using a feed as a variable

[ianraba]

Hi
On my test site: http://www.jobs4salons.co.uk/rss/rss_list.php
am pulling all my relvant feeds together and linking from that page to this:
http://www.jobs4salons.co.uk/rss/rss.php?id=2
am getting the error:

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in /home/jobs4sal/public_html/rss/rss.php on line 51

Line51 is: $MXRSS_rss1->getFeed("$row_rss['rss_feed']");

so i have replaced the actual hardcoded url with one from my database and i know its wrong and i need to encapsulate that call but am stumped as to how to achieve it so that the output between quotation marks is what is in DB

Thanks for any advice

[Fred]

Hi Ian,
Not sure if I understand you correctly.

You have a db with a list of feeds?
e.g. feed id=2 will be something like http://www.something.com/index.xml

On you list page leave it as it is.
On you rss.php?id=2 you will again query you db first to get the actual feed-url.

SELECT * from feeds WHERE id = url_paramter

I am using the MXRSS Recordset on a page where you normally hard_code the url

Just did a test "simulating" a variable feed-url

Code: Select all

$news = 'http://www.avweb.com/avwebbiz/index.xml';

$MXRSS_Recordset1 = new MX_RssReader();
$MXRSS_Recordset1->setConnection($database_fosterDb, $fosterDb);
$MXRSS_Recordset1->setCacheTime(60);		
$MXRSS_Recordset1->getFeed($news);