php - how to get the img url from a href using loadhtml - php

Question

how to get the img url from a href url using dom loadhtml ? i try using $link->nodeValue to get the img src url but is not working
Example url source:
<img src="www.google.com/test.jpg" />Photo NodeValue
My php code:
// -------------------------------------------------------------------------
// ----------------------- Get URLs From Source
// -------------------------------------------------------------------------
function getVidesURL($url) {
$web_source = $this->getSource($url);
if($web_source != '') {
$Data = $this->Websites_Data[$this->getHost($url)];
preg_match($Data['Index_Preg_Match'], $web_source, $Videos_Page);
$Videos_Page = $Videos_Page[$Data['Index_Preg_Match_Num']];
if($Videos_Page != '') {
$dom = new DOMDocument;
#$dom->loadHTML($Videos_Page);
$links = $dom->getElementsByTagName('a');
foreach ($links as $link) {
$Video_Status = "";
$Video_Error = "";
$Video = array(
"URL" => $link->getAttribute('href'),
"Title" => $link->getAttribute('title'),
"MSG" => $link->nodeValue,
);
// Get Image URL Start
$dom = new DOMDocument;
#$dom->loadHTML($Video['MSG']);
$Video_Image = $dom->getElementsByTagName('img');
foreach ($Video_Image as $Image) {
$Video = array(
"IMG" => $link->getAttribute('src'),
);
}
$Videos_URLs .= $Video['IMG'] . '<br />';
}
// Get Image URL Stop
return $Videos_URLs;
}
}
}
The only problem of my code is i don't know how to get the img url from a href

Answer 1

Here is a small function that can pull out image sources from an HTML input:
<?php
echo PHP_EOL;
var_dump(getImgSrcFromHTML('<img src="www.google.com/test.jpg" />Photo NodeValue<div><img src="www.google.com/test2.jpg" /></div><table><tr><td><img src="www.google.com/test3.jpg" /></td></tr></table>'));
echo PHP_EOL;
function getImgSrcFromHTML($html){
$doc = new DOMDocument();
$doc->loadHTML($html);
$imagepPaths = array();
$imageTags = $doc->getElementsByTagName('img');
foreach ($imageTags as $tag) {
$imagePaths[] = $tag->getAttribute('src');
}
if(!empty($imagePaths)) {
return $imagePaths;
} else {
return false;
}
}
Hope this helps.