错误从表单插入mysql中的数据

时间:2017-10-05 17:54:31

标签: php mysql ajax

请告诉我我的错误是什么。我写了一本留言簿。创建了一个数据库。在phpmyadmin她出现了。接下来,我创建了一个表单和一个发送数据的ajax脚本。发送警报时的消息写入数据已写入数据库,但实际上该表为空。我做错了什么?

创建数据库

define ("HOST", "localhost");

define ("DATABASE", "guests_db");   

define ("MYSQL_USER", "alekspvn");  

define ("MYSQL_PASS", "123");  


function create_db () {
$connect_db=mysql_connect(HOST, MYSQL_USER, MYSQL_PASS)   
    or die("No connection with SQL");  
mysql_query ("CREATE DATABASE IF NOT EXISTS ".DATABASE)   
    or die ("Error create database");  
mysql_select_db(DATABASE)  
    or die("Error connection database");  
mysql_query ("CREATE TABLE IF NOT EXISTS book  
             (id INT NOT NULL AUTO_INCREMENT PRIMARY KEY,  
              username VARCHAR (100),  
              email VARCHAR (100),
              message TEXT)")  
    or die ("Error creation table");
}

脚本ajax

<script type="text/javascript">
        $(document).ready(function() {
            $('#submit').click(function() {
                var username = $('#username').val();
                var email = $('#email').val();
                var message = $('#message').val();
                $.ajax({
                    type: "POST",
                    cache: false,        
                    url: 'index.php',
                    data: {username: username, email: email, message: message},
                    success: function(data) {
                        alert('data has been stored to database');
                    }
                });
            });
        });
    </script>

我的表格

<form action="index.php" method="POST">
<label>Name :</label>
<input id="username" type="text">
<label>Email :</label>
<input id="email" type="text">
<label>Password :</label>
<input id="message" type="text">
<input id="submit" type="button" value="Submit">
</form>

PHP插入

 <?php
$connect_db=mysql_connect(HOST, MYSQL_USER, MYSQL_PASS)
    or die("No connection with SQL");  

mysql_select_db("javascript", $connect_db);


function insert_db () {
$username=$_POST['username'];
$email=$_POST['email'];
$message=$_POST['message'];
    $query = "INSERT INTO `book` VALUES ('$username','$email','$message')";
    $resource = mysql_query($query) 
        or die (mysql_error());
}

?>

0 个答案:

没有答案