-1
我想用一个简单的网页来更新我的mysql表格的一个单元格。我不知道为什么它不会将数据发送到我的表格。我有两个网页。 form.php和process.php。以下是各自的代码。为什么我的代码不能将数据插入到mysql表中?
form.php的
<!DOCTYPE HTML PUBLIC
<html>
<head>
<title></title>
</head>
<body>
<!-- form to get key detail of record in database -->
<form method="POST" action="process.php">
<input type="text" name="inputtest" />
<input type="submit" name"submitButton" value="Submit!" />
</form>
<?php
$inputtest = $_POST["inputtest"];
?>
</body>
</html>
process.php
<?php
$con=mysqli_connect("localhost","root","********","allstate");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM producers");
?>
<?php
UPDATE producers
SET TEST={$inputtest}
[WHERE ID=1]
?>
@ user2593697另请参阅http://www.phptherightway.com/#databases,了解有关避免SQL注入以及其他有用的PHP建议的类似讨论。 – contrebis