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Retrieving values from 2 db tables and selecting the checkboxes.

Posted on 2003-03-12
3
Medium Priority
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Last Modified: 2009-07-29
Hi,
I currently have a problem and I've been working on it the whole day but I'm still getting no where.
The situation is...

I have 2 db tables: Tyres and Customers

Tyres
---------------------------
|     Brand     | tyreId |
---------------------------
| Yokohama   |   10     |
| GoodYear    |   11     |
| Michelin      |   12     |


Customers
---------------------------
| Name | custId | tyreId |
---------------------------
| John     |    J1     |  10     |
| John     |    J1     |  11     |
| Tom     |    T2     |  10     |




I'm displaying the Tyres table using checkboxes on a page.
Then, for a customer, say John, he can have more than 1 tyre order.
How do I select 2 checkboxes out of the 3?
I've tried using nested while loop but it seems that it only loop through once and selects one checkbox and that's it.
This was what I did:


<?  
$query2 = "SELECT * FROM Customer where custID='J1'";
$result2 = mysql_query($query2) or die( mysql_error() );
while ($r = mysql_fetch_array($result2)){
          $tyreID[ ] = $r["tyreID"];
}

echo "<tr>";
     echo "<td>";
        echo "Tyres:";
     echo "</td>";
     echo "<td>";
     $query = "SELECT * FROM Tyres";
     $result = mysql_query($query) or die( mysql_error() );
     while ($r = mysql_fetch_array($result)){
          $Brand = $r["Brand"];
          $tyreId = $r["tyreId"];

          if (is_array($tyreID)){
               while(list($key, $ID) = each($tyreID)){

                    echo "<tr>";
                         echo "<td>";
                         echo "</td>";
                         echo "<td>";
                              if ($tyreId == $ID){
                                   echo "<input type=checkbox name='custOrder' Value=$typeId selected> $Brand";
                              }
                              else{
                                   echo "<input type=checkbox name='custOrder' Value=$typeId> $Brand";
                              }
                         echo "</td>";
                    echo "</tr>";
               }
          }
     }
     echo "</td>";
echo "</tr>";
?>




I'm not sure if this logic is the right one to use.
If you have better suggestions, feel free to comment.
I need help as soon as possible. Thanks!
0
Comment
Question by:G-Force
3 Comments
 

Expert Comment

by:beauty_fool
ID: 8118472
try replacing the nested while with this:

/* code */
foreach($tireID as $ID){
/* end code */

just this one line, nothing else
0
 
LVL 11

Accepted Solution

by:
jausions earned 90 total points
ID: 8118488
Hi,

You need to do a SQL OUTER JOIN statement:

<?php

    $sSQLQuery = 'SELECT t.tyreId As tyreId, Brand, custID FROM tyres As t LEFT OUTER JOIN Customer USING (tyreId) where custID=\'J1\'';
    $iResult = mysql_query($sSQLQuery) or die( mysql_error() );

    echo "<tr>\r\n";
    echo "<td>Tyres:</td>\r\n";

    $iLabel = 0;

    while ($aRow = mysql_fetch_array($result)){
         $Brand = $aRow['Brand'];
         $tyreId = $aRow['tyreId'];
         $sSelected = ($aRow['custID']) ? ' selected="selected"' : '';
         echo '<td><input type="checkbox" name="custOrder[]" Value="' . $typeId . '"' . $sSelected . ' id="custOrder_' . $iLabel . '"> <label for="custOrder_' . $iLabel++ . '">' . htmlspecialchars($Brand) . "</label></td>\r\n";
    }

    echo "</tr>\r\n";
?>
0
 

Author Comment

by:G-Force
ID: 8125359
Oh, thanks to you both, but I've found a way to solve this problem here.
jausions, I'll give you the 30 points 'cause I see that you do put in effort.
Anyway, thanks!
0

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