using a feed as a variable
Posted: 2010-11-26 19:16
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
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