how to display db result in form_dropdown

you need to pass the dropdown options in an array like..

 $options = array( 'small' => 'Small Shirt', 'med' => 'Medium Shirt', 'large' => 'Large Shirt', 'xlarge' => 'Extra Large Shirt', ); 

so i created a new array .. added options in correct array format and displayed in the view

try this

Controller

public function view_dept() { $this -> load -> model('model_database'); $query = $this -> model_users -> deptlist_view(); $tempArray=array(); foreach($query as $row){ $tempArray[$row->var_staffdepartment]=$row->var_staffdepartment; } $data['department']=$tempArray; $this -> load -> view('signup', $data); } 

view

 <?php echo form_open('welcome/signup_validation'); ?> <table align="center"> <tr> <td> <p> Departments :</p> </td> <td><?php echo form_dropdown('staffdept', $department); ?></td> </table> <?php echo form_close(); ?> 

Controller:

public function view_dept() { $this->load->model('model_database'); $query = $this->model_users->deptlist_view(); $data['staffdept'] = $query->result(); // $this->load->view('signup', $data); } 

View:

<?php $options = array('Select a department'); foreach ($staffdept as $staff): $options[$staff->ID] = $staff->name; //insert the sql query results here endforeach; echo form_dropdown('staffdept', $options, $staff->ID); ?> 

To add a class or ID to the dropdown, edit the last 'echo form_dropdown' line in your code

$other = "id='staffdept', class='staffdept'"; echo form_dropdown('staffdept', $options, $staff->ID, $other);