CODEWITHSUNDEEP
shellconditional-statementsexit
Gecko
Gecko

Reputation: 74

shell test always exit

I'm begginner on shell, currently i wrote a small script and i got a problem without any error :/

This code always exit my script and i dont understand why :

[[ -x $PATH ]] || log_failure_msg "Binary file not found or not executable!"; exit 0

When $PATH is valid i got nothing and if the path is wrong i got my failure message.

If i remove log_failure_msg "Binary file not found or not executable!"; the script work perfectly -_-

Ho i can solve this problem without if/fi conditions?

Thank you for your help!

Upvotes: 2

Views: 73

Answers (2)

phlogratos
phlogratos

Reputation: 13924

[[ -x $PATH ]] || log_failure_msg "Binary file not found or not executable!"; exit 0

is equivalent to 

{ [[ -x $PATH ]] || log_failure_msg "Binary file not found or not executable!" } ; exit 0

What you need is 

[[ -x $PATH ]] || { log_failure_msg "Binary file not found or not executable!"; exit 0 }

I'm assuming you are using bash. The bash man page states:

..., && and || have equal precedence, followed by ; and &, which have equal precedence.

Upvotes: -1

Guido
Guido

Reputation: 2634

The issue is precedence, as explained by phlogratos. However, you can't use parenthesis as they spawn a sub-shell and you'll be exiting that shell. For this particular issue, curly braces exist. They have almost the same semantics but they spawn jobs in the current shell.

$ cat a.sh 
[[ -f file ]] || { echo error; exit 0; }
echo "ok"
$ touch file
$ ./a.sh 
ok
$ rm file 
$ ./a.sh 
error
$

Upvotes: 3

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