Problem with loop


 
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# 1  
Old 07-29-2016
Problem with loop

Hi all,
a problem with a loop. Imagine it starts with

Code:
var1="$(cat txtif.out )"

Code:
while  [ $var1 = x ]
  do
  echo file1
  echo file2
  echo y > txtif.out 
  if [ $var1 = y ]
  then
    break
  fi
done

exit  0

the problem is that if the file is changing during the loop seems to continue as it reads "x" state, while in the loop it should change to y and break, instead it returns from the beginnig.

Any idea how to break the loop?

Thanks!

Last edited by Board27; 07-29-2016 at 05:01 PM..
# 2  
Old 07-29-2016
What and why should change to "y"? The "control variable" var1 is NOT modified within the loop. So - either the loop is not entered at all, or it will loop forever.
# 3  
Old 07-29-2016
Thanks. Is there a way to modify a control variable within the loop?

Imagine the starting file has x inside. (txtif.out)
i would like to modify it so when, in the loop, i modify the file (ex with echo y) it will read the y and since it is different from x , will break the loop.

I hope it has some sense..
# 4  
Old 07-29-2016
You then need to read that file within the loop. The break is pointless then, as the while condition will be false and leave the loop.
# 5  
Old 07-29-2016
I don't understand.

If the first line of the file txtif.out is not x followed by a <newline> character, your script will exit without printing anything. Otherwise, you have an infinite loop printing three lines each time through the loop.

If you want to break out of the loop, you either need to change the value of var1 to something other than x inside the loop, or execute a break statement in the loop (which you code cannot do since the value of var1 never changes in your loop).

If something inside your loop is changing the contents of txtif.out (which nothing currently does) or in some external process is changing the contents of txtif.out and you want to the loop to stop when that happens, you would need something more like:
Code:
read var1 < txtif.out
while  [ "$var1" = x ]
do
  echo file1
  echo file2
  echo y txtif.out 
  read var1 < txtif.out
done

exit  0

If you think that the command:
Code:
  echo y txtif.out

would change the contents of txtif.out to the string y; it does not. And, if you think that changing the contents of txtif.out will affect your infinite loop; it will not. That would be done by:
Code:
  echo y > txtif.out

but, if you just want to execute a loop once, why have a loop. An if statement would be more appropriate:
Code:
read var1 < txtif.out
if  [ "$var1" = x ]
then
  echo file1
  echo file2
  echo y txtif.out  # or maybe you wanted echo y > txtif.out instead.
fi

exit  0


Last edited by Don Cragun; 07-30-2016 at 03:25 AM.. Reason: fix typos in if statement: s/do/then/ & s/done/fi/, and add missing quotes.
This User Gave Thanks to Don Cragun For This Post:
# 6  
Old 07-29-2016
Thanks!

i've done with

Code:
read var1 < txtif.out 
while  [ $var1 = x ] ; do   
echo file1   
echo  file2   
echo y  >  txtif.out
 read var1 < txtif.out
 done  
exit  0

thanks for help!
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