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我有三个文件...

第一个是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'];
?>

当我提交表单时，它会将我带到logincheck.php页面，当我单击链接时，它会将页面带到我的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……而且我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
4

1 回答 1

2

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() 不只是启动一个会话，如果它已经存在，它也会恢复任何现有的会话（参见http://php.net/manual/en/function.session-start.php）。

于 2013-10-20T12:58:58.600 回答