当我在一个页面上提交一个表单时,我遇到了一个问题,这导致会话结束,用户必须重新登录。我花了几个小时试着整理这件事,但什么也没做!这是表单所在页面的代码-
<?php
session_start(); /// initialize session
$title = 'Month Edit';
include("passwords.php");
$user_email = check_logged(1, $dbc);
?>表单代码只是
passwords.php页面是
<?php
require ('ct_db.php');//This is the database info
$query = "SELECT * FROM users";
$result = mysqli_query ($dbc, $query);
while($check_row = mysqli_fetch_array($result, MYSQLI_ASSOC))
{
$this_us = $check_row['user'];
$this_ps = $check_row['pass'];
$USERS[$this_us] = $this_ps;
}
function check_logged($page, $dbc)
{
global $_SESSION, $USERS;
if (!array_key_exists($_SESSION["logged"],$USERS))
{
header("Location: login.php");
}
else
{
$this_user = $_SESSION["logged"];
$query = "SELECT * FROM users WHERE user = '$this_user'";
$result = mysqli_query ($dbc, $query);
$row2 = mysqli_fetch_array($result, MYSQLI_ASSOC);
$user_email = $row2['email'];
$user_permission = $row2['permissions'];
if($page == 3 && ($user_permission=="admin" || $user_permission=="user"))
{
header("Location: login.php");
}
else
{
if($page == 2 && $user_permission=="user")
{
header("Location: login.php");
}
else
{
echo '<div id="logged_in">';
echo '<form action="login.php" method="post"><input type="hidden" name="ax" value="out"> ';
echo 'Logged in as <span style="color:blue">'.$_SESSION["logged"]."</span> ";
echo 'email '.$user_email.' ';
echo '<input type="submit" value="Logout" id="log_out"/></div><br>';
return $user_email;
}
}
}
};
?>我还有一个登录页面,其中包括以下内容
<?php
session_start();
$title = 'login';
require ('ct_db.php');
$query = "SELECT * FROM users";
$result = @mysqli_query ($dbc, $query);
while($check_row = mysqli_fetch_array($result, MYSQLI_ASSOC))
{
$this_us = $check_row['user'];
$this_ps = $check_row['pass'];
$USERS[$this_us] = $this_ps;
}
function check_logged($page)
{
global $_SESSION, $USERS;
if (!array_key_exists($_SESSION["logged"],$USERS))
{
header("Location: login.php");
}
else
{
$this_user = $_SESSION["logged"];
require ('ct_db.php');
$query = "SELECT * FROM users WHERE user = '$this_user'";
$result = mysqli_query ($dbc, $query);
$row2 = mysqli_fetch_array($result, MYSQLI_ASSOC);
$user_email = $row2['email'];
$user_permission = $row2['permissions'];
if($page == 3 && ($user_permission=="admin" || $user_permission=="user"))
{
header("Location: login.php");
}
else
{
if($page == 2 && $user_permission=="user")
{
header("Location: login.php");
}
else
{
echo '<div id="logged_in">';
echo '<form action="login.php" method="post"><input type="hidden" name="ax" value="out"> ';
echo 'Logged in as <span style="color:blue">'.$_SESSION["logged"]."</span> ";
echo 'email '.$user_email.' ';
echo '<input type="submit" value="Logout" id="log_out"/>';
return $user_email;
}
}
}
};
require ('ct_db.php');
if ($_POST["ac"]=="log")
{ /// do after login form is submitted
if ($USERS[$_POST["username"]]==$_POST["password"]) /// check if submitted username and password exist in $USERS array
{
$_SESSION["logged"]=$_POST["username"];
$_SESSION["email"] = 'email address';
echo 'email = '.$_SESSION["email"];
}
else
{
echo 'Incorrect username/password. Please, try again.';
};
};
if ($_POST["ax"]=="out")
{ /// do after login form is submitted
if ($USERS[$_POST["username"]]==$_POST["password"]) /// check if submitted username and password exist in $USERS array
{
session_destroy();
echo 'You are logged out';
//header("Location: login.php");
die;
}
};
if (array_key_exists($_SESSION["logged"],$USERS)) //// check if user is logged or not
{
echo "You are logged in.";
echo '<form action="login.php" method="post"><input type="hidden" name="ax" value="out"> ';
echo '<input type="submit" value="Logout" />';
echo '</form>';
header("Location: mainPage3.php");
//// if user is logged show a message
}
else
{ //// if not logged show login form
echo '<div id="login">';
echo '<h1>Connectivity</h1>';
echo 'Please login<br><br>';
echo '<form action="login.php" method="post"><input type="hidden" name="ac" value="log"> ';
echo 'Username: <input type="text" name="username" id="username"/><br>';
echo 'Password: <input type="password" name="password" id="password"/><br>';
echo '<input type="submit" value="Login" />';
echo '</form>';
echo '</div><!--End login-->';
};
?>在我尝试解决这个问题的过程中,我已经更改并添加了很多代码,我确信它会比实际情况更复杂。我可以在不注销的情况下使用站点从一个页面移动到另一个页面,并且可以刷新包含表单的页面。但是如果我按下submit,我就会返回到登录页面。我唯一能想到的可能是用户信息与站点信息在不同的数据库中,但我已经检查过了,以确保我有正确的代码。感谢您给予的任何帮助。
发布于 2014-03-02 20:07:39
我只能在您的代码中看到两种提交按钮,登录和注销(两次),并且两个表单都有action = login.php。
单击注销按钮时,将通过代码销毁会话。当已经登录的用户单击登录按钮(假设该按钮可见)时,它会将post中的值设置为空值,并且用户将被重定向到login.php页面,因为验证将抛出login == false。
我没有测试你的代码,但显然这似乎是问题所在。
https://stackoverflow.com/questions/22126596
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