Show/ hide custom source from XML-feed
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Hi,
I am a big fan of HungryFEED and I am sure it might also work for my current problem.
I tried to figure it out for hours, but I do not find the right way to configure the template.
I hope somebody can help me do this:The feed I want to insert in HungryFEED is:
https://www.scoop.it/t/socialmedia-umschau/rss.xmlHere is the live example:
https://socialmedia-doktor.de/socialmedia-umschau/My shortcode for this example:
[hungryfeed url=”https://www.scoop.it/t/socialmedia-umschau/rss.xml” feed_fields=”” item_fields=”title,description” template=”2″ max_items=”10″]My template for this example:
<div class=”hungryfeed_item”><div>{{title}}
{{content}}</div>My questions:
1.
Each item has a <link> that goes to scoop.it (from there visitors can go to the original source).
But the xml file shows also a <source url> that indicates the original post.Is there a way to use the <source url> for HungryFEED?
2.
When I import the description from the mentioned feed with HungryFEED, it shows a picture, the description and then two
and the part “See it on Scoop.it, via Socialmedia Umschau”Is there a way to get rid of at least the two
behind the image and description in the content before the part “See it on Scoop.it, via Socialmedia Umschau”?And even better: Is there a way to get rid of the entire part “See it on Scoop.it, via Socialmedia Umschau”?
I really hope there is somebody out there that could help me with this.
Best regards,
Sebastian
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