Hi There,
I am currently trying to integrate an RSS feed from my own site into another site and to that end I am attemtping to place the contents of the feed into a table in a MYSQL database so I can build up a searchable archive of news stories.
I have used the following code which only seems to insert 1 news item into the table, can anyone tell me how I can get the entire contents of the feed to the database? I have set the feed to contain 10 items of news so esentially
This is the code I have used so far:
<?php
require("MagicParser.php");
$link = mysql_connect("localhost", "xxxxxxxxxx", "xxxxxxxxxx") or die('Could not connect: ' . mysql_error());
mysql_select_db('xxxxxxxxxxxxxx') or die('Could not select database');
function myRSSRecordHandler($item)
{
$sql = "INSERT INTO rss SET title = '".mysql_escape_string($item["TITLE"])."', link = '".mysql_escape_string($item["LINK"])."', description = '".mysql_escape_string($item["DESCRIPTION"])."', pagebody = '".mysql_escape_string($item["PAGEBODY"])."', logo = '".mysql_escape_string($item["LOGO"])."', contactdetails = '".mysql_escape_string($item["CONTACTDETAILS"])."'";
mysql_query($sql);
}
$url = "http://edmundson-electrical.voltilink.co.uk/rss.jsp?partner=edmundson&password=xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx";
// now we call Magic Parser and let myRecordHandler load each item into the database
MagicParser_parse($url,"myRSSRecordHandler","xml|RSS/CHANNEL/ITEM/");
?>
Any help would be greatly appreciated.
Thanks,
Ray
I didn't auto-increment the primary key, seemed to be causing the problem. Thanks very much for your help
You are a gentleman and a scholar!!
Thanks,
Ray
Hello Ray,
Your code essentially looks correct, which means the problem is almost certainly down to one of:
i) There is only 1 item being returned in the feed. To check this, printout some debug code so
that you can see what is going on - for example add the following immediately after the mysql_query():
print "<p>".$item["TITLE"]."</p>";You should then see the 10 items displayed, which will rule out this problem. Otherwise,
ii) There could be a key problem with the database (duplicate key for example). To find out what might
be happening here, add some database debug code, so where you currently have:
mysql_query($sql);...change this to:
if (!mysql_query($sql)){
print "<p>".mysql_error()."</p>";
}
The error message displayed (if any) should help identify the cause of the problem...
Hope this helps!
Cheers,
David.