Question

我的问题是我总是得到“失败”状态，即使我在登录时输入了正确的值。但是，当我将login.php代码中的$sql修改为数据库中的正确值时，我会获得“成功”状态。我认为将数据从AJAX发送到PHP可能有问题。你能帮我解决一下吗？

这是login.php代码：

header('Access-Control-Allow-Origin: *'); 
include "database.php"; 
header("Content-Type: application/json");

$numer=$_POST['number'];
$pass=$_POST['pass'];
$sql="SELECT * FROM `Users` WHERE Phone = '$numer' AND Password = '$pass'";

if ($result=mysqli_query($db,$sql))
  {
  // Return the number of rows in result set
  $rowcount=mysqli_num_rows($result);

 if($rowcount==1)
 {
  while($row = $result->fetch_assoc()) {
       $dane = array("status" => "success",
    "ID" => $row["ID"],
    "Name"=> $row["Name"],
    "Surname"=> $row["Surname"],
    "Phone"=> $row["Phone"],
    "Password"=> $row["Password"],
    "Admin"=> $row["Admin"],
    "Class"=> $row["Class"],
    "Banned"=> $row["Banned"]
    );

 }
  echo json_encode($dane);
 }
 else
 {
  $dane = array("status"=>"failed");
  echo json_encode($dane);
 }

  mysqli_free_result($result);
  }

mysqli_close($db);

?>

这是AJAX代码：

$("#login").click(function(){
      var number=$("#number").val();
      var pass=$("#password").val();
       if($.trim(number).length>0 & $.trim(pass).length>0)
       {
            $.ajax({
            type: "POST",
            url: "http://smietana.mechanikrac.website.pl/cgi-bin/janeta/login.php",
            data: { number:number, pass:pass}, 
            contentType:"application/json; charset=utf-8", 
            dataType:'json', 
            crossDomain: true,
            cache: false,
            beforeSend: function(){ $("#login").html('Connecting...');
            alert(number+pass);},

            success: function(data){
                alert('test'+data["status"]);

            if(data["status"]=="success")
            {
                localStorage.login="true";
                localStorage.number=number;

                alert(data["status"]);

                window.location.href = "order.html";

                $('#witaj').html(number);

            }

            else if(data["status"]=="failed")
       {
                alert("Błędne Dane");

                $("#login").html('Login again');

                }
                }
      });
    }return false;
   });

Answer 1

将数据更改为

   data: { 'number':number, 'pass':pass}

然后移除contentType和dataType

将数据从AJAX发送到PHP

1 个答案: