mySQL problem

mp3guy · 04-10-2009 02:10PM #1

I'm having trouble with a database I'm setting up, notably my php script isn't returning the correct information from my table.

I'm using the following code:

   $userId = $_SESSION['user'];
   $sql = "SELECT feed, feed_id
           FROM feeds
           WHERE user_id = '$userId' ORDER BY feed_id";

   $result = mysql_query($sql)
             or die('Query failed. ' . mysql_error());
   $feeds = mysql_fetch_assoc($result);
   var_dump($feeds);

This worked fine first run, then I added a new entry to the table through the mysql prompt, but it's not showing up on my page as it should, just the original result is. If I run the query in the mysql prompt I get the correct results. Whats going on here?

feylya · 04-10-2009 02:16PM

What output are you getting when you var_dump? Check the query that it's running too by adding echo $sql;

mp3guy · 04-10-2009 02:21PM

Well, the output of echo $sql is:

SELECT feed, feed_id FROM feeds WHERE user_id = 'john' ORDER BY feed_id;

The output of var_dump is:

array(2) { ["feed"]=>  string(31) "http://www.rte.ie/rss/sport.xml" ["feed_id"]=>  string(1) "1" }

And then if I run it in the mysql prompt:

mysql> SELECT feed, feed_id FROM feeds WHERE user_id = 'john' ORDER BY feed_id;
+-----------------------------------------+---------+
| feed                                    | feed_id |
+-----------------------------------------+---------+
| http://www.rte.ie/rss/sport.xml         |       1 | 
| http://www.rte.ie/rss/entertainment.xml |       3 | 
+-----------------------------------------+---------+
2 rows in set (0.00 sec)

feylya · 04-10-2009 02:30PM

mysql_fetch_assoc only returns 1 row at a time

You need to wrap it in a while loop

while ($row = mysql_fetch_assoc($result))
{
echo $row['feed'] . " - " . $row['feed_id'];
}

mp3guy · 04-10-2009 02:32PM

Ahh, thanks. That did the trick, I'm new to actually implementing databases, so I may be back with more "queries" :P

feylya · 04-10-2009 02:35PM

No worries. PHP & MySQL is great but you'll start getting some stupidly obvious errors.

mySQL problem

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