因此,我尝试在我的网络应用的每个页面中执行登录部分 它看起来像附图,但我在会话上遇到了困难。
如果用户已登录,则“显示表格”>显示下拉列表 但我总是在没有注册的情况下得到下拉,
我想要一个简单的代码,这样做对我来说因为我很困惑 有没有一种简单的方法来指导我完成整个过程?谢谢
<?php
/*
*Create db in mydb for users
*connect
*recieve from login forum >> check with db if exists
*start session > save the id of that user >> display the forum with details of his profile
*/
/*
if(isset($_POST["submit"])){
$user= $_POST["username"];
$pass= $_POST["password"];
//$_SESSION["username"]=$user;
echo "username is clicked";
}else{
echo "nothing is clicked";
}
*/
$signin=false; // checking the concept
$_SESSION["username"] =$signin;
if(isset($_SESSION["username"])){ #if the user is signed in
echo $_SESSION["username"];
$query= "select * from users";
$select= mysqli_query($connection, $query);
while ($row = mysqli_fetch_assoc($select)){
$userid = $row["idusers"];
$username=$row["username"];
$password= $row["password"];
}
?>
<ul class="nav navbar-nav navbar-right">
<li>
<a href="help.html">Help</a>
</li>
<li class="dropdown">
<a href="#" class="dropdown-toggle" data-toggle="dropdown"><?php echo $username;?><strong class="caret"></strong></a>
<ul class="dropdown-menu">
<li>
<a href="profile.php?id=<?php echo "$userid";?>">Profile</a>
</li>
<li>
<a href="#">Contests</a>
</li>
<li class="divider">
</li>
<li>
<a href="signout.php">SignOut</a>
</li>
</ul>
</li>
</ul>
</div><?php
}else{
?>
<form class="navbar-form navbar-right" method="post" role="search">
<div class="form-group">
<input type="text" class="form-control" name="username" placeholder="Username">
</div>
<div class="form-group">
<input type="password" class="form-control" name="password" placeholder="Password">
</div>
<button type="submit" class="btn btn-default">Sign In</button>
</form>
<?php
}
?>
答案 0 :(得分:0)
假设您在有人登录并查看您使用的标记时设置了一个变量我在$_SESSION
投注
<?php
if (!empty($_SESSION['user_logged_in'])) {
echo " "; // HTML when the user is logged in
} else {
echo " "; // HTML when the user is logged out
}