如何在Codeigniter中創建從數據庫中刪除下拉菜單

Question

對不起，我想從數據庫中使用下拉菜單創建表單輸入。但我試圖創建和這樣的錯誤。

我在控制器

public function ajax_add() 
{ 
    $data = array(
    'date_man_activity_ra' => $this->input->post('date_man_activity_ra',TRUE), 
    'dd_user' => $this->mymodel->dd_user(), 
    'user_selected' => $this->input->post('id_user') ? $this->input->post('id_user') : '', 
    'id_user' => $this->input->post('id_user',TRUE), 
    'note' => $this->input->post('note',TRUE), 
     ); 
    $insert = $this->Man_activity->save($data); 
    echo json_encode(array("status" => TRUE)); 
} 

，併爲MyModel

public function dd_user() 
{ 
    // ambil data dari db 
    $this->db->order_by('id_user', 'asc'); 
    $result = $this->db->get('ops_user'); 

    // bikin array 
    // please select berikut ini merupakan tambahan saja agar saat pertama 
    // diload akan ditampilkan text please select. 
    $dd[''] = 'Please Select'; 
    if ($result->num_rows() > 0) { 
     foreach ($result->result() as $row) { 
     // tentukan value (sebelah kiri) dan labelnya (sebelah kanan) 
      $dd[$row->id_user] = $row->username; 
     } 
    } 
    return $dd; 
} 

代碼和我查看

<div class="modal fade" id="modal_form" role="dialog"> 
<div class="modal-dialog"> 
    <div class="modal-content"> 
     <div class="modal-header"> 
      <button type="button" class="close" data-dismiss="modal" aria-label="Close"><span aria-hidden="true">&times;</span></button> 
      <h3 class="modal-title">Person Form</h3> 
     </div> 
     <div class="modal-body form"> 
      <form action="#" id="form" class="form-horizontal"> 
       <input type="hidden" value="" name="id"/> 
       <div class="form-body"> 
        <div class="form-group"> 
         <label class="control-label col-md-3">date_man_activity_ra</label> 
         <div class="col-md-9"> 
          <input name="date_man_activity_ra" placeholder="yyyy-mm-dd" class="form-control datepicker" type="text"> 
          <span class="help-block"></span> 
         </div> 
        </div> 
        <div class="form-group"> 
         <label class="control-label col-md-3">Id User</label> 
         <div class="col-md-9"> 
         <?php 
         $dd_user_attribute = 'class="form-control select2"'; 
         echo form_dropdown('id_user', $dd_user, $user_selected, $dd_user_attribute); 
         ?> 
         </div> 
        </div> 
        <div class="form-group"> 
         <label class="control-label col-md-3">Note</label> 
         <div class="col-md-9"> 
          <input name="note" placeholder="note" class="form-control" type="text"> 
          <span class="help-block"></span> 
         </div> 
        </div> 
       </div> 
      </form> 
     </div> 
     <div class="modal-footer"> 
      <button type="button" id="btnSave" onclick="save()" class="btn btn-primary">Save</button> 
      <button type="button" class="btn btn-danger" data-dismiss="modal">Cancel</button> 
     </div> 
    </div> 
</div> 

如何牛逼解決它？或另一種方式？

謝謝

Answer 1

public function ajax_add() 
{ 
    $data = array(
    'date_man_activity_ra' => $this->input->post('date_man_activity_ra',TRUE), 
    'dd_user' => $this->mymodel->dd_user(), 
    'user_selected' => $this->input->post('id_user') ? $this->input->post('id_user') : '', 
    'id_user' => $this->input->post('id_user',TRUE), 
    'note' => $this->input->post('note',TRUE), 
     ); 
    $insert = $this->Man_activity->save($data); 
    // Load your view and pass the data that you use in dropdown 

    //echo json_encode(array("status" => TRUE)); // Remove it 
}