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How to find out if an error is causing an error - PHP

I am currently using imagickimage processing on my website. It is installed correctly for me and works fine. Heres my starting code is:

$image = new imagick($filename); $geo=$image->getImageGeometry();  
$image->setImageInterlaceScheme(2);  $image->setImageCompressionQuality(85);
$image->setImageBackgroundColor('white'); $image = $image->flattenImages(); 
$image->setImageFormat('jpg');   $image->stripImage();

After that I do the rest. Now suppose that this generates an error as I code for this. I tried to add if(!$image){ echo 'error' exit(); }after $image = new imagick($filename), but in vain.

Please help ... Thanks to everyone :)

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1 answer

If an error occurs, Imagick will throw ImagickException, which you can catch:

    try {
        $image = new Imagick($filename);
        $geo = $image->getImageGeometry();  
        $image->setImageInterlaceScheme(2);  
        $image->setImageCompressionQuality(85);
        $image->setImageBackgroundColor('white'); 
        $image = $image->flattenImages(); 
        $image->setImageFormat('jpg');   
        $image->stripImage();
    } catch (ImagickException $e) 
{
        var_dump($e);
    }

for further reference see here .

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