php 连接PHP源代码并提交表单到MySQL数据库
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/24244203/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Connecting PHP source code and submit form to MySQL Database
提问by user3717206
I'm trying to learn PHP and I'm trying to connect a MySQL database with my PHP code to make a submit form that lets me input data into the database. My problem is that the source code is connecting but the HTML isn't posting the variables to the PHP file. I could really use some help.
我正在尝试学习 PHP，并且我正在尝试将 MySQL 数据库与我的 PHP 代码连接起来，以制作一个提交表单，让我将数据输入到数据库中。我的问题是源代码正在连接，但 HTML 没有将变量发布到 PHP 文件。我真的可以使用一些帮助。
This is my HTML source code
这是我的 HTML 源代码
<html>
<head>
<title>Form Input Data</title>
</head>
<body>
<table border="1">
<tr>
<td align="center">Form Input Employees Data</td>
</tr>
<tr>
<td>
<table>
<form action="input.php" method="POST">
<tr>
<td>Name</td>
<td><input type="text" name="name" size="20">
</td>
</tr>
<tr>
<td>Address</td>
<td><input type="text" name="address" size="40">
</td>
</tr>
<tr>
<td></td>
<td align="right"><input type="submit"
name="submit" value="Sent"></td>
</tr>
</table>
</td>
</tr>
</table>
And this is my PHP source code
这是我的 PHP 源代码
<?php
$user_name = "fees0_14446440";
$password = "********";
$database = "fees0_14446440_addressbook";
$server = "sql107.0fees.net";
mysql_connect("$server","$user_name","$password");
mysql_select_db("$database");
$order = "INSERT INTO Trial
(name, address)
VALUES
('$name',
'$address')";
$result = mysql_query($order);
if($result){
echo("<br>Input data is succeed");
} else{
echo("<br>Input data is fail");
}
?>
采纳答案by aplewandowski
You can find the POST data in PHP's $_POSTvariable. It should hold all the values that were passed via the POST method.
您可以在 PHP 的$_POST变量中找到 POST 数据。它应该保存所有通过 POST 方法传递的值。
$name = $_POST["name"];
$name = $_POST["name"];
回答by Bibek Jana
put POST values into variable
将 POST 值放入变量中
$name=$_POST['name'];
$address=$_POST['address'];
before sql query and write sql query as
在 sql 查询之前并将 sql 查询写为
$order = "INSERT INTO Trial
$order = "插入试用版
(name, address)
VALUES
('".$name."',
'".$address."')";
回答by JustStarting
Try the following:
请尝试以下操作：
<?php
$user_name = "fees0_14446440";
$password = "********";
$database = "fees0_14446440_addressbook";
$server = "sql107.0fees.net";
mysql_connect("$server","$user_name","$password");
mysql_select_db("$database");
if (isset($_POST['submit'])) {
$name = $_POST['name'];
$address = $_POST['address'];
$order = mysql_query("INSERT INTO Trial (name, address) VALUES ('$name', '$address')");
if ($order) {
echo '<br>Input data is successful';
} else {
echo '<br>Input data is not valid';
}
}
?>
回答by Khushboo
Try this
尝试这个
<html>
<head>
<title>Form Input Data</title>
</head>
<body>
<form action="input.php" method="POST">
<table border="1">
<tr>
<td align="center">Form Input Employees Data</td>
</tr>
<tr>
<td>
<table>
<tr>
<td>Name</td>
<td><input type="text" name="name" size="20">
</td>
</tr>
<tr>
<td>Address</td>
<td><input type="text" name="address" size="40">
</td>
</tr>
<tr>
<td></td>
<td align="right"><input type="submit"
name="submit" value="Sent"></td>
</tr>
</table>
</td>
</tr>
</table>
</form>
In PHP code
在 PHP 代码中
$order = "INSERT INTO Trial
(name, address)
VALUES
('$_POST["name"]',
'$_POST["address"]')";
