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Home/ Questions/Q 8661783
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T16:34:52+00:00 2026-06-12T16:34:52+00:00

I have a form which contains a dropdown list of countries generated from database.

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I have a form which contains a dropdown list of countries generated from database. The values are stored in database. There is an option in which user can view or update the valuesinserted. For updating all the form values get fetched from database. What i require is that when form is loaded for updating the selected option of the country dropdown must be that stored in the database.
For eg: if from the following dropdown option2 is selected and inserted into database.

 Dropdown: |option1|<selected>
           |option2|
           |option3|

during update it should be like this

 Dropdown: |option1|
           |option2|<selected>
           |option3|

Here is the code i tried.

      $selected = $list["country_country_name"];

     <tr><td>Country</td><td><select onchange="getCountry(this.value);" name="country" id="country" ><?php  foreach( $query as $qry ) { 
     print '<option value="'.$qry["country_country_name"].'"'; 
     if( $qry["country_country_name"] == $selected ) print'selected'; 
     print '>'.$qry["country_country_name"].'</option>'."\n";} ?>
     </select></td></tr>
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  1. Editorial Team
    Editorial Team
    2026-06-12T16:34:53+00:00Added an answer on June 12, 2026 at 4:34 pm

    $selected = $list[“country_country_name”];

     <tr><td>Country</td>
     <td>
     <select onchange="getCountry(this.value);" name="country" id="country" >
     <?php  foreach( $query as $qry ) { 
        $sel = '';
       if( $qry["country_country_name"] == $selected ) 
        $sel = 'selected="selected"';           
    
        echo '<option value="'.$qry["country_country_name"].'" '.$sel.'>'.$qry["country_country_name"].'</option>'."\n";
        } ?>
        </select>
        <?php echo form_error('country'); ?>
        </td>
     </tr>
    
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