How can I display a picture from a field as the user picture rather than the drupal default with user-picture.tpl.php?

Question

Because my site requires user pictures to be approved, the default Drupal user picture is not suitable.

Instead, I want to show a picture from a field in each user's profile as the user's profile picture.

The field is "field_profile_picture".

The image style is "profile_picture".

Based on this tutorial, I put the following code in user-picture.tpl.php in my theme's templates folder:

if (!empty($user->field_profile_picture['und'][0]['filename'])) {
  echo theme('image_style', array('style_name' => 'profile_pic', 'path' => file_build_uri($user->field_profile_picture['und'][0]['filename'])));
}

This doesn't result in any PHP errors but no picture is displayed. It appears that field_acc_profile_picture is evaluating as empty because even if I put a print "Hello world"; inside the {} it doesn't appear. However, I am sure that the field has a value because I can see the picture displayed in views.

Answer 1

I don't think $user->field_profile_picture['und'][0]['filename'] would be set by default, you normally only get the fid column (this does vary depending on context, but you should always get the fid column either way).

This sort of code always works for me:

if (!empty($user->field_profile_picture[LANGUAGE_NONE][0]['fid'])) {
  // Load the file
  $file = file_load($user->field_profile_picture[LANGUAGE_NONE][0]['fid']);

  // Note the style name is "profile_picture" as explained in your question,
  // not "profile_pic" which was in your original code
  echo theme('image_style', array('style_name' => 'profile_picture', 'path' => $file->uri));
}

If you're worried about the extra overhead involved with potentially loading the file twice, don't be, Drupal caches the object the first time round.

It the above doesn't work it's likely that the $user object is not fully loaded and so the fields haven't been attached. If that's the case, load the user like this:

$account = user_load($user->uid);

And then use the $account object to test for the field value.