PHP/JS:无法获得

Question

我只想印上“赢”！如果他们登录数据库中的详细信息(正确工作)和“丢失”，如果由于某种原因，他们的信息在数据库中找不到。

因此，我的问题是，由于某种原因，我的代码行“echo$email；”无法工作。它似乎被设置为空。

目前，它只打印‘丢失’，不管我输入了什么，但是，如果我在数据库中添加了一个空白电子邮件和密码(email =“，password=”)的行，则php脚本返回'win!‘。

PHP代码：

<?php
// echo "php test";
//server info
$servername = "localhost";
$username = "root";
$dbpassword = "root";
$dbname = "personal_data";


//Establish server connection
$conn = new mysqli($servername, $username, $dbpassword, $dbname);


//Check connection for failure 
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

//Read in email & password
echo "reading in email & password...";
$email = mysqli_real_escape_string($conn, $_POST['email1']);
$password = mysqli_real_escape_string($conn, $_POST['password1']); 
echo $email; //this prints blank
echo $password; //this also prints blank

$sql = "SELECT Name FROM personal_data WHERE Email='$email' AND Password='$password' LIMIT 1";
$result = mysqli_query($conn, $sql);


if(mysqli_num_rows($result) > 0){
        echo "win!!";
    } else {
        echo "loss";
    }

mysqli_close($conn);
?>

JS代码：

$(document).ready(function(){
// alert("js working");

$('#login_button').click(function(){
    var email = $('#email').val(); //prints the correct value
    var password = $('#password').val(); //prints the correct value

var dataString = 'email1=' + email 
               + '&password1=' + password;

$.ajax({
                type: "POST",
                url: "http://localhost:8888/php/login.php",
                data: dataString,   //posts to PHP script
                success: success()
            });

});//eo login_button

function success(){
alert("success");
}

});//eof

Answer 1

除了这样一个事实，那是完全的，疯狂的无用和没有任何安全性，您可以只是交换$.ajax()为$.post()，并这样做：

var loginEmail = $('#email').val();
var loginPassword = $('#password').val();

$.post('login.php',{email:loginEmail,password1:loginPassword},function(data) {
console.log(data);
})