代码中的jquery ajax post方法错误

时间:2017-04-02 07:06:41

标签: php jquery html css ajax

此代码的问题是当我运行它并尝试在用户名字段中输入john并单击注册时,不会打印hello。在每种情况下都会打印hello again,这意味着数据不会发送到process.php文件。原因是$ .post无效

可能出现什么问题?

index.php文件

<!DOCTYPE html> 
<html>
<head>
<title>Login Registration System</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.0/jquery.min.js"></script>
<style>
label{
    display:block;
}
input{
    display:block;
}
</style>
<body>
    <h1>Login Form</h1>
    <form id="login">
        <label>UserName</label>
        <input type="text" id="username" name="username" placeholder="username" />
        <label>Password</label>
        <input type="password" id="pwd" name="password" placeholder="password"/>
        <input type="submit" name="submit" value="Login"/>
    </form>
    <h1>Registration form</h1>
    <form id="register">
        <label>UserName</label>
        <input type="text" id="username" name="username" placeholder="username" />
        <label>Password</label>
        <input type="password" id="pwd" name="password" placeholder="password" />
        <label>Email</label>
        <input type="email" id="email" name="email" placeholder="email" />
        <input type="submit" name="submit" value="Register" />
    </form>
    <p id="paragraph"></p>
    <script src="script.js"></script>
</head> 
</body>
</html>

script.js文件

$(document).ready(function(){

    $("#register").submit(function(){

        var username=$("#username").val();
        var pwd=$("#pwd").val();
        var email=$("#email").val();

        $.post("process.php",{username:username,pwd:pwd,email:email},function(data){

            $("#paragraph").html(data);

        });

        return false;

      });

});

process.php文件

<?php

if(isset($_POST['username']))
{

$username = $_POST['username'];

if($username == "john")
{
    echo "hello";
}
else{
    echo "hello again";
}

}


?>

0 个答案:

没有答案