我在插入数据时遇到问题,它没有向我的phpmyadmin输入任何数据,我有问题
<?php
if (isset($_POST['submit'])){
$LastName=$_POST['LastName'];
$FirstName=$_POST['FirstName'];
$MiddleInitial=$_POST['MiddleInitial'];
$Age=$_POST['Age'];
$Gender=$_POST['Gender'];
$Birthday=$_POST['Birthday'];
$Address=$_POST['Address'];
$EmailAddress=$_POST['EmailAddress'];
mysql_query("insert into studentsrecords(LastName,FirstName,MiddleInitial,Age,Gender,Birthday,Address,EmailAddress)
values('$LastName','$$FirstName','$MiddleInitial','$Age','$Gender','$Birthday','$Address','$EmailAddress',NOW())
")or die(mysql_error());
?>
<script type="text/javascript">
alert('You are Successfully Register Thank You');
window.location="index.php";
</script>
<?php
}
?>
这是我的配置.php
代码中的一些常见错误。修复这些。
-
在 mysql 查询中将
'$$FirstName'
更改为'$FirstName'
。 -
您没有建立与数据库的连接。由
antyrat
提议 -
从 mysql 查询中删除
NOW()
。由andrewsi
提议
将
"$$FirstName"替换为"$FirstName"检查您的表单方法是否为"发布",检查您的配置文件。您可以通过以下方法创建数据库连接
$user = "xxxx"; //Database Username here
$pass = "xxxx"; //Database Password here
$db = "xxxx"; //Database Name here
$localhost = "xxx"; //Database Server
$link = mysql_pconnect($localhost, $user, $pass);
if ( ! $link )
die( "Couldn't connect to MySQL" );
mysql_select_db( $db)
or die ( "Couldn't open $db: ".mysql_error() );