Nested dropdown with dynamic data

Question

I want to create a drop down list which is showing data from two tables. I have a CATEGORY table and a SUB_CATEGORY table. I have written the following code, but its only showing me items from CATEGORY table, and a blank space after each category. I want something like this in the drop down list.

Required Output

I have written the following code, but it's not giving me the required output. Kindly check it.

 <?php
include 'connect.php';

echo "<select name='category'>";

$select_query=          'Select * from category';
$select_query_run =     mysql_query($select_query);

  $sub_category_query="Select * from sub_categories where id='".$select_query_array['category_id']."'";

      $sub_category_query="Select * from sub_categories ";
      $sub_query_run=         mysql_query($sub_category_query);


while ($select_query_array=   mysql_fetch_array($select_query_run) ) {

     echo "<optgroup label='".$select_query_array['name']."' >".

              //    $sub_category_query="Select * from sub_categories where id='".$select_query_array['category_id']."'";

                    $sub_category_query="Select * from sub_categories";

                $sub_query_run=         mysql_query($sub_category_query);


     while   ($sub_query_run1=   mysql_fetch_array($sub_query_run) ) {
          echo "<option value='".$sub_query_run1['sub_category_id'] . "' >" .
               htmlspecialchars($sub_query_run1['sub_category_name']) . "</option>";
     }
     echo "</optgroup>";
 }
 echo "</br>";

 $selectTag= "</br><input type='submit' value='Insert Product'  /></select></form>";

 echo "</div></div>";

 echo $selectTag;





?>

Assuming you have a link key column between the 2 tables your code is almost fine just need to move the sub query into the while loop and put your sub output inside its own while — Dave
– Dave, Commented Jul 22, 2013 at 10:02

nickle · Accepted Answer · 2013-07-26 11:29:33Z

1

This will work only if u maintain forgien key relation between two tables.

echo "<select name='category'>";

$select_query=          'Select * from category';
$select_query_run =     mysql_query($select_query);

while ($select_query_array=   mysql_fetch_array($select_query_run) ) {

     echo "<optgroup label='".$select_query_array['category_id']."' >".
          htmlspecialchars($select_query_array["name"]).

     $sub_category_query="Select * from sub_categories where id='".$select_query_array['category_id']."'";

     $sub_query_run=         mysql_query($sub_category_query);

     while   ($sub_query_run1=   mysql_fetch_array($select_query_run) ) {
          echo "<option value='".$sub_query_run1['sub_category_id'] . "' >" .
               htmlspecialchars($sub_query_run1['sub_category_name']) . "</option>";
     }
     echo "</optgroup>";
 }
 echo "</br>";

 $selectTag= "</br><input type='submit' value='Insert Product'  /></select></form>";

 echo "</div></div>";

 echo $selectTag;

edited Jul 26, 2013 at 11:29

answered Jul 22, 2013 at 9:58

nickle

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2 Comments

Taha Kirmani Over a year ago

Parse error: syntax error, unexpected '}' in F:\xampp\htdocs\CMS\drop_down_list_test2.php on line 65

Taha Kirmani Over a year ago

I think its because of the missing semi colon at the end of </option> line, but now after rectifying error, its showing me empty drop down list in the output.

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