Convert Image hyperlink into <img> with PHP

I have this block of code that has a URL to an image. How can convert that into tag with the src? here is the sample code

    <div class="field-items">
        <div class="field-item odd"><a type开发者_StackOverflow中文版="image/jpeg; length=72486" href="http://www.xyz-website.com/files/263780_147936011949340_128540677222207_316406_533387_n.jpg">263780_147936011949340_128540677222207_316406_533387_n.jpg</a></div></div>

So from the above code, I want to convert that into something like this

<div><img src="http://www.xyz-website.com/files/263780_147936011949340_128540677222207_316406_533387_n.jpg"/><div>

I also want to do this for a particular domain. in the example above its xyz-website.com

Thank you.

Here is the code you need. You will want to modify it to be more specific to your needs. But this uses a regular expression to help extract the image path and saves all found images to an array that you can then use to output it.

$str = '<div class="field-items"><div class="field-item odd"><a type="image/jpeg; length=72486" href="http://www.xyz-website.com/files/263780_147936011949340_128540677222207_316406_533387_n.jpg">263780_147936011949340_128540677222207_316406_533387_n.jpg</a></div></div>';

preg_match('/href="(.)+"/', $str, $matches);
$image_extensions = array('.jpg', '.bmp', '.png');

foreach ($matches AS $match)
  foreach ($image_extensions AS $ext)
    if (stristr($match, $ext . '"')) //I add the double-quote to the string to signify the end of the href param and prevent string detection confusion
      $images[] = str_replace(array('href="', '"'), '', $match);

foreach ($images AS $image)
  echo "<div><img src='$image'/><div>";

use regex.

preg_replace('~<div class="field-item odd"><a type="image/jpeg; length=72486" href="(.+?)">(?:.+?)</a></div></div>~i', '<div><img src="$1"/><div>', $str);

something like this

You could explode the a tag and grab the info you want:

$all_html = '263780_147936011949340_128540677222207_316406_533387_n.jpg';
$tmp1 = explode('href="', $all_html);
$tmp2 = explode('"', $tmp1[1]);
$image_url = $tmp2[0];
if (strpos($image_url, 'xyz-website.com') > -1) {
    //only output image if the url of the image includes 'xyz-website.com'
    echo '';
}