2

According to the CreativeWork Schema, "license" is a valid property, but testing the below code with the Google Structured Data Testing Tool renders an error:

Error: Page contains property "license" which is not part of the schema

<span itemscope itemtype="http://schema.org/CreativeWork">
    <meta itemprop="genre" content="Travel">
    <meta itemprop="isFamilyFriendly" content="true">
    <link itemprop="license" content="http://creativecommons.org/licenses/by-nd-nc/1.0/">
</span

Am I to assume that this is a glitch in the testing tool, or that Google ignores the property and it is used elsewhere by other SEs?

GDP
  • 8,109
  • 6
  • 45
  • 82

1 Answers1

2

The license property was added in version 1.6 (2014-06-16).

The Google Structured Data Testing Tool doesn’t recognize many other properties from after and even before this release, so it’s likely that they did not catch up yet.

If Google makes use (or ever will make use) of this property can’t be answered. Neither Google nor other schema.org consumers necessarily make use of all properties and classes, and specific uses might be undocumented or secret.

(Side note: As you have a URL as value of the license property, you must use the link element instead of meta.)

Community
  • 1
  • 1
unor
  • 92,415
  • 26
  • 211
  • 360
  • So...if it's on schema.org, it's valid, whether Google uses/understands or not, right? Google just seems to be the de facto standard in these matters, so I wanted to be sure. Thx for the meta/link, was a typo while asking the question. – GDP Oct 14 '14 at 19:02