【问题标题】：Session if else statement会话 if else 语句
【发布时间】：2016-11-21 05:30:40
【问题描述】：

如果以用户身份登录，我正在尝试显示信息 A，如果以管理员身份登录，则显示信息 B。这里的问题是，我只能执行第一个if statement 和else statement，这意味着我可以登录。但是，它永远不会进入else if statement。谁能帮帮我，我不确定这里有什么问题。

<?php
$dbhost = "localhost";
$dbuser = "root";
$dbpassword = "";
$dbdatabase = "test";
$db = mysql_connect($dbhost, $dbuser, $dbpassword);
mysql_select_db($dbdatabase, $db);
?>

somewhere.php

<?php
if(!isset($_SESSION)){
    session_start();
}
include("configuration.php");
?>
<html>
<body>
<?php

    if (isset($_SESSION['SESS_EXIST']) == true && isset($_SESSION['SESS_TYPE']) == 'A' ){ ?>
            //this is the user html information form
        <?php } 
    else if (isset($_SESSION['SESS_EXIST']) == true && isset($_SESSION['SESS_TYPE1']) == 'B' ){ ?>
            //this is the admin html information form
    <?php } else { ?>
            //ask user/admin to login html form
            <form action ="login.php">
               <button type="submit" name="login" class="btn-primary">Sign Up</button>
            </form>
    <?php } ?>
</body>
</html>

login.php

<?php
session_start();
require("configuration.php");
if(isset($_SESSION['SESS_EXIST']) == TRUE) {
header("Location: somewhere.php");
die();
}
$email = $_POST['Email'];
$pass =  $_POST['Pass'];

$sql = "SELECT * FROM user WHERE Email='$email' and Pass ='$pass' " ;
$res = mysql_query($sql);
$rows = mysql_num_rows($res);

$sql1 = "SELECT * FROM admin WHERE Email='$email' and Pass ='$pass' " ;
$res1 = mysql_query($sql1);
$rows1 = mysql_num_rows($res1);

if($rows == 1)
{
$row = mysql_fetch_assoc($res);
$_SESSION['SESS_EMAIL'] = $row['Email'];
$_SESSION['SESS_NAME'] = $row['Name'];
$_SESSION['SESS_PASS'] = $row['Pass'];
$_SESSION['SESS_TYPE'] = 'A';
$_SESSION['SESS_LOGGED'] = 1;
header("Location: somewhere.php");
die();

}
else if($rows1 == 1)
{
$row1 = mysql_fetch_assoc($res1);
$_SESSION['SESS_EMAIL1'] = $row['Email'];
$_SESSION['SESS_NAME1'] = $row['Name'];
$_SESSION['SESS_PASS1'] = $row['Pass'];
$_SESSION['SESS_TYPE1'] = 'B';
$_SESSION['SESS_LOGGED'] = 1;
header("Location: somewhere.php");
die();
}
else {
echo '<script language = "javascript">';
echo 'alert("Fail login")';
echo '</script>';
echo "<script>window.location.assign('somewhere.php')</script>";
die();
}
?>

【问题讨论】：

isset($_SESSION['SESS_TYPE']) 返回 true（有时为 false）。而'A' 始终是true
使用准备好的语句来防止 SQL 注入
我可以知道为什么 'A' 总是返回 true ，因为我设置了 'B' 也是 true 吗？
你用的是什么版本的php？
somewhere.php 脚本末尾有一个额外的 ?>（php 结束标记）

标签： php html session isset

【解决方案1】：

你好，你的情况永远不会成真

if (isset($_SESSION['SESS_TYPE']) && $_SESSION['SESS_TYPE'] == 'A') {
   // your   code here
}


  `isset` — Determine if a `variable` is set and is not `NULL`. `Returns` `TRUE` if var exists and has value other than NULL, FALSE otherwise.

【讨论】：

【解决方案2】：

我认为您在注销时不会重置 SESSION。试试这个：
Login.php

<?php
session_start();
require("configuration.php");
if(isset($_SESSION['SESS_EXIST']) == TRUE) {
header("Location: somewhere.php");
die();
}
session_unset();
$email = $_POST['Email'];
$pass =  $_POST['Pass'];

$sql = "SELECT * FROM user WHERE Email='$email' and Pass ='$pass' " ;
$res = mysql_query($sql);
$rows = mysql_num_rows($res);

$sql1 = "SELECT * FROM admin WHERE Email='$email' and Pass ='$pass' " ;
$res1 = mysql_query($sql1);
$rows1 = mysql_num_rows($res1);

if($rows == 1)
{
$row = mysql_fetch_assoc($res);
$_SESSION['SESS_EMAIL'] = $row['Email'];
$_SESSION['SESS_NAME'] = $row['Name'];
$_SESSION['SESS_PASS'] = $row['Pass'];
$_SESSION['SESS_TYPE'] = 'A';
$_SESSION['SESS_LOGGED'] = 1;
header("Location: somewhere.php");
die();

}
else if($rows1 == 1)
{
$row1 = mysql_fetch_assoc($res1);
$_SESSION['SESS_EMAIL1'] = $row['Email'];
$_SESSION['SESS_NAME1'] = $row['Name'];
$_SESSION['SESS_PASS1'] = $row['Pass'];
$_SESSION['SESS_TYPE1'] = 'B';
$_SESSION['SESS_LOGGED'] = 1;
header("Location: somewhere.php");
die();
}
else {
echo '<script language = "javascript">';
echo 'alert("Fail login")';
echo '</script>';
echo "<script>window.location.assign('somewhere.php')</script>";
die();
}
?>

您的 If 语句也不正确。正如其他答案中提到的，它应该是这样的：
somewhere.php Edit

<?php
if(!isset($_SESSION)){
    session_start();
}
include("configuration.php");
?>
<html>
    <body>
    <?php
    if(isset($_SESSION['SESS_EXIST'])){
        if ($_SESSION['SESS_EXIST'] == true && $_SESSION['SESS_TYPE'] == 'A' ){
            ?> //this is the user html information form <?php
        } 
        else if ($_SESSION['SESS_EXIST'] == true && $_SESSION['SESS_TYPE1'] == 'B' ){
            ?> //this is the admin html information form <?php
        }
    } else { ?> //ask user/admin to login html form
        <form action ="login.php">
            <button type="submit" name="login" class="btn-primary">Sign Up</button>
        </form>
    <?php
    }
    ?>
    </body>
</html>

【讨论】：

我得到一个错误 undefined index on else if ($_SESSION['SESS_EXIST'] == true && $_SESSION['SESS_TYPE1'] == 'B' ) 即使第一个是正确的
我用 isset() 添加了一个外部 If 语句。
您不需要带有外部检查的$_SESSION['SESS_EXIST'] == trues。不过，$_SESSION['SESS_TYPE'] 也会发出通知。
if ($_SESSION['SESS_EXIST'] == true && $_SESSION['SESS_TYPE'] == 'A') 和 else if ($_SESSION[' SESS_EXIST'] == true && $_SESSION['SESS_TYPE1'] == 'B'){
这是一个通知，而不是一个错误。而且这种类型的通知不会影响您的脚本。