我试图在ajax发布上获得响应,但总是返回NULL。
我以为已经尝试了所有可能的方法...但是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函数不会向JQuery ajax方法返回任何数据-此链接对我没有帮助
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(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"}"
在响应头上给我Content-Length:0(正在localhost上进行测试...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给我:
null --->ajaxPostProfile.js:28
Uncaught TypeError: Cannot read property 'content' of null --->ajaxPostProfile.js:31
代替使用return ($result);
,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();
}
并再次测试。
本文收集自互联网,转载请注明来源。
如有侵权,请联系[email protected] 删除。
我来说两句