从AJAX调用PHP中检索POST数据

Three days had passed and still having problems to get this things work. This AJAX Call on my js file seems working when it comes to sending JSON data:

三天过去了，要让这些东西运转起来还存在问题。在发送JSON数据时，AJAX调用我的js文件似乎很有效:

 var _lname = $('#ptLastName').val();
 var _fname = $('#ptFirstName').val();
 var _mname = $('#ptMiddleName').val();
 var _gender = $('#ptGender').val();
 var _bday = $('input[name="birthdate"]').val(); // $('#ptBirthDate').val();
 var _ssn = $('#ptSSN').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;
          } 
})

but in my PHP file:

但是在我的PHP文件中:

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

mysql_connect ("*****", "****") or die ('Error: ' . mysql_error());
mysql_select_db ("********");

$query = "SELECT Lastname, Firstname, MiddleName FROM tbl_people WHERE Lastname = '$lastname' || Firstname = '$firstname' || MiddleName = '$middlename'";

$result = mysql_query($query);

$row = mysql_fetch_array($result);

    if ($row) {     
        $response = json_encode(array('d' => true, 'test' => $lastname)); 
    }
    else { 
    $response = json_encode(array('d' => false, 'test' => $lastname));
    }
echo $response;
print json_encode($_POST);

some error from firebug console says:

firebug控制台的一些错误提示:

<br />
<b>Notice</b>:  Undefined index: lastName in <b>C:\xampp\htdocs\..\CheckPerson.php</b> on line <b>2</b><br />
<br />
<b>Notice</b>:  Undefined index: firstName in <b>C:\xampp\htdocs\..\CheckPerson.php</b> on line <b>3</b><br />
<br />
<b>Notice</b>:  Undefined index: middleName in <b>C:\xampp\htdocs\..\CheckPerson.php</b> on line <b>4</b><br />
{"d":false,"test":null}[]

i believe that json_decode() is working fine in my php file but $_POST[''] can't recognize my posted data from my ajax call w/c variables had been declared:

我相信json_decode()在我的php文件中运行良好，但是$_POST["]无法识别我的从ajax调用w/c变量中发布的数据:

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

I believe I am doing right with my codes seems i had read many questions here and done what they had said but don't know why the error occurred. Is there any problem/bug you had seen? please tell me.

我相信我对我的代码做得很好，似乎我在这里读了很多问题，做了他们所说的，但不知道为什么会发生错误。你看到什么问题了吗?请告诉我。

2 个解决方案

#1

Can u see the ajax request data using the firebug console ?

可以使用firebug控制台查看ajax请求数据吗?

You cannot get the lastname, firstname from $_POST. Its inside the json string. So first you have to get the data using

您无法从$_POST获得lastname、firstname。它位于json字符串内。首先，你需要使用数据

 $data = $_POST['data'] or $_REQUEST['data']

Then decode the $data using json_deocde and access your attributes.

然后使用json_deocde解码$数据并访问属性。

json_decode($data);

#2

$post = file_get_contents('php://input');

#1