, Export is not working in bash shell , creating tmp file is not allowed.

Any suggestion to get the solution.

Upvotes: 1

Views: 150

Answers (2)

Mihai
Mihai

Reputation: 2155

The problem with count is that you process it inside a new bash child process (bash creates it automatically when you pipe input into while). This is why you get back the initial value.

The solution for you would be to output the value in that child process. This should work (braces are forcing bash to create a child process that includes the output part):

count=0
currDate=`date +%Y%m%d`
lpstat -o| (while read line
do
  jobid=`echo $line|awk '{print $1}'|cut -d"-" -f2` 
  jobDate=`echo $line|awk -F ' ' 'BEGIN{OFS="-";} {print $5,$6,$7;}'`
  formattedDate=`date -d"${jobDate}" +%Y%m%d`

  if [ `expr $currDate - $formattedDate` -gt 1 ]
  then 
  count=`expr $count + 1`
      echo " cancelling printer job with jobid $jobid "
  cancel $jobid
  fi;
done
[ $count -gt 0 ] && echo "NO of Printer jobs pending more than 1 days are $count")

Upvotes: 1

William Pursell
William Pursell

Reputation: 212248

The simplest solution is to reference the variable in the same shell you are setting it. Syntactically, the easiest thing to do is to add a block:

count=0
currDate=`date +%Y%m%d`
lpstat -o| { while read line   # Add opening brace here
do
  jobid=`echo $line|awk '{print $1}'|cut -d"-" -f2` 
  jobDate=`echo $line|awk -F ' ' 'BEGIN{OFS="-";} {print $5,$6,$7;}'`
  formattedDate=`date -d"${jobDate}" +%Y%m%d`

  if [ `expr $currDate - $formattedDate` -gt 1 ]
  then 
  count=`expr $count + 1`
      echo " cancelling printer job with jobid $jobid "
  cancel $jobid
  fi;
done
echo "Number of Printer jobs pending more than 1 day: $count"
}    # Add closing brace here

This is not a code review site, but I feel compelled to address other elements of this code. echo $line | awk '{print $1}' | cut -d"-" -f2 is an atrocity. Rather than reading in an entire line and then parsing it with echo/awk/cut, let read do the parsing for you by assigning IFS appropriately. Stop using backticks. $() notation is better. The last line of this code [ $count -gt 0 ]; echo ... completely ignores the evaluation of $count. It would be better written [ $count -gt 0 ] || echo ... >&2 or test "$count" = 0 && echo ... >&2 (redirect the error message to file descriptor 2, since error belong on stderr).

Upvotes: 0

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