Question

我们在尝试登录时遇到问题。我们是否通过XMLhttprequest发送我们的用户名和密码，因为这些参数似乎没有与他们一起发送，因此我们无法登录。

代码如下： Javascript文件

 $("#submit").click(function(){   
console.log("click");

     usernm= document.getElementById("username").value;
      passwd= document.getElementById("password").value;


      var send2 = "username=" + usernm + "&password=" + passwd;
      var request = new XMLHttpRequest;

        request.open('POST' , "myurl.com/login.php",true);
        request.dataType=('jsonp');
        request.setRequestHeader("Content-type","application/x-www-form-  urlencoded");


 request.onreadystatechange = function() {//Call a function when the state changes.
 if(request.readyState == 4 && request.status == 200) {
 alert(request.responseText);
 }
 }
 request.send(send2);

login.php就是这个

 require_once 'connect.php';

 session_start();
 $uName = ($_GET['username']);
 $pWord = ($_GET['password']);
 $login = "SELECT Username,Password FROM User WHERE Username = '$uName' and Password='$pWord'";
 $res = mysql_query($login);
 $num_row = mysql_num_rows($res);
 $row=mysql_fetch_assoc($num_row);

 if( $num_row == 1 ) {
  echo "true";


 }
 else {
 echo "false";

 }

Answer 1

您折叠了POST和GET方法，

修改这个，

 $uName = ($_POST['username']);
 $pWord = ($_POST['password']);

Answer 2

 $uName = ($_POST['username']);
 $pWord = ($_POST['password']);

Ajax发布请求登录问题

2 个答案: