Question

对不起我的英语不好。当我获得与数据库匹配并插入选择菜单的父类别ID类别时，我想更新。看图像吼叫：

我的数据库是：

==========================================
kat_id | parent_kategori |    kategori   |
==========================================
1      |     0           | wanita        |
2      |     1           | sandal wanita |
3      |     1           | sepatu wanita |

我的观点是：

<div class="input group">
        <label>Kategori :</label>
        <input type="text" name="kat" class="form-control" id="kat" placeholder="Nama Kategori" value="<?php echo $updatedata['kategori'];?>" required>
        <input type="hidden" name="id" class="form-control" id="id_kat" value="<?php echo $updatedata['kat_id'];?>" required>
        <br>
    </div>
    <div class="input group">
        <label>Parent kategori :</label>
        <?php echo form_dropdown('parent',$jabatan,$updatedata['kat_id']); ?>
        <br>
    </div>

我的控制器是：

$update = $this->kategori_adm->get_categorie($id);
    $update = $this->kategori_adm->ambil_data_employee($id);

    $dd_jabatan = array();
    foreach ($this->kategori_adm->master_jabatan() as $data_jabatan) 
    {
    $dd_jabatan[$data_jabatan['kat_id']] = $data_jabatan['kategori'];
    }

    $this->data['jabatan']=$dd_jabatan;
    $this->data['updatedata'] = $update;

    $this->load->view('user_log_as_admin/header');
    $this->load->view('user_log_as_admin/kategori/edit',$this->data);
    $this->load->view('user_log_as_admin/footer');

和我的模型是

class kategori_adm extends CI_Model{

function master_jabatan(){
$this->db->order_by('kat_id');
$sql_jabatan=$this->db->get('kategori');
if($sql_jabatan->num_rows()>0){
    return $sql_jabatan->result_array();
    }
}

function ambil_data_employee($id){
$this->db->where('kat_id', array('parent_kategori' => 0));
$sql_employee=$this->db->get('kategori');
if($sql_employee->num_rows()>0){
        return $sql_employee->row_array();
    }
}

function get_categorie($id){
  $query = $this->db->query("SELECT * FROM kategori WHERE kat_id='$id'");
  return $query->row_array();
}

Answer 1

让我们假设这是表：

+----+----------------+-----------------+
| id | category_name  | parent_category |
+----+----------------+-----------------+
| 1  | Main           | 0               |
+----+----------------+-----------------+
| 2  | Under Main     | 1               |
+----+----------------+-----------------+
| 3  | Under Main too | 1               |
+----+----------------+-----------------+

然后选择所有包含parent_category = 0的类别，您可以使用AJAX获取parent_category所有类别等于所选类别的ID。

试试这个： Populate another select dropdown from database based on dropdown selection

如何在codeigniter中更新父类别时获取id类别

1 个答案: