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On my views-view-field template, I would like to print out 'field_images' if 'field_cover_image' is empty.

I tried this solution here but it throws me this error message.

Notice: Undefined variable: node in include() (line 45 of /sites/all/themes/mytheme/templates/views-view-fields--incubator.tpl.php).

EntityMalformedException: Missing bundle property on entity of type node. in entity_extract_ids() (line 7539 of /includes/common.inc).

Is there any other ways to check if a field is empty or not?

Thanks!

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    please paste your code, what code have you tried?
    – Ahmad
    Jul 17, 2012 at 12:49
  • This is the code that I tried: $field = field_get_items('node', $node, 'field_cover_image'); if($field){ print 'yes there is'; }else{ print 'no there isn't'; }
    – kyooriouskoala
    Jul 19, 2012 at 7:10

4 Answers 4

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You get that error because $node is not one of the variables passed to views-view-fields.tpl.php, which receives:

  • $view: The view in use.
  • $fields: an array of $field objects.
  • $row: The raw result object from the query, with all data it fetched.

Basing on the description, $row could be used to retried the node object. I cannot tell you the content of that variable, but using print_r() from your template file can help to understand its content.

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  • Hi, I've tried all three variables ($view, $fields, $row) that you mentioned and there's no error. However, it doesn't seem to be working as one of the entry that has a Cover Image still prints out "no". Here's the code: $field = field_get_items('row', $row, 'field_cover_image'); if($field){ print 'yes'; }else{ print 'no'; }
    – kyooriouskoala
    Jul 19, 2012 at 7:22
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Use views preprocess API and check if(isset(field['name'])) and then do things as per your needs. Put this in a module to achieve your needs. This will be a long term solution.

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  • I have 2 entries, 1 without cover image and another one with a cover image. So I tried this code below but on the entry without cover image it still prints out "yes". if(isset($fields["field_cover_image"]->content)) { print 'yes'; } else { print 'no'; } PS: I'm no programmer. I would be happy if I could use GUI to set things up to achieve this; which is working for some other content type but not for this particular content type.
    – kyooriouskoala
    Jul 19, 2012 at 7:13
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    I am not sure about all the variables in your code. But i would say there will be a way to differentiate the variable with cover and with out cover image. Put print_r in the module to get the list of variables and then use isset condition. I think you are almost half way done, keep trying.
    – Gladiator
    Jul 19, 2012 at 7:16
  • I think Gladiator is correct. There must be a way to differentiate the things. Try some other fields instead of the one mentioned above in ur coment.
    – user4081
    Jul 19, 2012 at 7:18
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Been awhile but thanks again for sharing your solutions. I ended up using Views if Empty module and that saves me from creating a views .tpl

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The easiest solution for this is to add both fields but add field_images first and mark is as 'exclude from display'. Then add the field_cover_image and in the 'NO RESULTS BEHAVIOR' section add the token for field_images (can be found in replacement patterns). Done!

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