Question

I have returned a result_array() in $categories variable and now I want to create a dropdown menu. But if I directly use the variable, all the elements of the array are going to be displayed. However I just want to display the name and keep its id as value. How do i do it?

$options = $categories;
echo form_dropdown('category', $options);

I want something like

<select name="category">
<option value="1"> ABC </option>
..... 
<option value="10"> NNNC </option>
</select>

Answer 1

Keys in your $options array are not actual ID’s of database rows.

So your $options array looks like this:

Array(
'0' => array('id'=>'10', 'value'=>'someval1'),
'1' => array('id'=>'22', 'value'=>'someval2'),
'2' => array('id'=>'36', 'value'=>'someval3'))

To see this, put print_r($options); before echo call.

To make real dropdown you should make helper function that looks like this:

function my_form_dropdown($name, $result_array){
    $options = array();
    foreach ($result_array as $key => $value){
        $options[$value['id']] = $value['value'];
    }
    return form_dropdown($name, $options);
}