PHP 表单 提交数据到mysql
来源:百度知道 编辑:UC知道 时间:2024/06/25 22:04:20
从网站上,摘下了一段代码,自己修改了一些地方,可是该程序无法运行。
<html>
<body>
<?php
if ($submit) {
// 处理表格输入
$db = mysql_connect("localhost:3318", "root","123456");
mysql_select_db("syetem",$db);
$sql = "INSERT INTO lighting
(ID,Name,Luminous,Power)
VALUES
('$ID','$Name','$Luminous','$Power',)";
$result = mysql_query($sql);
echo "Thank you! Information entered.\n";
} else{
// 显示表格内容
?>
<form method="post" action="<?php echo $PATH_INFO?>">
ID:<input type="Text" name="ID"><br>
Name:<input type="Text" name="Name"><br>
Luminous:<input type="Text" name="Luminous"><br>
<html>
<body>
<?php
if ($submit) {
// 处理表格输入
$db = mysql_connect("localhost:3318", "root","123456");
mysql_select_db("syetem",$db);
$sql = "INSERT INTO lighting
(ID,Name,Luminous,Power)
VALUES
('$ID','$Name','$Luminous','$Power',)";
$result = mysql_query($sql);
echo "Thank you! Information entered.\n";
} else{
// 显示表格内容
?>
<form method="post" action="<?php echo $PATH_INFO?>">
ID:<input type="Text" name="ID"><br>
Name:<input type="Text" name="Name"><br>
Luminous:<input type="Text" name="Luminous"><br>
PHP 提交表单,然后保存数据库示例:
1.sql脚本:
create database company;
use company;
create table employee(
id int(11) not null primary key auto_increment,
emp_name varchar(20) not null,
emp_no varchar(30) not null,
emp_job varchar(50)
);
2.index.php代码:
<?php
header("Content-type:text/html;charset=utf-8;");
//判断是否提交表单
if(isset($_POST['btn'])){
//连接数据库
$conn=mysql_connect("localhost","root","root");
if(!$conn){
die("数据库连接错误!".mysql_error());
}
mysql_select_db("company")