在弹出窗口中显示信息

Question

我有这个排行榜: 排行榜

单击排行榜中的名称时,将显示一个弹出窗口. 弹出式快照

现在,我想显示弹出窗口中已点击其名称的任何人的信息.因为,您可以看到弹出窗口中不包含任何数据.

我一直在使用一个活动目录,从那里我可以获取数据,如配置文件pic和我正在维护的数据库中的其他信息.

问题是如何链接活动目录和数据库,并在单击名称时在弹出窗口中显示所需信息.请帮助

弹出窗口涉及的Javascript:

$(document).ready(function() {
        $('.tab a').on('click', function(e) {
            e.preventDefault();
            var _this = $(this);
            var block = _this.attr('href');
            $(".tab").removeClass("active");
            _this.parent().addClass("active");
            $(".leadboardcontent").hide();
            $(block).fadeIn();
        });


        /**
         * Fade in the Popup
         */
        $('.leaderboard li').on('click', function () {
            $('#popup').fadeIn();
            var mark = $(this).find('name').text();
            var small = $(this).find('points').text();
            $('#popup-name').text('Name: ' + name);
            $('#popup-score').text('Score: ' + points);
        });

});

对于活动目录我正在使用此变量并在登录用户的任何地方回显它:

<?php
$username   = $_POST['username'];
$password   = $_POST['password'];
$server = 'ldap:xxxxx';
$domain = 'xxx'
$port       = xxx;

$ldap_connection = ldap_connect($server, $port);
if (! $ldap_connection)
{
    echo '<p>LDAP SERVER CONNECTION FAILED</p>';
    exit;
}

// Help talking to AD
ldap_set_option($ldap_connection, LDAP_OPT_PROTOCOL_VERSION, 3);
ldap_set_option($ldap_connection, LDAP_OPT_REFERRALS, 0);

$ldap_bind = @ldap_bind($ldap_connection, $username.$domain, $password);

if (! $ldap_bind)
{
    echo '<p>LDAP BINDING FAILED</p>';
    exit;
}
else
{
	echo 'login successful <br/>';

}
$base_dn = "OU=Employees,OU=Accounts,OU=India,DC=asia,DC=manh,DC=com";
$filter ="(&(objectClass=user)(samaccountname=$username))";

$result = ldap_search($ldap_connection,$base_dn,$filter);

$rescount = ldap_count_entries($ldap_connection,$result);

$data = ldap_get_entries($ldap_connection,$result);

if ($data["count"] > 0)
{
for ($i=0; $i<$data["count"]; $i++)
{

if(isset($data[$i]["employeeid"][0]))
{
$user= $data[$i]["employeeid"][0];
session_start();
$_SESSION['id']=$user;
}


if (isset($data[$i]["thumbnailphoto"][0]))
{
$photo=$data[$i]["thumbnailphoto"][0];
$_SESSION['photo']=$photo;

}


if (isset($data[$i]["title"][0]))
{
$title=$data[$i]["title"][0];
$_SESSION['Employeetitle']=$title;
}


if (isset($data[$i]["department"][0]))
{
$department=$data[$i]["department"][0];
$_SESSION['Employeedepartment']=$department;
}



}
}
else
        {
            echo "<p>No results found!</p>";
        }


if(isset($_SESSION['id']))
{
echo "session set";
header('location:Profile/index.php');
}


?>

Answer 1

您是否尝试过这样...定义 li 元素的类如下..

<div class="leaderboard">
<ul>
   <li class="name">value</li>
   <li class="points">value</li>
</ul>
</div>

然后像这样获取值

var name= $(".leaderboard").find('.name').text();
var points= $(".leaderboard").find('.points').text();

$('#popup-name').text('Name: ' + name);
$('#popup-score').text('Score: ' + points);