I've to create a custom function to display images because I've to add values from custom fields.

I've added a function to template php, like this:

function mytheme_image($variables) {  

The problem is that in $variables I haven't the fid of the file, but only the path. I have to load the file to get values from field, how should I do?

I'm looking for functions to get fid from path but I haven't found something..

5 Answers 5


@pvorb, you can also use EntityFieldQuery. See the first comment on https://api.drupal.org/api/drupal/includes%21file.inc/function/file_load_multiple/7 for an example. Something along the lines of:

$filename = 'myfirstpony.png';
$query = new EntityFieldQuery();
$result = $query->entityCondition('entity_type', 'file')
  ->propertyCondition('filename', $filename)
// if you are guaranteed there is exactly one result, then:
$file_object = reset($result['file']);
$fid = $file_object->fid;
// but beware you might get 0 or many results

If you have more information -- such as the path (from which you can generate a URI) -- then change the query condition to something like ->propertyCondition('uri', 'public://path/to/myfirstpony.png').

  • Hello and welcome. Please remember: this is not a forum. Answer the question, not other answers or comments.
    – Mołot
    Commented Dec 15, 2014 at 15:49
  • @molot, the original question was: "The problem is that in $variables I haven't the fid of the file, but only the path. I have to load the file to get values from field, how should I do?" Commented Dec 15, 2014 at 15:52
  • If you are addressing arrubiu's question, don't start your answer with "@pvorb, you can also" - that's misleading. If you are addressing pvorb's post, you shouldn't post it under arrubiu's question at all. Simple as that. You may comment @ someone else to tell him your answer solves their doubts, that's OK, but only answer to the question.
    – Mołot
    Commented Dec 15, 2014 at 15:55
  • @molot, why are you being so pedantic? pvorb asked on dman's answer how to answer arrubiu's original question but using EntityFieldQuery. I did that. Commented Dec 15, 2014 at 16:12
  • @ElephantJim - I believe Molot is just trying to keep the q&a format clean and readable for future visitors. Sure no offence meant Commented Dec 15, 2014 at 17:31

Assuming you want to change the image field output - you might want to look at the theme_image_formatter function that you can override and add your variables to the $image var, that will be later sent to theme_image or theme_image_style.

Be aware, that theme_image is working with the paths, so it is possible that the function is not called from within the theme_image_formatter.

  • I've tried it, but using netbeans the system doesn't enter in the function while shows images.
    – arrubiu
    Commented Nov 12, 2012 at 13:16
  • so are the images coming through the image field and displayed as such? Commented Nov 12, 2012 at 13:45
  • Images are "file" field, with "media selector" as widget, and displayed as "rendered entity"
    – arrubiu
    Commented Nov 12, 2012 at 13:50

Suggestions about formatting are helpful, but to answer the actual question, the answer at https://drupal.org/comment/6493610#comment-6493610 looks good.

// Take a look at: file.inc::file_load_multiple
$files = file_load_multiple(array(), array('uri' => $uri));
$file = reset($files); 
// If empty, $file will be false, otherwise will contain the required file

A two-liner instead of one, but should get the result we need.

  • $conditions is marked deprecated in file_load_multiple() so what to do instead?
    – pvorb
    Commented Oct 8, 2014 at 10:29

If using the file_entity module, you can use file_uri_to_object as it can load an existing file object (if exists) or create a new one.


If not using this module, just copy this function from "Source" at the link above.

This is a core issue and should be fixed in D8 sometime: https://www.drupal.org/node/685818


This seems to be the way to do it In Drupal 8

$files = \Drupal::entityTypeManager()->getStorage('file')->loadByProperties(['uri' => $uri]);

Where $files will be an array of file objects, keyed by the file id.

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