如何在CodeIgniter的編輯頁面中顯示選定的多個下拉值

Question

如何在CodeIgniter的編輯頁面中顯示選定的多個下拉值

編輯頁面上的多個下拉列表中未顯示值

 <div class="form-group col-md-6">
                        <label for="admin_id"><?php echo get_phrase('Assign User'); ?>
                            <span class="text-danger">*</span></label>
                        <select class="form-control selectpicker" name="admin_idd[]" id="admin_id"
                            placeholder="Assign User" required multiple>

                            <option value="" hidden><?php echo get_phrase('Select User'); ?></option>
                            <?php
                        $system_usertable = $this->db->get('system_usertable')->result_array();
                        foreach($system_usertable as $row2):
                        ?>
                            <option value="<?php echo $row2['admin_id'];?>"
                                <?php if($assignuserstable['admin_id'] == $row2['admin_id'])echo 'selected';?>>
                                <?php echo $row2['first_name'];?>
                            </option>
                            <?php
                        endforeach;
                        ?>
                        </select>
                    </div>

這是將多個數組添加到數據庫的方式

模型

  public function addclientdetails(){

$data['business_name']         = html_escape($this->input->post('business_name'));
$data['legal_name']         = html_escape($this->input->post('legal_name'));
$data['status']         = html_escape($this->input->post('status'));
$data['rating']         = html_escape($this->input->post('rating'));
$data['SU_id'] =html_escape($this->input->post('admin_id'));
date_default_timezone_set("Asia/Kolkata");
$data['created_at'] =  Date('Y-m-d h:i:s');   
$data['created_by']         = $this->session->userdata('admin_id');

  $this->db->insert('user_table', $data);

    $insertId = $this->db->insert_id
    $admin_idd =html_escape($this->input->post('admin_idd'));
    $result = array();
        foreach($admin_idd AS $key => $val){
             $result[] = array(
              'user_id'   => $insertId,
              'admin_id'   => $_POST['admin_idd'][$key],
              'user_type'   => html_escape($this->input->post('user_type')),
             );
        }    

        $this->db->insert_batch('assignuserstable', $result);  
        $insertId = $this->db->insert_id();
    return  $insertId;
   
}

Answer 1

如果您不共享代碼，則無法識別問題。 也請分享您的模型和控制器。 首先請echo $system_usertable 和 $row2 並檢查是否有數據。 你的選項應該是這樣的

<option value="<?php echo $row2['admin_id'];?>" <?php echo (isset($assignuserstable['admin_id']) && $assignuserstable['admin_id'] == $row2['admin_id'])? 'selected' : NULL; ?>><?php echo $row2['first_name'];?></option>