Codeigniter使用where_not_in获取数据

Question

这是我想要实现的目标 我有两张桌子 投放和付款

所以我想显示不在付款表中但存在于交货表中的数据，因为您可以看到付款表有[del_id] 2和3。所以我试图获取[delivery_id] 1.但有些如何获取[delivery_id] 2.

这是型号代码

public function noPayment_tables(){

  $query = $this->db->query('SELECT del_id FROM payments');
  foreach ($query->result_array() as $row)
  {
      echo  $row['del_id'];
  }

    $this->db->select('*');
    $this->db->from('delivery');
    $this->db->where_not_in('delivery_id', $row);
    return $this->db->get()->result();
}

和控制器

public function status(){


        $data['mixs'] = $this->time_model->noPayment_tables();


          $data['main_view'] = "status_view";
            $this->load->view('header', $data);

 }

并查看

<table class="table">
  <thead>
    <tr>
      <th>consignee</th>
      <th>airway id</th>
      <th>status</th>
    </tr>
  </thead>
  <tbody>
<?php foreach ($mixs as $ta): ?>
<tr>
  <td><?php echo $ta->consignee ?></td>
  <td><?php echo $ta->airway_id ?></td>
  <td><?php echo $ta->delivery_id ?></td>


</tr>
  <?php endforeach; ?>
</tbody>
</table>

和当前的结果

Answer 1

我相信你需要一行才能成为一个简单的数组。在你的情况下，如果你var_dump行，你可能会看到类似的东西（因为你在每个循环上都取代它的值）：

array
    'del_id' => string '3' (length=1)

您需要做的是，在迭代的每一行上获取每个ID并将这些键存储在单独的数组中，如下所示：

$query = $this->db->query('SELECT del_id FROM payments');
$deleted = [];
foreach ($query->result_array() as $row)
{
  echo  $row['del_id'];
  $deleted = $row['del_id'];
}

$this->db->select('*');
$this->db->from('delivery');
$this->db->where_not_in('delivery_id', $deleted);
return $this->db->get()->result();

这应该有用。

Answer 2

public function noPayment_tables(){
$query = $this->db->query('SELECT del_id FROM payments');
$array=array();
foreach ($query->result_array() as $row)
  {
        array_push($array,$row['del_id']);
  }

$this->db->select('*');
$this->db->from('delivery');
$this->db->where_not_in('delivery_id', $array); //need to pass array 
return $this->db->get()->result();

}