javascript - 在 php 中找不到值

Question

对于登录，我从 javascript 文件传递​​邮件 ID 和密码，并且我已经通过 console.log 检查值是否已打印。但是当我在 php 中回显这两个值时，只显示密码而不是邮件。但是我找不到任何错误。我在这里粘贴 php 文件。

<?php
require_once('DBconnection.php');

ini_set('display_errors', 1);
ini_set('log_errors', 1);

$datamail = $_GET["mailID"];
$datapass = $_GET["psw"];
//$datamail = isset($_GET["mailID"]) ? $_GET["mailID"] : '';

echo $datamail;
echo $datapass;

$login_query = "SELECT * FROM student_table where mail_id = '$datamail' AND password='$datapass'";

//echo $login_query;

$login_res = $db->query($login_query);

if( $login_res->num_rows == 1 ){
//if( $login_res == true ){
    echo "success";
}
else {
    //echo $login_res;
    echo mysqli_error($db);
    exit;
}
$db->close();
?>

Javascrit 文件在这里

    function globalLogin() {
    checkLogInMail();
    //pageEntry();
}
function checkLogInMail() {
    var mailET = document.getElementById("mailID");
    var mailIdError = document.getElementById("mailIdErr");
    mailID = mailET.value;
    var regex = /^(([^<>()\[\]\.,;:\s@\"]+(\.[^<>()\[\]\.,;:\s@\"]+)*)|(\".+\"))@(([^<>()[\]\.,;:\s@\"]+\.)+[^<>()[\]\.,;:\s@\"]{2,})$/i;
    if (!regex.test(mailID)) {
        mailIdError.innerHTML = "Enter a valid Email id";
        //loginFlag = 1;
    }
    else{
        checkmailPass();
    }
}


function checkmailPass() {
    var passET = document.getElementById("psw");
    var passError = document.getElementById("pswErr");

    psw = passET.value;

    console.log(mailID);
    console.log(psw);

    var xhttp = new XMLHttpRequest();
    xhttp.onreadystatechange = function () {
        console.log(this.readyState);

        if(this.readyState == 4 && this.status == 200)
        {
            console.log(this.status);
            var response = xhttp.responseText;
            alert(response);
            if(!response.localeCompare( "success" )){
                document.getElementById("loginErr").innerHTML = "Mail or Password is correct";
                //alert("Successfully logged in :)");
                //window.location.href = "index.html";

            }
            else{
                document.getElementById("loginErr").innerHTML = response;

            }
        }

    }
    xhttp.open("GET", "passwordChecker.php?psw="+psw+"&mailID"+mailID, true);
    xhttp.send();

}

Answer 1

你在 mailID 的 get 请求中错过了 =

 xhttp.open("GET", "passwordChecker.php?psw="+psw+"&mailID="+mailID, true);