Ajax调用php文件不起作用

Question

I am trying to implement a simple form which will eventually connect to a database and make entries in it. 我正在尝试实现一个简单的表单，该表单最终将连接到数据库并在其中进行输入。 In the tag,I am calling the php file which will connect me to the database in the back-end. 在标签中，我正在调用php文件，它将把我连接到后端的数据库。

index.html index.html

<html>
<head>
<script>
function submitForm(formId){

    //var formData= $.(formId).serialize();
    $.ajax({
        url:'new-user.php',
        type:'POST',
        data:{
            user_name=$("#user_name").val(),
            password=$("#password").val();
        }
        success:function(response){
            alert(response);
        }
    });
}
</script>
</head>

<body>
<form onsubmit="submitForm('#myForm');" id='myForm'>
User Name: <input type="text" name="user_name" id="user_name" />
Password: <input type="text" name="password" id="password" />
<input type="submit"/>
</form>
</body>

</html>

new-user.php new-user.php

<?php include 'database.php';?>

<?php 
mysqli_query($connect,"create table login(User_name varchar(50) NOT NULL,Password varchar(50) NOT NULL)");
$user_name=$_POST['user_name'];
$password=$_POST['password'];
if(empty($user_name)){
    $name_error="name is required";
}

mysqli_query($connect,"Insert into login(User_name,Password) values('$user_name','$password')");
if(mysqli_affected_rows($connect)>0){
    echo "<p>Credentials added</p>";
    echo "<a href='index.html'>Go back</a>";
}else{
    echo "<p>Error</p>";
    echo mysqli_error($connect);
}
?>

database.php database.php

<?php
$connect=mysqli_connect('localhost','root','','testdb');
if(mysqli_connect_errno($connect)){
    echo 'failed to connect';
}
?>

The above is not creating any table in the testdb database.Neither,it is generating any alert messages.The Url however changes after clicking the submit button as http://localhost/try2/?user_name=aayushi&password=ded but after that nothing happens. 上面没有在testdb数据库中创建任何表。也没有生成任何警报消息。但是，在单击“提交”按钮后，URL更改为http://localhost/try2/?user_name=aayushi&password=ded但此后什么也没有发生。 This is my first php code, so I don't really know what's the meaning of this exactly. 这是我的第一个php代码，所以我真的不知道这到底是什么意思。

Answer 1

Okay, since no one seems to actually be reading your code, there's a couple of syntax errors that I missed until I threw it into PhpStorm 好的，因为似乎没有人真正在阅读您的代码，所以在将其放入PhpStorm之前，我遗漏了一些语法错误

Change your function to this: 将功能更改为此：

function submitForm(formId){

    $.ajax({
        url:'/new-user.php',
        type:'POST',
        data:{
            user_name: $("#user_name").val(),
            password: $("#password").val()
        }
    })

        .complete(function (response) {
            alert(response)
        })

   return false; // Prevents the form from submitting the standard way
}

EDIT: Change the form to this: 编辑：将窗体更改为此：

<form onsubmit="return submitForm('#myForm');" id='myForm'>

Answer 2

In your ajax method, the success property is wrong 在您的ajax方法中，成功属性是错误的
It is written as suceess , when it was supposed to be success 当success时候，它被写成suceess 。

Also, to avoid refreshing the page, insert return false; 另外，为避免刷新页面，请插入return false; at the end of the function submitForm 在函数submitForm的末尾

Ajax调用php文件不起作用

问题描述

2 个解决方案

解决方案1
1 已采纳

解决方案2
0 2017-04-23 19:54:12

Ajax调用php文件不起作用

问题描述

2 个解决方案

解决方案1 1 已采纳

解决方案2 0 2017-04-23 19:54:12

解决方案1
1 已采纳

解决方案2
0 2017-04-23 19:54:12