I am creating an RSS feed that outputs all posts from two views: one is the men's blog, and the other is the women's. I found out that, when I specify my filters, I need to select both custom content types, leaving me with no results as if one cancels out the other. How do I add these two together, so I get all posts in the feed?

I noticed in contextual filters there is "month + year." I need something like this for two content types.

3 Answers 3


Are you creating two separate filters? For example in your filter list, does it say:

Content: Type (= MensBlog)

Content: Type (= WomensBlog)

This essentially becomes an AND filter, which means it's looking for a content type that is both MensBlog AND WomensBlog.

You could instead change this to have the data in one filter. Remove one of the filters and click to edit the other one. On this screen, make sure the "Is one of" radio box is selected under Operator and then select MensBlog and WomensBlog. This filter will be an or filter, rather

  • I have tried both ways you suggest but both result in an empty view. Jan 7, 2012 at 0:40

I ended up adding a new field to both content types labeled RSS enabled, and I was able to filter them together that way. I am not sure if this is the best solution, but it works.

  1. Add filter: "Content: Type"
  2. Make the operator "Is one of"
  3. Select the Women's Blog & Men's blog content type names under "Content Types" enter image description here

As mentioned by @Chapabu, you have to use only one "Content Type" filter with the "Is one of" filter. If you use two, you're asking for content type that is both Women's and Men's, which is impossible. That's why it's returning none.

Note: Make sure you remove the other "Content Type" filters for this to work. You should have only one in the end. Having more will display nothing.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.