我正在使用来自bootstrap 3的模态对话框创建一个登录系统,我正在添加一个功能,以便在用户选中模式对话框中的复选框时保持用户登录。 我希望让用户登录,即使他/她刷新或关闭浏览器。
的index.php
<div class="container" id="myLogin">
<div class="row">
<div class="modal fade" id="myModal" tabindex="-1" role="dialog" aria-hidden="true">
<div class="modal-dialog">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-hidden="true">×</button>
<h5 class="modal-title">PLEASE ENTER YOUR EMAIL ADDRESS AND PASSWORD TO LOG IN.</h5>
</div>
<div class="modal-body">
<div id="show" class="lalert lalert-warning"></div>
<div class="form-horizontal">
<div class="form-group">
<label for="email" class="col-sm-2 control-label">Email</label>
<div class="col-md-9">
<div class="input-group">
<span class="input-group-addon"><i class="fa fa-envelope-o fa-fw"></i></span>
<input type="text" name="lemail" id="lemail" value="<?php echo $unm ?>" class="form-control" placeholder="Enter Email Address..." />
</div>
</div>
</div>
<div class="form-group">
<label for="password" class="col-sm-2 control-label">Password</label>
<div class="col-md-9">
<div class="input-group">
<span class="input-group-addon"><i class="fa fa-key fa-fw"></i></span>
<input type="password" name="lpassword" id="lpassword" value="<?php echo $pwd ?>" class="form-control" placeholder="Enter Password..." />
</div>
</div>
</div>
<div class="form-group">
<div class="col-sm-offset-2 col-sm-10">
<input type="checkbox" name="chkbox" value="staylogged" class="checkbox-inline" />
<label> Keep me logged in</label> <b>|</b>
<a href="" style="text-decoration:none"> Forgot your password?</a>
</div>
</div>
<div class="form-group">
<div class="col-sm-offset-2 col-sm-10">
<button type="submit" id="login" name="login" class="btn btn-primary"><span class="glyphicon glyphicon-user"></span> Login</button>
<button type="button" class="btn btn-info show-page modal-btn" data-page="Signup" data-dismiss="modal" aria-hidden="true"><span class="glyphicon glyphicon-list-alt"></span> Register</button>
</div>
</div>
</div>
</div>
</div>
</div>
</div>
</div>
</div>
脚本:
jQuery(function () {
$("#lemail").val($.cookie("unm"));
$("#lpassword").val($.cookie("pwd"));
});
$("#email").val($.cookie("unm") || "");
$("#lpassword").val($.cookie("pwd") || "");
PHP:
<?php
session_start();
$unm = $_POST['lemail'];
$pwd = $_POST['lpassword'];
?>
答案 0 :(得分:0)
请确保您的Cookie保存或不保存。
在PHP中设置coockies,例如:
setcookie(
'pageVisits', // Name of the cookie, required
$visited, // The value of the cookie
time()+7*24*60*60, // Expiration time, set for a week in the future
'/', // Folder path the cookie will be available for
'demo.tutorialzine.com' // Domain to which the cookie will be bound
);
阅读JS中的coockies:
$(document).ready(function(){
// Getting the kittens cookie:
var str = $.cookie('pageVisits');
// str now contains the value of $visited
});
答案 1 :(得分:0)
为了获得session_start();上班, 您需要将变量添加到$ _SESSION数组中。将PHP代码更改为:
<?php
//Start the session
session_start();
//ALWAYS check if a $_POST variable is set before using it.
if(isset($_POST['lemail'])){
$unm = $_POST['lemail'];
} else {
//Error Code Goes Here
}
if(isset($_POST['lpassword'])){
$pwd = $_POST['lpassword'];
} else {
//Error Code Goes Here
}
//Set SESSION vars
$_SESSION["unm"] = $unm;
$_SESSION["pwd"] = $pwd;
如果要检查他们是否已登录,请使用此代码
session_start();
if(isset($_SESSION["unm"]) && isset($_SESSION["pwd"])){
$loggedIn = true;
//At this point you can also check your database if they are actually a user for extra security
} else {
$loggedIn = false;
}