php get URL of first img src inside of specified html element

Question

I'm trying to find the simplest way using PHP to get the URL of the first image inside a specified element/class, to insert into Open Graph og:image and Twitter twitter:image

<figure class="pictures">
<img src="/pictures/1.jpg"> <!-- it would be this here -->
<img src="/pictures/2.jpg">
<img src="/pictures/3.jpg">
</figure>

<meta property="og:image" content="<?php echo $og_image; ?>"/>

Also wondering if I can have it get the domain in front of the relative URL through the echo, instead of hardcoding it in front of the echo in the meta tags.

score 0 · Answer 1 · answered Apr 01 '14 at 06:57

0

Try This:

function gsb($string, $start, $end)
{
    $string = " " . $string;
    $ini    = strpos($string, $start);
    if ($ini == 0)
        return "";
    $ini += strlen($start);
    $len = strpos($string, $end, $ini) - $ini;
    return substr($string, $ini, $len);
}
$pic=gsb($string,'src="','"');

answered Apr 01 '14 at 06:57

here `$pic` contains required link. – Apr 01 '14 at 07:00

retrohacker · Answer 2 · 2014-04-01T07:38:34.007

0

I don't have PHP installed, so I can't actually test this code. But take it as psuedo-code that expresses the idea.

<?php
    echo "<figure class='pictures'>";
    $URLS = array("/pictures/1.jpg","/pictures/2.jpg","/pictures/3.jpg");
    for($i=0;$i<count($URLS);i++) {
        echo "<img src='$URLS($i)'>";
    }
    echo "</figure>";
    echo "<meta property='og:image' content='$URLS(0)' />";
?>

This will give you your desired result, make adding/removing imgs easier, and allow you to algorithmically generate the list of imgs if your code builds out in that direction.

For getting the domain of your server, checkout $_SERVER['HTTP_HOST'] and $_SERVER['SERVER_NAME']

edited Apr 01 '14 at 07:38

answered Apr 01 '14 at 07:23

retrohacker

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I don't think that will work. I'm still very much a beginner with programming, but I think it'd have to be something without any specified URLs inside like that, since it has to apply to many pages with different content, but inside the same element/class. Pseudo-code (obviously doesn't work, lol): ``. It has to be something that will get and echo the first URL of an img src inside the specified element/class, which in this case is `
`
– Orange Apr 01 '14 at 07:41
In which case you will have to process the document as a DOM. From my limited understanding, php doesn't understand the context of the HTML surrounding it, so you will have to pass the whole page through PHP, parse out what you need, generate the extra code, and then place it in the response using echo. This is what HaRsH recommended. – retrohacker Apr 01 '14 at 07:45

HaRsH · Answer 3 · 2014-04-01T07:10:05.707

    with javascript like this BUT
    **by using php FOLLOW THIS LINK**
    [http://stackoverflow.com/questions/10130858/get-img-src-with-php][1]

        //this is jquery code for finding the SRC of first imahe 
        <script>
        //$( document ).ready(function() {//here we are playing with image so use
         $( window ).load(function() {


        var image_url = $( ".pictures img:first-child" ).attr("src");
        //in variable image_url the url of the image
        alert(image_url);

        });


        </script>


      [1]: http://stackoverflow.com/questions/10130858/get-img-src-with-php

    USING PHP:-
<?php     
//i just copy that code from upper link and make small change


    //if you are generating these image by loop then generate $html variable and use this code
    $html = '<img src="/arrow_down.png"><img src="/arrow_down1.png"><img src="/arrow_down2.png">';

    $doc = new DOMDocument();

    $doc->loadHTML($html);

    $xpath = new DOMXPath($doc);

    echo $src = $xpath->evaluate("string(//img/@src)"); # "/images/image.jpg"
?>

It has to be PHP since JavaScript is not compatible with the scraping. However, I don't understand the PHP example since it doesn't display any way to target a specific element/class. — Orange, Apr 01 '14 at 06:33
you want the 1st image so make condition where you generate image in .php file — HaRsH, Apr 01 '14 at 06:35
may be this code helps you '; $doc = new DOMDocument(); $doc->loadHTML($html); $xpath = new DOMXPath($doc); echo $src = $xpath->evaluate("string(//img/@src)"); # "/images/image.jpg" ?> — HaRsH, Apr 01 '14 at 06:49

php get URL of first img src inside of specified html element

3 Answers3