CODEWITHSUNDEEP
php
StudioTime
StudioTime

Reputation: 23989

Find images but only jpeg

I am using the following to parse an html page and return the images:

$html = file_get_html($url);

foreach($html->find('img') as $e){ 

    $image = $e->src;

            }

Ideally though I only want to show files with .jpg extension - I know I can probably use preg_match but is there a more efficient way?

Upvotes: 0

Views: 123

Answers (3)

DaveRandom
DaveRandom

Reputation: 88657

NOTE: Just because a file has a JPEG-like extension does not mean that the content of the referenced file is actually JPEG data!

pathinfo() is the native PHP function you are looking for. Yes you can manipulate the string however you want, but if you are looking for the "right" way to do it then here it is:

$html = file_get_html($url);

// file extensions we are looking for
$allowedExtensions = ['jpg', 'jpeg', 'jpe'];

foreach($html->find('img') as $el){ 
    // file extensions are usually treated as case insensitive, but the
    // implied comparison operator used with in_array() is case sensitive
    // so normalise the data to lower-case.
    $extension = strtolower(pathinfo($el->src, PATHINFO_EXTENSION));

    if (in_array($extension, $allowedExtensions)) {
        // File has a JPEG-like file extension
    }
}

Upvotes: 3

Baba
Baba

Reputation: 95121

Your code is fine. This would work for your case

// Create DOM from URL or file
$html = file_get_html('http://www.yahoo.com/');
// Find all images
foreach($html->find('img') as $element)
{
    $img = strtolower($element->src);
    if(pathinfo($img,PATHINFO_EXTENSION) == "jpg")
        echo $element->src . '<br>';
}

Upvotes: 3

MakuraYami
MakuraYami

Reputation: 3428

$split = explode(".", $image);
$extention = $split[count($split)-1];

Upvotes: 0

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