Question

我想根据他们的名字获取user_ID，并将其插入到不同的表中。这是我的控制者：

    $customer = strtoupper($this->input->post('cust_name'));
    $address = $this->input->post('address');
    $category = $this->input->post('category');

    $sales = ($this->input->post('sic'));              
    $salesID = $this->user_model->getUserName($sales);       

    $result = $this->db->insert('customer', [        
        'cust_name' => $customer,
        'cust_address' => $address,
        'category' => $category,
        'user_id' => $salesID
    ]);

我的模特：

public function getUserName($sales) {
    $this->db->select("user_id");
    $query = $this->db->get_where('user', ['name' => $sales]);
    return $query->row();
}

当我执行查询时，它总是说＆＃34;类stdClass的对象无法转换为字符串＆＃34;

这里是var_dump：

object（stdClass）＃22（1）{[＆＃34; user_id＆＃34;] =＆gt;字符串（2）＆＃34; 15＆＃34; }

的print_r：

stdClass对象（[user_id] =＆gt; 15）

如何解决这个问题？谢谢

Answer 1

希望这会对您有所帮助：

使用row这样的对象：

  $salesID = $this->user_model->getUserName($sales);       

   $result = $this->db->insert('customer', [        
    'cust_name' => $customer,
    'cust_address' => $address,
    'category' => $category,
    'user_id' => $salesID->user_id  /* here is the change*/
  ]);

替代方式：像这样返回user_id：

    public function getUserName($sales) {
       $this->db->select("user_id");
       $query = $this->db->get_where('user', ['name' => $sales]);
       return $query->row()->user_id;
    }
   /*then use it as you are using currently*/

类stdClass的对象无法转换为字符串，CodeIgniter

1 个答案: