PHP | Imagick valid() Function

The Imagick::valid() function is an inbuilt function in PHP which is used to check if the current item is valid.

Syntax:

bool Imagick::valid( void )

Parameters: This function doesn’t accept any parameter.



Return Value: This function returns TRUE on success.

Exceptions: This function throws ImagickException on error.

Below given programs illustrate the Imagick::valid() function in PHP:

Program 1:

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
try {
  
    // Create a new imagick object with invalid image
    $imagick = new Imagick('undefined_source');
    if ($imagick->valid()) {
        echo 'Image is valid';
    }
} catch (exception $e) {
    echo 'Image is not valid';
}
?>

chevron_right


Output:

Image is not valid

Program 2:

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
try {
  
    // Create a new imagick object with valid image
    $imagick = new Imagick(
    if ($imagick->valid()) {
        echo 'Image is valid';
    }
} catch (exception $e) {
    echo 'Image is not valid';
}
?>

chevron_right


Output:

Image is valid

Reference: https://www.php.net/manual/en/imagick.valid.php




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.


Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.