Image as link using l() function

bbcode maybe

try http://drupal.org/project/bbcode

there is an optional parameter you can pass to tell drupal you are passing html in the link. By default this is set to false.

l($text, $path, $attributes = array(), $query = NULL, $fragment = NULL, $absolute = FALSE, $html = FALSE)

I want to see the coding of this l function, can any one tell me?

Thanks

Elvis Lam

For Drupal 6, this became something like:

print l($text,'node/'.$node->nid, array('html'=>TRUE));

In D6 you can also do this:

Simple:

l('<img src="path/to/image.jpg"/>', $filepath, $options = array('html' => TRUE));

With ImageCache module:

l(theme('imagecache', $preset, $filepath, $alt, $title, $attributes = array()), $filepath,  $options = array('html' => TRUE));

With ImageCache module and Lightbox2 module:

l(theme('imagecache', $preset, $filepath, $alt, $title, $attributes = array()), $filepath, array('attributes' => array('rel' => 'lightbox[roadtrip]'), 'html' => TRUE));
#ImageCache provides a formatter function which does the same. To my mind this one here is more simple...

Can somebody tell me how to convert in function

l($text,'node/'.$node->nid, array('html'=>TRUE));

$node->nid to original picture location?

I use image module and there is something like:
l(image_display($image, IMAGE_THUMBNAIL), 'node/'. $image->nid, array('html'=>true));

and it produce link to a node - I wanna make link to original location of image on disk. Any idea?

My snippet is like

    $src = drupal_get_path('module', yourmodule').'/images/icon.png';
    $img = theme_image($src);
    $path =  "yourpath"
    l($img,$path, array('attributes' => array('class' =>'link-image'),'html' => true,'query' => drupal_get_destination()));

Hope it helps you

Snippet from my template.php

 l(theme_image_style(array('path' => $variables['node']->field_logo['und']['0']['uri'], 'style_name' => '100x100')), 'node/' . $variables['node']->nid, array('html' => TRUE));