PHP if(!@include_once()) doesn't work

Question

This was working and then it stopped. That was in version 7.2.6 the latest.

if(!@include_once('config.php')) {
    echo 'failed';
}

So we've downgraded to 7.1.9 but it doesn't work there as well.

There is no error thrown, nothing. Just a blank screen. It's as if it's not even there...

If I echo something before that, it works. If I echo something after this, nothing happens.

Why is this happening?

So what does your http servers error log file reveal what is happening? — arkascha, Jun 23 '18 at 10:46
@arkascha Nothing in the PHP log file `/var/log/php_errors.log`. HTTP returns 200. Unbelievable. Not seen anything like this before with PHP. — Nikk, Jun 23 '18 at 10:48
@pradeep `Fatal error: Constant expression contains invalid operations`. — Nikk, Jun 23 '18 at 10:49
@GordonM Okay...I need to throw an error in case the file cannot be included. If the @ is removed it doesn't echo. — Nikk, Jun 23 '18 at 10:51
So maybe the call works, but the error is thrown in the included file.. Doesn't the error show a file and line number? — GolezTrol, Jun 23 '18 at 10:53
Also, if you want to thrown an error if a file cannot be included, you can use `require` or `require_once`. — GolezTrol, Jun 23 '18 at 10:53
https://stackoverflow.com/questions/40827870/constant-expression-contains-invalid-operations — Pradeep, Jun 23 '18 at 10:56

Andrew Chart · Answer 1 · 2018-06-23T11:40:12.953

Maybe you could try the advice around "example #4" on this page and see if that makes a difference? http://php.net/manual/en/function.include.php

E.g.

if ((@include_once 'config.php') === FALSE) { 
    echo "failed";
}

By the way: as other's have said, @ suppresses PHP warnings/errors for the line of code where it's applied.

My interpretation of your question is that you have used @ deliberately to suppress the engine's warning and echo your own message instead. If you removed the @ and the include failed, I'd expect to see the warning message output if errors are set to show, AND the word 'failed' should still be echoed afterwards.

score 0 · Answer 2 · answered Jun 23 '18 at 11:35

what, exactly, do you think @ does? it suppresses errors. so when you say There is no error thrown, nothing, this is totally expected, because if there is any errors, @ probably suppress them. remove the @, and if you can't trust your php.ini settings, do an ini_set call too, also set in some debug prints, eg

ini_set("error_reporting","-1");
echo "before include.";
if(!include_once('config.php')) {
    echo 'failed';
}
echo "after include";

then check your error logs after it executed.

score 0 · Answer 3 · answered Jun 23 '18 at 12:00

This is how I solved it.

if (!(@include 'config.php')) { die(json_encode(array('error'=>true))); }

I need the error to be suppressed because I wan't to handle the error without breaking my script—in a way that the other end know's it's an error and will inform the front-end user.

PHP if(!@include_once()) doesn't work

3 Answers3