2017-01-02 112 views
0

我有下面的代码调用PHP文件将数据插入到MySQL数据库,但不能似乎得到了PHP文件运行。有什么建议么?将数据插入MySQL数据库从Javascript调用PHP文件

我是很新的剧本,所以我想这个注册表单中。数据库'大学',其表'注册',似乎很好,但没有执行PHP代码。它一定是非常简单的东西,但我似乎无法解决它?

$(document).ready(function() { 
 
$("#register").click(function() { 
 
var name = $("#name").val(); 
 
var email = $("#email").val(); 
 
var password = $("#password").val(); 
 
var cpassword = $("#cpassword").val(); 
 
if (name == '' || email == '' || password == '' || cpassword == '') { 
 
alert("Please fill all fields...!!!!!!"); 
 
} else if ((password.length) < 8) { 
 
alert("Password should atleast 8 character in length...!!!!!!"); 
 
} else if (!(password).match(cpassword)) { 
 
alert("Your passwords don't match. Try again?"); 
 
} else { 
 
\t //alert(name); //This line reads fine 
 
\t $.post("register.php", 
 
\t { 
 
\t name1: name, 
 
\t email1: email, 
 
\t password1: password 
 
}, 
 
function(data) { 
 
\t if (data == 'You have Successfully Registered.....') { 
 
\t $("form")[0].reset(); 
 
} 
 
alert(data); 
 
}); 
 
} 
 
}); 
 
});
<?php 
 
$connection = mysql_connect("localhost", "root", "Winchester12"); // Establishing connection with server.. 
 
$db = mysql_select_db("college", $connection); // Selecting Database. 
 
$name=$_POST['name1']; // Fetching Values from URL. 
 
$email=$_POST['email1']; 
 
echo $email; 
 
$password= sha1($_POST['password1']); // Password Encryption, If you like you can also leave sha1. 
 
// Check if e-mail address syntax is valid or not 
 
$email = filter_var($email, FILTER_SANITIZE_EMAIL); // Sanitizing email(Remove unexpected symbol like <,>,?,#,!, etc.) 
 
if (!filter_var($email, FILTER_VALIDATE_EMAIL)){ 
 
echo "Invalid Email......."; 
 
}else{ 
 
$result = mysql_query("SELECT * FROM registration WHERE email='$email'"); 
 
$data = mysql_num_rows($result); 
 
if(($data)==0){ 
 
$query = mysql_query("insert into registration(name, email, password) values ('$name', '$email', '$password')"); // Insert query 
 
if($query){ 
 
echo "You have Successfully Registered....."; 
 
}else 
 
{ 
 
echo "Error....!!"; 
 
} 
 
}else{ 
 
echo "This email is already registered, Please try another email..."; 
 
} 
 
} 
 
mysql_close ($connection); 
 
?>
<!DOCTYPE html> 
 
<html> 
 
<head> 
 
<title>Squash Registration Form</title> 
 
<meta name="robots" content="noindex, nofollow"> 
 
<!-- Include CSS File Here --> 
 
<link rel="stylesheet" href="style.css"/> 
 
<!-- Include JS File Here --> 
 
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script> 
 
<script type="text/javascript" src="registration.js"></script> 
 
</head> 
 
<body> 
 
<div class="container"> 
 
<div class="main"> 
 
<form class="form" method="post" action="#"> 
 
<h2>Squash Registration Form</h2> 
 
<label>Name :</label> 
 
<input type="text" name="dname" id="name"> 
 
<label>Email :</label> 
 
<input type="text" name="demail" id="email"> 
 
<label>Password :</label> 
 
<input type="password" name="password" id="password"> 
 
<label>Confirm Password :</label> 
 
<input type="password" name="cpassword" id="cpassword"> 
 
<input type="button" name="register" id="register" value="Register"> 
 
</form> 
 
</div> 
 
</body> 
 
</html>

+0

可能重复的[PHP脚本中的增量值不工作](http://stackoverflow.com/questions/40573476/incrementing-value-in-php-script-not-working) – e4c5

回答

0

这可能是一个路径相关的问题。

您可能想要查看register.php文件的路径以确保其位置正确。

+0

我检查过,所有的文件都是在同一目录 –

+0

我需要一个PHP编译器吧? :) –

+0

然后在你的电脑上安装xamp – user1382875

0

似乎什么都没有传递给register.php文件。请使用'ajax'而不是jquery post。