Need help on parse error

I m using the code below to extract the first image from a wordpress post

function catch_that_image() {
global $post, $posts;
$first_img = '';
ob_start();
ob_end_clean();
$output = preg_match_all('/<img.+src=[\'"]([^\'"]+)[\'"].*>/i', $post->post_content, $matches);
$first_img = $matches [1] [0];

// no image found display default image instead
if(empty($first_img)){
$first_img = "/images/default.jpg";
}
$first_img = "<img src="' .  $first_img . '"  />";
return $first_img;
}

PHP:

<?php echo catch_that_image() ?>

PHP:

The output , i m getting as

<img src='http://mysiteurl.com/wp-content/upload/2011/3.gif'  />

PHP:

i want to change the single (') quote to double quote (")

Like

<img src="http://mysiteurl.com/wp-content/upload/2011/3.gif"  />

PHP:

How this can be achieved ..

Many thanks in Advance!

 

I would take a look at PHP's preg_replace method. You can create a simple regex to find a URL with single quotes (if one exists) and then replace it with the appropriate double quotes.

Regards,
Dennis M.

 

You have a syntax error in

$first_img = "<img src="' .  $first_img . '"  />";

PHP:

Try this, it should work:

$first_img = '<img src="' .  $first_img . '"  />';

PHP: