Question

我试图在ajax帖子上获得响应但总是返回NULL。

我认为已经尝试了所有我能做到的事情......但希望有人能够解决它。首先十分感谢。如果需要更多信息，请告诉我。

注意：PHP function not returning any data to JQuery ajax method - 此链接无法帮助我

这是jquery代码：

var post_id=3;
        $.ajax({

            type: "POST",
            url: "profile/ajax_get_new_posts",
            data: { post_id : post_id },
            dataType: 'json',
            success: function(data){

                console.log(data);
                var current_content = $("ol#posts_container").html();

                $("ol#posts_container").html(data.content + current_content);
            }


        });

profile / ajax_get_new_posts（它是.php）

function ajax_get_new_posts(){

        header('Content-Type: application/json');
            $post_id=$this->input->post('post_id');
            $chat_messages_html='testing';
            $result = array('status' => 'ok', 'content' => $chat_messages_html);
            $result=json_encode($result);
            return ($result);
            exit();
        }

如果我尝试使用var_dump（$ result）; die（）;给我正确的结果：

string(38) "{"status":"ok","content":"samoletite"}"

在响应标题上给我Content-Length：0（它在localhost上测试...而不是www.mywebsite）

 Remote Address:[::1]:80
    Request URL:http://www.mywebsite/bg/profile/ajax_get_new_posts
    Request Method:POST
    Status Code:200 OK
    Request Headersview source
    Accept:application/json, text/javascript, */*; q=0.01
    Accept-Encoding:gzip,deflate
    Accept-Language:en-US,en;q=0.8
    Connection:keep-alive
    Content-Length:9
    Content-Type:application/x-www-form-urlencoded
    Host:localhost
Origin:http://localhost
Referer:http://www.mywebsite/bg/profile
User-Agent:Mozilla/5.0 (Windows NT 6.3; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/38.0.2125.111 Safari/537.36
X-Requested-With:XMLHttpRequest
Form Dataview sourceview URL encoded
post_id:3
Response Headersview source
Connection:Keep-Alive
Content-Length:0
Content-Type:application/json
Date:Fri, 31 Oct 2014 13:24:56 GMT
Keep-Alive:timeout=5, max=97
.......

Console.log给了我：

null --->ajaxPostProfile.js:28
Uncaught TypeError: Cannot read property 'content' of null --->ajaxPostProfile.js:31

Answer 1

要使用return ($result);，请使用echo。像这样：

function ajax_get_new_posts(){

        header('Content-Type: application/json');
            $post_id=$this->input->post('post_id');
            $chat_messages_html='testing';
            $result = array('status' => 'ok', 'content' => $chat_messages_html);
            $result=json_encode($result);
            echo $result;
            exit();
        }

再次测试。

Answer 2

您可以使用以下语法在PHP中返回json数据：

  header('Content-Type: application/json');
  echo json_encode($result);

JSON使用jQuery $ .ajax在响应时返回null

2 个答案: