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PHP表单传递参数与获取参数有关问题

时间:2021-07-01 10:21:17 帮助过:3人阅读

PHP 表单传递参数与获取参数问题

Register

if (!isset($_POST['register1']) || ( $_POST['register1'] != 'Register'))
{
#echo "test1";
?>

Registration



}
else
{
?>
E-mail:
echo $_POST['email'];
?>


Name:
echo $_POST['last_name'];
?>


Password:
echo $_POST['password'];
?>
}
?>



初学 PHP 问个问题 在这里 获取
E-mail:
echo $_POST['email'];
?>


Name:


Password:

这些信息时 只输出一个没问题
当我输出多个的时候像这个样子的
E-mail:
echo $_POST['email'];
?>


Name:
echo $_POST['last_name'];
?>


Password:
echo $_POST['password'];
?>
他就出错了 - -!!!
总是出现
An error occurred.

Sorry, the page you are looking for is currently unavailable.
Please try again later.

If you are the system administrator of this resource then you should check the error log for details.

Faithfully yours, nginx.
这个东西 只输出一个时就可以
各位大神请指教
------解决方案--------------------

Register

if (!isset($_POST['register1'])
------解决方案--------------------
( $_POST['register1'] != 'Register'))
{
#echo "test1";
?>

Registration


















人气教程排行

E-mail address:
First Name:
Last Name:
PassWord: