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Help with MySQL / PHP problem

Hello,

I have come across a problem that I believe that I need to use a for loop but I don't know how to use it here.

Here is the code:

PHP Code:


		
			
                $q60 = mysql_query("SELECT VotingSiteType FROM votingsite WHERE CheckedForVoting=1 AND userid='$uid'");

                        $q61 = mysql_fetch_array($q60);

                        $usid3 = $q61['VotingSiteType'];

                        

                        $q1 = mysql_query("SELECT * FROM site_statuses WHERE id='$usid3'");

                        while ($q1d = mysql_fetch_array($q1)) {

                        

                        

                        $status = $q1d['status'];

                        if ($status == 1) { $input = ' class="i-16-sonline"'; $txt = 'Working'; } 

                        if ($status == 3) { $input = ' class="i-16-soffline"'; $txt = 'Broken'; } 

                        if ($status == 2) { $input = ' class="i-16-smaintain"'; $txt = 'Maintainence'; } 

                        echo '<a href="/panel-vote&id='.$q1d['id'].'" rel="tooltip-top" original-title="'.$txt.'"><li'.$input.'>'.$q1d['site'].'</li></a>

                            

                        ';

                        }

The output is only 1 list of the site, instead there should be seven. The problem consists here.

PHP Code:


		
			
$q60 = mysql_query("SELECT VotingSiteType FROM votingsite WHERE CheckedForVoting=1 AND userid='$uid'");

                        $q61 = mysql_fetch_array($q60);

                        $usid3 = $q61['VotingSiteType'];

                        

                        $q1 = mysql_query("SELECT * FROM site_statuses WHERE id='$usid3'");

                        while ($q1d = mysql_fetch_array($q1)) {

Instead of it just being like that it should get all the values not just the first one... I don't know how to explain it well but please help as much as you can.

Thanks!

Can you do a var_dump of the results?

If the var_dump only shows 1 result, then it is a problem within your query statement or you don't have your data set right in your database.

Yea, no. The problem is the fetch_array is suppose to fetch more then 1 row. But It doesn't since there is no loop for it

So then what's your question? You seem to already know the answer

Yea but I don't know how to do it

Ah! I understand what you're asking..

PHP Code:


		
			
$q60 = mysql_query("SELECT `VotingSiteType` FROM `votingsite` WHERE `CheckedForVoting` = 1 AND `userid` = '".$uid."'"); 
while($q61 = mysql_fetch_array($q60)){
    $usid3 = $q61['VotingSiteType']; 

    $q1 = mysql_query("SELECT * FROM `site_statuses` WHERE `id` = '".$usid3."'"); 
    while($q1d = mysql_fetch_array($q1)){
        $status = $q1d['status'];
        switch($status){
            case 1:
                $input    = ' class="i-16-sonline"';
                $txt      = 'Working';
            break;
            case 2:
                $input    = ' class="i-16-smaintain"';
                $txt      = 'Maintainence';
            break;
            case 3:
                $input    = ' class="i-16-soffline"';
                $txt      = 'Broken';
            break;
        }
        echo '<a href="/panel-vote&id='.$q1d['id'].'" rel="tooltip-top" original-title="'.$txt.'"><li'.$input.'>'.$q1d['site'].'</li></a>';
    }
}

Maybe I understand what you're asking...

: squint :
Why you have two queries there?

Code:

SELECT ss.status, ss.id, ss.site

FROM site_status ss

RIGHT JOIN votingsite v ON v.VotingSiteType = ss.id

WHERE v.CheckedForVoting = 1 AND v.userid = $uid

I don't think he wants a RIGHT JOIN there.

I think just an INNER JOIN.

If you look at his while loop on the votingsite records, the only thing he does with it is get and ID to be used in the inside while loop. So he doesn't really care about any values from the votingsite table.