Question

I’m trying to use TimThumb to display image on my blog

<?php the_title(); ?>'><img src="<?php bloginfo('template_directory'); ?>/timthumb.php?src=<?php echo catch_that_image() ?>&w=290&h=160

the above is the code I use

The problem is that the image doesn’t display because the URL come out something like this.

http://myurl.com/wp-content/themes/ThemeName/timthumb.php?src=&w=290&h=160

In which is suppose to be like this for example

http://myurl.com/wp-content/themes/ThemeName/timthumb.php?src=http://myurl.com/wp-content/uploads/i0n1c.png&w=290&h=160

As you can see above is missing the part from src= and that missing part is this code <?php echo catch_that_image() ?> I’ve tested this code separately and it works just fine but when I combine them together then it doesn’t display it properly?

Is there something I have to add to make it works?

UPDATE:

Here is the function I use to grab the image

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];

  if(empty($first_img)){ //Defines a default image
    $first_img = "/images/default.jpg";
  }
  return $first_img;
}

and here is the code to call the URL path for the image.

<?php echo catch_that_image() ?>

Answer 1

I’ve tested this code separately and it works just fine

That code should output the variable, so you did it in another context, where the value is properly set.

Update:

have you checked the generated source code? You have exta backticks here:

?src='<?php echo $image[0]; ?>'&w=
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