Imagick will not catch Notices or Warnings

Question

When a Notice or Warning occurs, the image fails. However, I am unable to catch the notice or warning.

<?php

    $image = new Imagick($resource);

    try {
        $image->setImageCompressionQuality("Should be a Number Here not String");
    }
    catch ( ImagickException $e ) {
        echo "This is a catch"; // should catch here but nope!
    }
?>

The code above should catch because of the string passed when should be INT. The image fails but catch does not execute. I still get this message:

Notice: Use of undefined constant Should be a Number Here not String - assumed 'd' in /var/www/class.php on line 15 Warning: Imagick::setimagecompressionquality() expects parameter 1 to be long, string given in /var/www/class.php on line 15

I also tried ( Exception $e )

because that's just a warning/notice. It's NOT an exception. An exception would only occur if something actually `throw`s it. — Marc B, Oct 28 '14 at 14:08
@MarcB Ok, makes sense but the image output still fails so what is one to do? — JavaScript_is_4_Real_men, Oct 28 '14 at 14:09
@MarcB Lol, yes. One would hope not, but when you do not have control of your parameter values then you've got a serious issue! — JavaScript_is_4_Real_men, Oct 28 '14 at 14:14
easy enough to test if the provided value is a number or not. — Marc B, Oct 28 '14 at 14:15
@MarcB yeah but the program has 20 to 30 methods that accept parameters and accepted values range so its possible but much work. — JavaScript_is_4_Real_men, Oct 28 '14 at 14:20

score 1 · Answer 1 · answered Oct 28 '14 at 15:08

Because the method doesn't throw an exception in case on invalid input. You should do something like:

$result = @$image->setImageCompressionQuality("Should be a Number Here not String");
if (!$result) {
    throw new \Exception('Operation has failed');
}

score 1 · Accepted Answer · edited May 23 '17 at 12:13

1

As said above, you should also check return value of $image->setImageCompressionQuality and can hide notices with @.

But also you can convert notices in your code to exceptions like described in this post. It is interesting solution, but i dont recommend to follow it. Checking correctness is better.

edited May 23 '17 at 12:13

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answered Oct 28 '14 at 16:42

Vavilen T

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Actually the return value of the method is always 1 even if it fails (aka a notice or warning) so that won't work. – JavaScript_is_4_Real_men Oct 28 '14 at 23:13

Imagick will not catch Notices or Warnings

2 Answers2