如何從PHP中的Alt代碼中獲取HTML代碼中的圖像URL

Question

實際上，我想通過PHP中的image alt從HTML代碼中獲取圖像URL [src]，但是我沒有選擇。

<?php 
$html = ' 
<img border="0" src="/images/image11.jpg" alt="Image11" width="100" height="100" /> 
<img border="0" src="/images/image22.jpg" alt="Image22" width="100" height="100" />'; 

$doc = new DOMDocument(); 
$doc->loadHTML($html); 
$xpath = new DOMXPath($doc); 
$src = $xpath->evaluate("string(//img/@src)"); 

// will return /images/image.jpg 
echo $src; 
?> 

此代碼只是獲取具有alt =「Image22」的圖像的src。

Answer 1

$html = '<img border="0" src="/images/image11.jpg" alt="Image11" width="100" height="100" /><img border="0" src="/images/image22.jpg" alt="Image22" width="100" height="100" />'; 

$dom = new DomDocument(); 
$dom->loadHTML($html); 

$images = $dom->getElementsByTagName('img'); 

foreach ($images as $key => $image){ 

    if ($image->hasAttributes()){ 

     if ($image->hasAttribute("src")){ 

      if ($image->hasAttribute("alt")){ 

       $alt = $image->getAttribute("alt"); 

       if ($alt == 'Image22'){ 

        $src = $image->getAttribute("src"); 

        echo "<br>" . $src . "<br>"; 

       } 

      } 

     } 

    } 

} 

Answer 2

<?php 
$html = ' 
<img border="0" src="/images/image11.jpg" alt="Image11" width="100" height="100" /> 
<img border="0" src="/images/image22.jpg" alt="Image22" width="100" height="100" />'; 

$doc = new DOMDocument(); 
$doc->loadHTML($html); 
$xpath = new DOMXPath($doc); 

$res = $xpath->query('//img'); 
foreach($res as $img) { 

    echo $img->getAttribute('alt'); 
} 


?>