页面无法看到代码写在isset函数语句下的登录表单。我已经正确地编写了代码并执行了很多次，但是现在isset语句内编写的代码不起作用了。代码如下:-
<?php
session_start();
echo "<p style='"font-color: #ff0000;'"> Catogoies </p>";
echo '<link href="var/www/html/sample.css" rel="stylesheet">';
require_once('../html/conn.php');
$query = "select * from catogories";
mysqli_select_db($dbc, 'odit');
$retrieve = mysqli_query($dbc, $query);
if(!$retrieve)
{
die(mysqli_error($query));
}
while($row=mysqli_fetch_array($retrieve, MYSQL_ASSOC)){
echo "<p style='"font-color: #ff0000;'">".'<a href="cats.php? catogory='.urldecode($row["Name"]).'">'.$row["Name"].'</a>'."</p>";
$_SESSION['cat']=$row["Name"];
}
if(!($_SESSION)) {
session_start();
}if(isset($_SESSION['lgout']))//the variable logout intialization line
{
if($_SESSION['lgout']!=1||$_SESSION['signup']){
echo "Hello : ".'<a href = "profile.php">'.$_SESSION['unme'].'</a>'; echo "<br><br>";
echo '<a href="logout.php">'."Logout";}
else {
include 'lform.php'; echo "<br><br>";
echo '<a href="Sign_up.php">'."Sign up"."<br>";
} }
mysqli_close($dbc);
//include 'lform.php';
?>
<br>
<a href = 'adding_catogory.php'>Create a New Catogory</a><br><br>
<a href = 'Log_in.php'></a>
<?php
$db = @mysqli_connect("localhost", "oddittor", "Odit@123", "odit");
if(isset($_POST['login'])){
$username=mysqli_real_escape_string($db, $_POST['l_id']);
$password=mysqli_real_escape_string($db, $_POST['pswd']);
$sql="SELECT * from users where usrName='$username' and pswrd = '$password'";
$result = mysqli_query($db, $sql) or die(mysqli_error($db));
$count=mysqli_num_rows($result) or die(mysqli_error($db));
if($count>0) {
$_SESSION['unme']=$username; //This is the global session variable...used for storing the variables across the pages.
$_SESSION['lgout']=0;
header('Location : session.php'.$_SESSION['unme']);
header("Location : Homepage.php".$_SESSION['unme'].$_SESSION['lgout']); header( "refresh:0;url=Homepage.php" );
$_SESSION['unme']=$username;
}
else {
$error = "Invalid Details! Please Renter them"; }
}
?>
问题在
if(isset($_SESSION['lgout']))
行，如果我删除这行，我可以看到登录页面表单，但通过这样做，我得到未定义的变量登出的错误，每当我第一次打开页面。
这是注销脚本
<html>
<?php
session_start();
$_SESSION['lgout']=1;
$_SESSION['signup']=0;
echo ' You have been successfully logged out';
header('Location : Homepage.php'.$_SESSION['lgout']);header( "refresh:0;url=Homepage.php" );
?>
</html>
你需要把
session_start();
全局设置在页面开头。因为它不能得到$_SESSION对象
直接删除
session_destroy();
可以访问所有的$_SESSION值。
您的查询不安全。使用Prepared Statements代替所有查询。http://php.net/manual/en/mysqli.quickstart.prepared-statements.php