LDAP身份验证失败时无法显示模式弹出窗口
LDAP身份验证失败时无法显示模式弹出窗口
提问于 2019-05-15 20:52:35
我正在通过PHP验证来自LDAP的用户。我已经尝试了一些代码，现在我可以完美地验证用户。但在登录失败的情况下，我希望显示一个带有一些自定义消息的模式弹出窗口。这是我的代码-
<?php
if ($_SERVER["REQUEST_METHOD"] == "POST") {
// username and password sent from form
$username = $_POST['username'];
//print($username);
$password = $_POST['password'];
//print($password);
// using ldap bind
$ldaprdn = 'uid=' .$username. ',ou=People,ou=AP,o=abcd.com'; // ldap rdn or dn
$ldappass = $password;
// connect to ldap server
$ldapconn = ldap_connect("ad.abcd.com") or die("Could not connect to LDAP server.");
if ($ldapconn) {
try {
$ldapbind = @ldap_bind($ldapconn, $ldaprdn, $ldappass);
} catch (Exception $e) {
//echo "<script type='text/javascript'>$('#myModal').modal('show');</script>";
//echo "<script type='text/javascript'>$('#myModal').fadeIn('show');</script>";
echo "<script>
$(window).load(function(){
$('#myModal').modal('show');
});
</script>";
}
if ($ldapbind) {
$filter = '(sAMAccountName='.$username.')';
$result = ldap_search($ldapconn, $ldaprdn, "(cn=*)") or exit("Unable to search LDAP server");
$entries = ldap_get_entries($ldapconn, $result);
$userDN = $entries[0]["ikealegacyuid"][0];
echo ('<p style="color:green;">I have the user DN: '.$userDN.'</p>');
$url = 'Location: logpage01.html?uid=' .$userDN;
echo $url;
header($url);
} else {
//echo "<script type='text/javascript'>$('#myModal').modal('show');</script>";
//echo "<script type='text/javascript'>$('#myModal').fadeIn('show');</script>";
echo "<script>
$(window).load(function(){
$('#myModal').modal('show');
});
</script>";
}
}
}
?>
<html>
<head>
<title>Login Page</title>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
<style type = "text/css">
body {
font-family:Arial, Helvetica, sans-serif;
font-size:14px;
}
label {
font-weight:bold;
width:100px;
font-size:14px;
}
.box {
border:#666666 solid 1px;
}
</style>
</head>
<body bgcolor = "#FFFFFF">
<div align = "center">
<div style = "width:300px; border: solid 1px #333333; " align = "left">
<div style = "background-color:#333333; color:#FFFFFF; padding:3px;"><b>Login</b></div>
<div style = "margin:30px">
<form action = "" method = "post">
<label>UserName :</label><input type = "text" name = "username" class = "box"/><br /><br />
<label>Password :</label><input type = "password" name = "password" class = "box" /><br/><br />
<input type = "submit" value = " Submit "/><br />
</form>
</div>
</div>
</div>
<div class="container">
<!-- Modal -->
<div class="modal fade" id="myModal" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true" data-backdrop="static" data-keyboard="false">
<div class="modal-dialog">
<!-- Modal content-->
<div class="modal-content">
<div class="modal-header">
<!-- <button type="button" class="close" data-dismiss="modal">×</button> -->
<h4 class="modal-title">System Message : Failure</h4>
</div>
<div class="modal-body">
<p>Sorry !! Something went wrong. Please contact your administrator.</p>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal" onclick="window.location.href = 'logpage01.html';">Close</button>
</div>
</div>
</div>
</div>
</div>
</body>
</html>我尝试了这里找到的许多解决方案。但都不管用。我需要一些帮助。
更新：也检查了页面源代码。看起来还不错。如果没有，请让我知道问题所在。
我也试着在页面加载时调用modal。它工作得很完美。
<script type="text/javascript">
$(window).on('load',function(){
$('#myModal').modal('show');
});
</script>回答 2
Stack Overflow用户
回答已采纳
发布于 2019-05-16 03:25:14
仅仅回显脚本在这里是行不通的，在加载加载jQuery的html之前，您的浏览器最多只能运行一个引用$的脚本，所以即使它运行您的脚本，它也会抛出Uncaught ReferenceError: $ is not defined。
相反，您可以使用一个标志来告诉浏览器是否显示模式。例如，您可以在else (和catch)语句中设置一个变量：
if ($ldapbind) { /*...*/ }
else {
$access_denied=1;
}并将其插入到html中，以便浏览器知道它必须在access_denied===1时加载模式，为此，您还需要添加仅在该条件下加载模式的脚本：
<div id="myModal" access_denied="<?php print $access_denied ?>" > ... </div>
<script>
$('#myModal[access_denied=1]').modal('show');
</script>在另一个使用ajax请求的上下文中，您可能只需回显一些json或直接输出模态html。
Stack Overflow用户
发布于 2019-05-15 21:00:17
如果我没看错的话，你缺少popper.js，这在引导中是必需的。请检查此链接https://getbootstrap.com/docs/4.3/getting-started/introduction/，您将获得一个您可以使用的popper的在线cdn。或者简单地复制我的代码片段。还请注意，您在版本3中使用的是bootstrap。我强烈建议您使用V4。
在php中，当你想要显示模式时，你只需要回显下面的内容。
echo "<script>
$('#myModal').modal('show');
</script>";
$('#myModal').modal('show');<link rel="stylesheet" href="https://stackpath.bootstrapcdn.com/bootstrap/4.3.1/css/bootstrap.min.css" integrity="sha384-ggOyR0iXCbMQv3Xipma34MD+dH/1fQ784/j6cY/iJTQUOhcWr7x9JvoRxT2MZw1T" crossorigin="anonymous">
<script src="https://code.jquery.com/jquery-3.3.1.slim.min.js" integrity="sha384-q8i/X+965DzO0rT7abK41JStQIAqVgRVzpbzo5smXKp4YfRvH+8abtTE1Pi6jizo" crossorigin="anonymous"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/popper.js/1.14.7/umd/popper.min.js" integrity="sha384-UO2eT0CpHqdSJQ6hJty5KVphtPhzWj9WO1clHTMGa3JDZwrnQq4sF86dIHNDz0W1" crossorigin="anonymous"></script>
<script src="https://stackpath.bootstrapcdn.com/bootstrap/4.3.1/js/bootstrap.min.js" integrity="sha384-JjSmVgyd0p3pXB1rRibZUAYoIIy6OrQ6VrjIEaFf/nJGzIxFDsf4x0xIM+B07jRM" crossorigin="anonymous"></script>
<div class="modal fade" id="myModal" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true" data-backdrop="static" data-keyboard="false">
<div class="modal-dialog">
<!-- Modal content-->
<div class="modal-content">
<div class="modal-header">
<!-- <button type="button" class="close" data-dismiss="modal">×</button> -->
<h4 class="modal-title">System Message : Failure</h4>
</div>
<div class="modal-body">
<p>Sorry !! Something went wrong. Please contact your administrator.</p>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal" onclick="window.location.href = 'logpage01.html';">Close</button>
</div>
</div>
</div>
</div>
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/56149845
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