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How can I request the url for a channel index page from another page. I got this message asked on twitter towards craft, but they misunderstood my question.

The came back to me using craft.request.getSegment(x), but this will only work if I would like the index page of a channel when I'm on the entry page of this channel.

But how would I get the url for the "news" channel index page, when I'm in the "projects" channel section (either the index or an entry) of the site?

News is an entry-field of projects btw, so that we can show news related to a project on the site.

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  • The channel index page or section landing page will be whatever you define it as in the channel's settings. If your channel is news then it will likely be simply <a href='news'>...<a/>. Is there something in your question that I am missing? Commented Nov 12, 2014 at 14:31
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    Thanks for the clarification. However, this seems like a duplicate of your previous question then. Check out that answer and see if it doesn't answer this question also. Commented Nov 12, 2014 at 14:55
  • I don't think these questions are related, as the other one is about creating subpages. This question is about requesting the slug from craft when I know the channel name, as these 2 obviously don't have to be related or have the same name. Commented Nov 13, 2014 at 8:14

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There is no predefined 'index' page for channels. Channels are simply a collection of entries. If you want an index page then you will need to create it: either as a single, as an entry within another structure, or as a stand alone template in your templates directory (i.e. craft/templates/news/index.html or craft/templates/news.html). In any case, the url (or uri) is going to be whatever you have defined it as. (i.e. mysite.com/news/). See craft's routing documentation for more info on how requests are routed.

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