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11:10 pm
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June 16, 2022
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I am passing an seoslug in my rss feed, I would like to know how a user can access that variable to create the URL as opposed to using the title.
Sample RSS:
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2:26 am
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June 16, 2022
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8:17 pm
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June 16, 2022
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6:58 pm
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June 16, 2022
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Thank you so much for adding this functionality, you all are amazing!
I am not quite sure how to implement this. I first added the custom feed to pull ds:seoslug from the RSS feed . I based this on the code I use to pull the meta description
I then added the %seoslug% to the post slug
I am doing something wrong here (obviously) because when I pull the feed and publish I get this as the URL for the post: "https://cstesting.dataskrive.com/2022/09/09/seoslug/"
You don't need the "Custom fields" box to do so. You should use templates instead. If you want to have <ds:metadescription> as a post excerpt, you have to add this line into the "Post excerpt" box:
Login to see the code
Actually you can put any HTML code there. E.g.:
Login to see the code
So your post excerpt will have the post title, meta description, excerpt and even... a YouTube video for your post title!
If you want to use <ds:seoslug> as a post slug, you should put this lint into the "Post slug" box:
Login to see the code
You will find more info about CyberSEO Pro templates here: https://www.cyberseo.net/content-syndicator/#templates
The "Custom fields" box is indented to automatically generate new custom fields.
Please find more here: https://www.cyberseo.net/content-syndicator/#custom-fields
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