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I'm new to drupal 7. I am trying to configure a side menu that has links in it from a block within a View I made.

All I want to do is to make it so that the external links in that menu go the actual page themselves (WITHOUT USING A MODULE SPECIFICALLY FOR LINKS).

Is this possible? Is it simply a matter of settings for the fields specific for the links. Here is a screenshot of my view from my daskboard, any help is appreciated

enter image description here

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    The view has nothing to output, you've set Content style instead of Fields. If the fields are on the node, you also need a contextual filter for the node id so the View knows what to load.
    – Kevin
    May 2, 2018 at 20:41
  • Can you give me and example of what you are talking about, which contextual filter should I use? Content: Website (field_website_link:attributes) ?? Content: Link to Full News Article (field_website_url:url)??
    – Greg
    May 3, 2018 at 13:35
  • The contextual filter only needs to be the node id (content id). The argument should be fetched from the URL.
    – Kevin
    May 3, 2018 at 14:04
  • But when I do that it removes the links and that entire sidebar completely off the page. There is a second page that gives me options and I am honestly not sure what to pick You mean select this one Content: Nid ?? Then what? I have two sub-sections that require input Section 1 When the filter value is NOT available Then Section 2 When the filter value IS available or a default is provided After spending hours playing around I have no idea which to put for each. Any ideas? What am I doing wrong?
    – Greg
    May 3, 2018 at 15:15
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    Provide default value, Content ID from url.
    – Kevin
    May 3, 2018 at 16:58

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