我有三个文件...
第一是index.php
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Cyber Boy Test App</title>
</head>
<body>
Hello
<br/> This is Login System Test
<br/>
<form method="post" action="logincheck.php">
<input type="text" name="username" />
<input type="password" name="password" />
<input type="submit" name="submit" />
</form>
</body>
</html>
第二个是logincheck.php
<?php
session_start();
$finaluser=$_POST['username'];
if($_SESSION['auth']=="yes")
{
header('Location: name.php');
}
else
{
$connection=mysql_connect('localhost','root','')or die("Could not Connect to Database ");
$db=mysql_select_db('test', $connection) or die(" Check the Database wrong database entered ");
$sql="SELECT name from users WHERE username='$finaluser'";
$result=mysql_query($sql) or die("Coudl not Execute the Query");
$num=mysql_num_rows($result);
$result=mysql_query($sql) or die("Coudl not Execute the Query2");
$row=mysql_fetch_array($result);
if($num==1)
{
echo "Login SuccesFull";
$_SESSION['auth']="yes";
$_SESSION['name']=$row['name'];
echo '<a href="name.php">To Check the Name of the User Click Here';
}
else
{
echo "Login Failed";
}
}
?>
第三个文件是name.php
<?php
echo $_SESSION['name'];
?>
当我提交表格时，它会带我进入logincheck.php页面，当我点击链接时，它会带我name.php页面。问题是，名称没有被打印出来。我很了解 PHP，但我是 Google App Engine 的新手......我的app.yaml看起来像这样
application: testcboy
version: 1
runtime: php
api_version: 1
threadsafe: yes
handlers:
- url: /(.*)'.php
script: '1.php
- url: /.*
script: index.php
您需要在所有使用会话变量的 PHP 页面（例如顶部）上调用 session_start（）。
尽管它的名字session_start（）不仅启动会话，如果它已经存在，它也会恢复任何现有会话（请参阅 http://php.net/manual/en/function.session-start.php）。