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为什么我不能在我的PHP文件中使用AJAX发布我的JSON数据?

[英]Why can't I get my JSON Data posted using AJAX in my PHP file?

I have an AJAX script that post data in one of my PHP file: 我有一个AJAX脚本,在我的一个PHP文件中发布数据:

     var _lname = $('#ptLastName').val();
    var _fname = $('#ptFirstName').val();
    var _mname = $('#ptMiddleName').val();
$.ajax({
                type: "POST",
                url: ".././CheckPerson.php",
                data: "{'lastName':'" + _lname + "','firstName':'" + _fname + "','middleName':'" + _mname + "'}",
                contentType: "application/json; charset=utf-8",
                dataType: "json",
                success: function (response) {
                    var res = response.d;
                    if (res == true) {
                        jAlert('Person Name already exists!', 'Error');
                        return;
                    }

It works fine and I can see the JSON Data posted in the Firebug console. 它运行正常,我可以看到Firebug控制台中发布的JSON数据。 The problem is with this PHP code: 问题出在这个PHP代码上:

$firstname = json_decode($_POST['firstName']);
$lastname = json_decode($_POST['lastName']);
$middlename = json_decode($_POST['middleName']);
$response = array();

The above PHP code seems that it can't recognize the 'firstName' , 'lastName' , and 'middleName' as a posted JSON parameter, and return an Undefined index: firstName in C:... something like that for all the posted parameters. 上面的PHP代码似乎无法将'firstName''lastName''middleName'为已发布的JSON参数,并返回一个Undefined index: firstName in C:...就像所有发布的那样参数。

I also tried using $data = $_POST['data'] and $_REQUEST['data'] to get all the JSON parameters and decode it using json_decode($data); 我还尝试使用$data = $_POST['data']$_REQUEST['data']来获取所有JSON参数并使用json_decode($data);对其进行解码json_decode($data); but didn't work. 但没有奏效。

I've also used the AJAX shortened code for post $.post('.././CheckPerson.php', {data: dataString}, function(res){ }); 我还使用了AJAX缩短代码来发布$.post('.././CheckPerson.php', {data: dataString}, function(res){ }); , it works great with my PHP file and my PHP file can now read lastName , firstName , and middleName , but i think it is not a JSON data but only a text data because firebug can't read it as JSON data. 我的PHP文件很好用,我的PHP文件现在可以读取lastNamefirstNamemiddleName ,但我认为它不是JSON数据而只是文本数据,因为firebug无法将其作为JSON数据读取。 Now,i'm confused how will my PHP file read the JSON data parameters.Do you guys have any suggestions about this? 现在,我很困惑我的PHP文件将如何读取JSON数据参数。你们有什么建议吗?

The problem is that dataType: "json" doesn't mean that you're posting json, but that you're expecting to receive json data from the server as a result of your request. 问题是dataType: "json"并不意味着你发布了json,而是因为你的请求而希望从服务器接收json数据。 You could change your post data to: 您可以将帖子数据更改为:

data: {myPostData : "{'lastName':'" + _lname + "','firstName':'" + _fname + "','middleName':'" + _mname + "'}"}

and then parse it on your server like 然后在你的服务器上解析它

$myPostData = json_decode($_POST['myPostData']);
$firstname = $myPostData["firstName"];
$lastname = $myPostData["lastName"];
$middlename = $myPostData["middleName"];

One issue- you're using single quotes for your json. 一个问题 - 你正在为你的json使用单引号。 You should be using double quotes (according to spec). 你应该使用双引号(根据规范)。

{"lastName":"Smith", "firstName":"Joe"}

instead of 

{'lastName':'Smith', 'firstName':'Joe'}

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