试图获取非对象的属性“join_id”

Question

请就如何解决这个问题向我提出建议

错误尝试获取非对象的属性“join_id”

询问

Select * from join_chat where
(user_1 = '24' and user_2 = '26') or
(user_1 = '26' and user_2 = '24')

result of query
+---------+--------+--------+
| join_id | user_1 | user_2 |
+---------+--------+--------+
|      20 | 26     | 24     |
+---------+--------+--------+
1 row in set (0.01 sec)

从结果中，您可以看到 join id 返回为 20，但我得到了错误

Trying to get property 'join_id' of non-object

请参阅下面的 controller 方法

 public function get_join_user()
    {
        $this->load->library('form_validation');
        $this->form_validation->set_rules('post_user_id', 'post_user_id', 'trim|required');
        $this->form_validation->set_rules('login_user_id', 'login_user_id', 'trim|required');
        if ($this->form_validation->run() == false) {
            $data["responce"] = false;
            $data["error"] = 'field is required';
        } else {
            $this->load->model("common_model");
            $q = $this->db->query("Select * from join_chat where
                      (user_1 = '".$this->input->post("login_user_id")."' and user_2 = '".$this->input->post("post_user_id")."') or
                      (user_1 = '".$this->input->post("post_user_id")."' and user_2 = '".$this->input->post("login_user_id")."')");
            //$row = $q->row();
            $row = $q->result();
            if (count($row) > 0) {
                $data["responce"] = true;
                //*****error causing line below***
                $data["data"] = $row->join_id;
            } else {
                $add_chat = array(
                                   "user_1"=>$this->input->post("post_user_id"),
                                   "user_2"=>$this->input->post("login_user_id")
                                   );

                $this->db->insert("join_chat", $add_chat);

                $insertid = $this->db->insert_id();
                if ($insertid) {
                    $data["responce"] = true;
                    $data["data"] = $insertid;
                }
            }
        }
        echo json_encode($data);
    }

导致错误的行

$data["data"] = $row->join_id;

提前致谢

Answer 1

再次@Mlike Eps，现在您已将代码从$q->row(); 到$row = $q->result(); 参考： prev_question 。 现在它是一个 object 数组，因此您必须相应地获取数据。

$data["data"] = $row[0]->join_id; //because, now array[0] contains the previous information. 

如果它有多个数组，请使用foreach()检索数据。 希望它可以帮助你。 :)