Get image source from html dom element [duplicate]

Question

I am querying image using getElementsByTagName("img") and printing it using image->src , it does not work. I also tried to use image->nodeValue this to does not work.

require('simple_html_dom.php');

$dom=new DOMDocument();
$dom->loadHTML( $str);       /*$str contains html output */

$xpath=new DOMXPath($dom); 
$imgfind=$dom->getElementsByTagName('img');  /*finding elements by tag name img*/

foreach($imgfind as $im)
{
    echo $im->src;        /*this doesnt work  */   
    /*echo $im->nodeValue;  and also this doesnt work (i tried both of them             separately ,Neither of them worked)*/

    // echo "<img src=".$im->nodeValue."</img><br>"; //This also did not work
}

/*the image is encolsed within div tags.so i tried to query value of div and print but still image was not printed*/

$printimage=$xpath->query('//div[@class="abc"]'); 
foreach($printimage as $image)
{
    echo $image->src;   //still i could not accomplish my task
}

use echo $im->getAttribute('src'); instead of echo $im->src; — abhij89
– abhij89, Commented Apr 17, 2013 at 5:31
Also, since you're using DOM, there doesn't seem to be necessary to require('simple_html_dom.php'). — Passerby
– Passerby, Commented Apr 17, 2013 at 5:46
@saveATcode i tried using echo $im->getAttribute('src'); it just prints src value of image and not the image.i.e.<img src='a.jpg'> it prints a.jpg as text and it does not print the actual image . — CodeDada
– CodeDada, Commented Apr 17, 2013 at 6:18

abhij89 · Accepted Answer · 2013-04-17 08:50:57Z

7

Okay, use this to display your image:

foreach($imgfind as $im)
{
  echo "<img src=".$im->getAttribute('src')."/>"; //use this instead of echo $im->src;
}

and it will surely display your image. Make sure path to the image is correct.

answered Apr 17, 2013 at 8:50

abhij89

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1 Comment

CodeDada Over a year ago

,Thanks the code worked ,there is one small correction i made to ur code ,i.e,a space before ending slash in echo "<img src=".$im->getAttribute('src')." />"; rather than echo "<img src=".$im->getAttribute('src')."/>";(it considers image to be a directory without space ..otherways its perfectly fine.thanks a lot buddy.

Raphael Christian Valencia Die · Accepted Answer · 2021-09-17 05:11:48Z

0

Espero te sirva

  $dom = new DOMDocument();
  $filename = "https://www.amazon.com/dp/B0896WB9XD/";
  $html = file_get_contents($filename);

  @$dom->loadHTML($html);

$imgfind=$dom->getElementsByTagName('img');
foreach($imgfind as $im)
{
    $ids= $im->getAttribute('id'); 

    if ($ids == 'landingImage') {
        $im2 = $im->getAttribute('src'); 
       echo '<img src="'.$im2.'">';
    }
    else{
        
    }
}

para amazon.

answered Sep 17, 2021 at 5:11

Raphael Christian Valencia Die

1

2 Comments

ponsfrilus Over a year ago

Please read stackoverflow.com/help/how-to-answer and use english.

Community Over a year ago

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