CODEWITHSUNDEEP
bash
Krøllebølle
Krøllebølle

Reputation: 3028

Only first of two identical if statements is executed in bash

I have a small bash test script like this:

$ cat test
#!/bin/bash
if ! grep --silent "$1" "$3"; then echo "$1 not matched."; fi; exit 1;
if ! grep --silent "$2" "$3"; then echo "$2 not matched."; fi; exit 1;

and a test file like this:

$ cat foobar
FOO
BAR

Now I do the following:

$ ./test FOO BAR foobar
$ ./test FOOH BAR foobar
FOOH not matched.
$ ./test FOO BARG foobar
$

In my mind the last run, ./test FOO BARG foobar should produce the output BARG not matched. and the script stop executing. What's the reason this does not work as I expect it to? Is there a better way of achieving the same logic?

Edit:

Also, if I comment out the second statement, like this:

$ cat test
#!/bin/bash
#if ! grep --silent "$1" "$3"; then echo "$1 not matched."; fi; exit 1;
if ! grep --silent "$2" "$3"; then echo "$2 not matched."; fi; exit 1;

The second if statement is executed:

$ ./test FOO BARG
$ BARG not matched.

So it seems that only the first of the consecutive if statements are executed. This also happens if I extend the script to have three similar if statements.

Upvotes: 0

Views: 57

Answers (1)

fejese
fejese

Reputation: 4628

Your code looks like this with bit better formatting:

#!/bin/bash
if ! grep --silent "$1" "$3"; then
  echo "$1 not matched.";
fi;

exit 1;

if ! grep --silent "$2" "$3"; then
  echo "$2 not matched.";
fi;

exit 1;

Probably you wanted to have the exit statements within the ifs:

#!/bin/bash
if ! grep --silent "$1" "$3"; then
  echo "$1 not matched.";
  exit 1;
fi;

if ! grep --silent "$2" "$3"; then
  echo "$2 not matched.";
  exit 1;
fi;

Upvotes: 3

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