Posted by Harrie Verveer on 07/17/07 11:55

Hi Max,

mysql_fetch_array($result) returns only 1 record from your result. Try:

while($row = mysql_fetch_array($result))
{
print_r($row); // row is 1 record from the memberships table
}

Kind regards,

Harrie Verveer
--
http://www.ibuildings.nl/blog/authors/Harrie-Verveer

J_K9 wrote:
> Hi,
>
> I'm having a problem with a web app I am developing. The code in
> question is:
>
> $sql = "SELECT groupID FROM memberships WHERE userID = ".$uid;
> $result = mysql_query($sql);
> $result = mysql_fetch_array($result);
> print_r($result);
>
> Now, print_r($result) displays the following (when $uid = 1):
>
> Array ( [0] => 1 [groupID] => 1 )
>
> HOWEVER, that's not what it should be. If I run the same command at
> the MySQL CLI I get the following:
>
> +---------+
> | groupID |
> +---------+
> | 1 |
> | 2 |
> +---------+
>
> So, if I'm not mistaken, the $result array should contain ([0] => 1,
> [1] => 2), NOT what PHP says it does. Can someone please tell me why
> this is happening and how to correct it? I need to use the data
> outputted by MySQL in my PHP script but if PHP is not receiving the
> same data then I have no way of using it.
>
> Thanks,
>
> -Max
>

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