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initialize a variable only once

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# 1  
Old 02-08-2012
initialize a variable only once


i have a script which needs to be run every 10 minutes.This is achieved using crontab utility,
it is checking Number of calls on a service... if number of calls are not increasing then it will stop the service else do nothing.

Now in the script, i fetch the current value in variable myval and then compare it with variable COUNT_DEPOSIT

Here is the code:-
myval=$(ls -l service | grep 449 | awk -F"|" '{print $10}')

if [ "$myval" -gt "$COUNT_DEPOSIT" ]; then
      do nothing
     stop the service

At the end, i reinitialize the variable COUNT_DEPOSIT with the new value so that next time COUNT_DEPOSIT variable has a new value..
But problem is that COUNT_DEPOSIT is everytime initialize to value 0

I want to initialize the this variable only once and pick the new value each time.

Can anyone please help?

thanks in advance
# 2  
Old 02-08-2012
Write the variable (value) to a file, then read the file every time the script runs.
if [ ! -r $fname ]; then

  COUNT_DEPOSIT=$(< $fname )

# at the end of the script:
echo $COUNT_DEPOSIT > $fname

This has issues - at reboot of your app or the system, the file needs to be removed. Otherwise the code will "think" it has some previous value.
try that for a start
This User Gave Thanks to jim mcnamara For This Post:
# 3  
Old 02-08-2012
Bug thanks Jim it works...

it is working for me.

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