[SOLVED] Too many rows outputed!

[techiefreak05]

I have a widget system on my site Im trying to make, where users can choose which widgets they want to display on their home page.

I have the following code that selects which "widgets" the user has, and then the second query in the code is supposed to pick out the widgets that the user doesnt already have..

 

<?php
	$queryFlips = mysql_query("SELECT flipID  FROM userFlips WHERE `userID` = '$_SESSION[id]'") or die(mysql_error()); //Get FLIPS user has.
	$db = mysql_fetch_array($queryFlips);
	$fid=$db[flipID];
			$queryFlips2 = mysql_query("SELECT * FROM flips WHERE `id` != '$fid'") or die(mysql_error()); // Get info of flips that user doesnt have.
			while ($db2 = mysql_fetch_array($queryFlips2)){
				$flipName=$db2['name'];
				$flipId=$db2['id'];
				echo '<tr>
		<td><form action="" method="post"><input type="submit" name="add" value="Add"><input type="hidden" name="addID" value="' . $flipId . '"></form></td><td>' . $flipName . '</td>
				</tr>';
			}

?>

 

But.. it is displaying widgets that I already have... I'm not great at joining queries or what ever its called, so id love some help!

[Barand]

[pre]

user          userflips      flips

--------      ---------      --------

userID  ---  userID    +--- flipID

username      flipID  ---+    name

[/pre]

 

To get user's flips

 

SELECT f.name FROM flips f

INNER JOIN userflips uf ON uf.flipID = f.flipID

WHERE uf.userID = '$user';

 

flips user doesn't have

 

SELECT f.name FROM flips f

LEFT JOIN userflips uf ON uf.flipID = f.flipID AND uf.userID = '$user'

WHERE uf.flipID IS NULL

[techiefreak05]

should that work right away? anything I need to change?

 

Also, what is "f.name" and "uf" ??

[Barand]

table aliases

 

SELECT f.name FROM flips f

LEFT JOIN userflips uf ON uf.flipID = f.flipID AND uf.userID = '$user'

WHERE uf.flipID IS NULL

 

 

$user = $_SESSION['id'];

[techiefreak05]

Oh ok, thanks a lot, I'll give it a try and return with the results.

[techiefreak05]

It works great! Thanks a lot! I needed to change the "f.flipID" to "f.id" because in the "flips" table, the id column is called "id". and I added a while loop:

 

<?php
$queryFlips = mysql_query("SELECT f.name FROM flips f
LEFT JOIN userFlips uf ON uf.flipID = f.id AND uf.userID = '$_SESSION[id]'
WHERE uf.flipID IS NULL");
while($db = mysql_fetch_array($queryFlips)){
$flipId=$db['id'];
$flipName=$db['name'];
echo '<tr>
	<td><form action="" method="post"><input type="submit" name="add" value="Add"><input type="hidden" name="addID" value="' . $flipId . '"></form></td><td>' . $flipName . '</td>
</tr>';
}
?>

[Barand]

You'll need to select f.id as well as f.name.

[techiefreak05]

I was JUST going to ask about that Thanks again.

[techiefreak05]

Uh oh, wait. it's not quite working yet.

 

I had this, which selected the flip name just fine, but not the ID..

 

<?php
while($db = mysql_fetch_array($queryFlips)){
$flipId=$db['id']; // this didnt work.
$flipName=$db['name']; // this DID work.
?>

 

Then I changed it to this, and now name doesnt work as well as ID.

 

<?php
while($db = mysql_fetch_array($queryFlips)){
$flipId=$db['f.id']; //this doesnt work..
$flipName=$db['f.name']; ///this doesnt work either.
?>

[Barand]

did you change the query to

 

SELECT f.id, f.name FROM flips f

LEFT JOIN userFlips uf ON uf.flipID = f.id AND uf.userID = '$_SESSION[id]'

WHERE uf.flipID IS NULL

[techiefreak05]

Ah, didnt catch that. Ill do it now.