Very simple bash script ... However can't figure it out

[Almendrico]

Hello all,

I am writing a very simple bash script but I can't figure out why it is not working. I am just searching a file for a string and evaluating the return values using if - elsif. There are three possible values 0 (string found), 1 (string not found) and 2 (file not found).

However, for some reason, when the return code is 2, i.e. the file hello.txt does not exist (and I have checked that the return code is 2 in the terminal), the script is not getting into the second elsif as expected, but in the first (return code 1).

I believe the problem is that evaluating the conditions is making the $? to change its value. Actually, assigning $? to a variable and then comparing using this variable does the trick ... (i just found the solution myself ) Anyway, why $? works in that way ? Is this the only workaround to this problem ?

Cheers

#!/bin/sh

grep hello hello.txt > /dev/null

if [ $? -eq 0 ]; then
  echo "Text found"
elif [ $? -eq 1 ]; then
  echo "Text Not found"
elif [ $? -eq 2 ]; then
  echo "File does not exist"
else
  echo "other error"
fi

[Piyu_86]

that is because $? hold the return status of immediate executed command
after grep hello hello.txt > /dev/null
is executed the $? value is 2 if the file doesn't exist . but immediately after that the script is executing if command which results the change of value of $? where you will get the return status of if statement but not grep statement .

[Almendrico]

yes, that makes sense to what I found
Thanks

[sgosnell]

You can only evaluate whether the text is found if the file is found. Thus you need to check for the file first, and only check whether the text is found if the file exists. Obviously the text won't be found if the file isn't found.