While loop with if bash

Question

This maybe a duplicate of Bash loop control while if else with a return. The method that I am inquiring about might be different. Here's my challenge.

I want my bash script to look for a string in a file. If the string is not found, I want it to say so, wait 10 seconds, and keep looking for the string. Once the string is found, I want it to exit and do other functions. I am having trouble placing the logic of while and if. Code below:

 while ! grep -q Hello "filename.txt"; do
  echo "String Hello not found. Waiting 10 seconds and trying again"
  sleep 10
   if grep -q Hello "filename.txt"; then
    echo "String Hello found in filename.txt. Moving on to next procedure"
    sleep 2
    return 0
   fi
   #do other functions here
 done
exit

Superficially, you remove the if…fi and the do other functions here from the body of the loop. The if you discard. The 'do other functions' code goes after the loop. The loop won't exit until the string is seen; when the string is seen, the other functions will be executed. You may decide you want some sort of timeout on the loop; count the iterations and if the string hasn't appeared inside an hour (360 iterations), give up with an error message and exit. — Jonathan Leffler
– Jonathan Leffler, Commented Nov 5, 2015 at 16:51

choroba · Accepted Answer · 2015-11-05 16:53:34Z

4

The only way how the while loop can be exited is by finding the string. So, don't check for the presence again inside the loop, and don't check for it even after it, it's guaranteed by the fact the loop ended:

while ! grep -q Hello filename.txt ; do
    echo "String Hello not found. Waiting 10 seconds and trying again"
    sleep 10
done
echo "String Hello found in filename.txt. Moving on to next procedure"

answered Nov 5, 2015 at 16:53

choroba

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1 Comment

ToofanHasArrived Over a year ago

Thanks @choroba. That fixed it.

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