我想尝试使用xampp server开发简单的网页。这是我的html编码
<html>
<body>
<form action="registration2.php" method="post">
id: <input type="text" name="id">
pass: <input type="text" name="pass">
Email: <input type="text" name="email">
<input type="submit">
</form>
</body>
</html>
这是我的php编码连接到服务器并从html form.my数据库名称接收信息是项目,我的表名是registration.attributes在我的表上是reg_id,reg_pass,reg_email。
<?php
//the example of inserting data with variable from HTML form
mysql_connect("localhost","user_name","password");//database connection
mysql_select_db("project");
// Get values from form
$id = $_POST['id'];
$pass = $_POST['pass'];
$email = $_POST['email'];
//inserting data order
$order = "INSERT INTO registration
(reg_id, reg_pass,reg_email)
VALUES
('$id','$pass','$email')";
//declare in the order variable
$result = mysql_query($order); //order executes
if($result)
{
echo("
Input data is succeed");
}
else
{
echo("
Input data is fail");
}
?>
我收到了错误。我无法连接到mysql。结果没有显示出来。有人告诉我哪里错了?我的英语不好。
答案 0 :(得分:0)
尝试添加
$con=mysql_connect("localhost","user_name","password")or die('could not connect'.mysql_error()); //database connection
mysql_select_db("project",$con);
//all other lines
$order = "INSERT INTO registration
(reg_id, reg_pass,reg_email)
VALUES
('$id','$pass','$email')";
//declare in the order variable
$result = mysql_query($order);
if(!$result) {
die();}