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I have a rss feed for my demo blog here.

http://clickquickcash.com/feed/?user=john

It passes the 'user' variable 'john' into the feed posts.

I need to add the 'user' variable to the link.

Note - I had to replace my domain with 'websiteurl' below as this site wouldn't let me post more than 1 link as new user

websiteurl/this-is-a-test-2/

becomes...

websiteurl/this-is-a-test-2/john websiteurl/this-is-a-test-2?user=john

Please can someone tell me what I need to change in the function.php or feed.php files.

Thanks

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How does your site pass "the 'user' variable 'john' into the feed posts" Post that code, please. –  s_ha_dum Jan 18 '13 at 15:15

1 Answer 1

add the following to your functions.php file:

add_filter('feed_link','my_site_feed', 10, 2);
function my_site_feed($url, $feed_type ){
    $user = my_get_the_username(); //this function should return the username.
    return add_query_arg( array('user' => $user ), $url );
}

And get your rss feed url using:

echo get_bloginfo('rss2_url');

or

bloginfo('rss2_url');
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Thanks - but I John is a variable. I added the following code - WILL IT WORK? add_filter( 'query_vars', 'addnew_query_vars', 10, 1 ); function addnew_query_vars($vars) { $vars[] = 'user'; // var1 is the name of variable you want to add return $vars; } add_filter('feed_link','my_site_feed', 10, 2); function my_site_feed($url, $feed_type ){ return add_query_arg( array('user' => '$vars' ), $url ); } –  Dom Bowkett Jan 18 '13 at 12:37
    
I don't think you need to filter query_vars and at any rate, that's the wrong filter ;) if John is a variable then you just need to add it within the my_site_feed() function. I've updated my answer to reflect this. –  WP Themes Jan 18 '13 at 14:01

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