Session if Session语句
Session if Session语句
提问于 2016-07-18 14:41:05
我试图显示信息A如果登录为用户和信息B如果登录作为管理。这里的问题是，我只能执行第一个if statement和else statement，这意味着我能够登录。然而，它永远不会进入else if statement。有人能帮我吗，我不知道这里有什么问题。
<?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';
$_SESSION['SESS_LOGGED'] = 1;
header("Location: somewhere.php");
die();
}
else {
echo '<script language = "javascript">';
echo 'alert("Fail login")';
echo '</script>';
echo "<script>window.location.assign('somewhere.php')</script>";
die();
}
?>回答 2
Stack Overflow用户
回答已采纳
发布于 2016-07-18 14:59:31
我想当你注销的时候你不会重设你的会话。试试这个：
Login.php
<?php
session_start();
require("configuration.php");
if(isset($_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';
$_SESSION['SESS_LOGGED'] = 1;
header("Location: somewhere.php");
die();
}
else {
echo '<script language = "javascript">';
echo 'alert("Fail login")';
echo '</script>';
echo "<script>window.location.assign('somewhere.php')</script>";
die();
}
?>您的If-语句也不正确。正如其他答案所述，应该是这样的：
somewhere.php编辑
<?php
if(!isset($_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 && $_SESSION['SESS_TYPE1'] == 'B' ){
?> //this is the admin html information form <?php
}
} else { ?> //ask user/admin to login html form
<form action ="login.php">
<button type="submit" name="login" class="btn-primary">Sign Up</button>
</form>
<?php
}
?>
</body>
</html>Stack Overflow用户
发布于 2016-07-18 14:58:42
嗨，你的情况永远不会变成真的。
if (isset($_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. 页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/38439575
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