Question

我正在尝试从数据库获取数据这是查询

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<table>
  <tr>
    <th>Method</th>
    <th>Expires</th>
    <th>Actions</th>
  </tr>
  <tr>
    <td>Card Info</td>
    <td>Expiration</td>
    <td>
      <form class="sg-inline-form" method="post" action="">
        <input type="hidden" name="sg_customer_id" value="customerID1">
        <input type="hidden" name="sg_card_id" value="cardID1">
        <a href="#" class="set-default">Set As Default</a>
        <a href="#" class="delete-card">Delete</a>
      </form>
    </td>
  </tr>
    <tr>
    <td>Card Info</td>
    <td>Expiration</td>
    <td>
      <form class="sg-inline-form" method="post" action="">
        <input type="hidden" name="sg_customer_id" value="customerID2">
        <input type="hidden" name="sg_card_id" value="cardID2">
        <a href="#" class="set-default">Set As Default</a>
        <a href="#" class="delete-card">Delete</a>
      </form>
    </td>
  </tr>
</table>

但不是返回数组，而是返回以下内容：

     public function updating($id)
    {
        $uquery=$this->db->query(' SELECT * FROM `form` WHERE id='.$id);
        print_r($uquery);
        //return $request_data;

    }

为什么这个字符串要来？

Answer 1

使用result_array（）;

public function updating($id)
    {
        $uquery=$this->db->query(' SELECT * FROM `form` WHERE id='.$id);
        $array =$uquery->result_array();
        print_r($array);


    }

Answer 2

您必须使用

  1) ->result_array() to get result data in array format
  2) ->result() to get result data in object format
  3) ->row() to get single result data in object format
  4) ->row_array() to get single result data in array format

  e.g    $data = $this->db->query(' SELECT * FROM `form` WHERE id='.$id)->result_array();

谢谢

查询返回不良数组联合

2 个答案: