来自 html 输入的值未发布到 html 文件答案

【问题标题】：Values from html input are not posting to html file来自 html 输入的值未发布到 html 文件
【发布时间】：2015-07-20 15:58:43
【问题描述】：

所以我有一个网站，我正在尝试测试新的用户注册功能。我的 LogInHelper.php 文件中没有任何回声的输出。目标是能够为新用户输入信息，将其发布到 LogInHelper.php，查看输入的密码是否相等，如果不相等，则将结果 div 更新为错误消息。

这是我的包含 html 的 LogIn.php 文件：

<!DOCTYPE html>
<html>
    <head>
        <title>Ticket Log In</title>
        <script src = "../../jquery.js"></script>
<style>
body{
    padding:0;
    margin:0;
}
#box {
    background:blue;
    text-align:center;
    padding:10px;
    color:white;
    width:500px;
    height:300px;
    margin: 0 auto;

}
#LogIn {
    background:blue;
    text-align:center;
    padding:10px;
    color:white;
    width:500px;
    height:300px;
    margin: 0 auto;
}
</style>
<script>
    $(document).ready(function(){
        $("#sub").click(function(){
            var fist_name = $("#firstname").val();
            var last_name = $("#lastname").val();
            var user_email = $("#email").val();
            var department_id = $("#department").val();
            var user_pass = $("#pass").val();
            var user_pass2 = $("#pass2").val();
            $.post("LogInHelper.php",{firstname:first_name,lastname:last_name,email:user_email,department:department_id,pass:user_pass,pass2:user_pass2},function(data){
                $("#result").html(data);
            });
        });
        $("#sub2").click(function(){
            var loginuser_email = $("#loginemail").val();
            var loginuser_pass = $("#loginpass").val();
            $.post("LogInHelper.php",{loginemail:loginuser_email,loginpass:loginuser_pass},function(data){
                $("#result2").html(data);
            });
        });
        $('#box').hide();
        $('.new').click(function() {
            $('#box').toggle();
            $('#LogIn').toggle();
});
    });
</script>
    </head>
<body>
    <div id="box">
        <input type ="submit" class = "new" value = "New User Register">
        <h2>New User Register Here:</h2>
        <input type ="text" name="firstname" id="firstname" placeholder="Enter Your First Name"/></br>
        <input type ="text" name="lastname" id="lastname" placeholder="Enter Your Last Name"/></br>
        <input type ="text" name="email" id="email" placeholder="Enter Your email"/></br>
        <form action="LogInHelper.php" method="post">
        <select id = "department" name = "department">
        <?php 
                    $servername = "localhost";
                    $username = "quantco_Ted";
                    $password = "Quantum1";
                    $database = "quantco_Interns";

                    $con = mysqli_connect($servername,$username,$password,$database);
                        if($con->connect_error){
                        die("Connection failed " . $con->connect_error);
                    } 
                    $sql = "select Department_name,id from Department";
                    $result = mysqli_query($con,$sql);
                    while ($row = mysqli_fetch_array($result)) {
                        $department = $row['Department_name'];
                        $id = $row['id'];
                        echo "<option value = '$id'>$department</option>";
                    }
        ?></select></form>
        <input type ="password" name="pass" id="pass" placeholder="Enter Your Password"/>
        <input type ="password" name="pass2" id="pass2" placeholder="Re-Enter Your Password"/>
        </br></br>
        <input type ="submit" name = "sub" value = "Register" id = "sub"/>
        <div id="result"></div>
    </div>
    <div id="LogIn">
        <input type ="submit" class = "new" value = "New User Register">
        <h2>Registered User Log In:</h2>
        <input type ="text" name="loginemail" id="loginemail" placeholder="Enter Your Email"/></br>
        <input type ="password" name="loginpass" id="loginpass" placeholder="Enter Your Password"/>
        </br></br>


        <input type ="submit" name = "sub2" value = "Submit" id = "sub2"/>
        <div id="result2"></div>
    </div>
</body>
</body>
</html>

这是接收帖子的 LogInHelper 文件：

<?php
    $servername = "localhost";
    $username = "user";
    $password = "pass";
    $database = "db";

    $con = mysqli_connect($servername,$username,$password,$database);

    if($con->connect_error){
        die("Connection failed " . $con->connect_error);
    }

    $firstname = $_POST['firstname'];
    $lastname = $_POST['lastname'];
    $id = $_POST['department'];
    $email = $_POST['email'];
    $pass = md5($_POST['pass']);
    $pass2 = md5($_POST['pass2']);

    echo $firstname;
    echo $lastname;
    echo $id;
    echo $email;
    echo $pass;
    echo $pass2;


    $sel = "select * from Employee where email='$email'";
    $run = mysqli_query($con,$sel);
    $check_email = mysqli_num_rows($run);

    if(!($pass==$pass2)){
        echo "<h2>Your emails do not match, please try again!</h2>";
        exit();
    }
    else if($check_email==1){
        echo "<h2>This email is already registered, please try another!</h2>";
        exit();
    }
    else{
        $insert = "insert into Employee (email, first, last, department_id,pass,) values ('$email','$firstname','$lastname','$id', '$pass')";
        $run_insert = mysqli_query($con,$insert);
        if($run_insert){
            echo "<h2>Registration Successful, Thanks!</h2>";
        }
    }
    mysqli_close($con);

?>

【问题讨论】：

var fist_name?除非你是职业拳击手，否则命名你的拳头只是……很奇怪。哦，还有可爱的sql injection attack 漏洞...
* 职业拳击手....和@lazersquids mcgee，您应该阅读有关面向对象的mysqli 和准备好的语句：php.net/manual/en/mysqli.prepare.php
@MarcB 哪里有漏洞？

标签： php jquery html mysqli

【解决方案1】：

我检查了你的$("#sub").click(function(){，发现你的var fist_name = $("#firstname").val();有typo错误，所以应该是var first_name = $("#firstname").val();

请更新它之后它会起作用

【讨论】：