Question

当下拉通知栏打开时，我希望点击此按钮。我希望它应该在那里，当我点击该按钮下拉菜单打开并将值0发送到SQL。

<li class="dropdown notifications-menu">
    <a href="#" class="dropdown-toggle" data-toggle="dropdown">
        <i class="fa fa-bell-o"></i>
        <span class="label label-warning">10</span>
    </a>
    <ul class="dropdown-menu">
        <li class="header">You have 10 notifications</li>
        <li>
            <!-- inner menu: contains the actual data -->
            <ul class="menu">
                <li>
                    <a href="#">
                        <i class="ion ion-ios-people info"></i> Notification title
                    </a>
                </li>
                ...
            </ul>
        </li>
        <li class="footer"><a href="#">View all</a></li>
    </ul>
</li>

Answer 1

<li class="dropdown notifications-menu">
              <a href="#" class="dropdown-toggle" data-toggle="dropdown" id="updatevalue>
                <i class="fa fa-bell-o"></i>
                <span class="label label-warning">10</span>
              </a>
              <ul class="dropdown-menu">
                <li class="header">You have 10 notifications</li>
                <li>
                  <!-- inner menu: contains the actual data -->
                  <ul class="menu">
                    <li>
                      <a href="#">
                        <i class="ion ion-ios-people info"></i> Notification title
                      </a>
                    </li>
                    ...
                  </ul>
                </li>
                <li class="footer"><a href="#">View all</a></li>
              </ul>
            </li>


<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.2/jquery.min.js"></script>
<script>
$(document).ready(function(){
    $("#updatevalue").click(function(){
        //Do your ajax works here.
        $.ajax({url: "Your url", success: function(result){
        //After success what you want to do. Do here

        }});
    });
});
</script>

希望你知道ajax。

Answer 2

这是php Mysql的脚本在哪里把它放在ajax中???

<?php
    $host = 'localhost';
    $dbname   = 'project';
    $username = 'root';
    $password = '1234';
    $charset = 'utf8';

     $d1 = $row7['userID'];
    try {
        $pdo = new PDO("mysql:host=$host;dbname=$dbname", $username, $password);


         $usid = ($row['userID']);
        $sql1 = 
     "SELECT 
        follow.followmak, follow.followrec, follow.date, follow.status , follow.time , tbl_users.userID , tbl_users.userName ,tbl_users.Fname , tbl_users.Lname 
     FROM 
          follow , tbl_users 
     WHERE 
          follow.followmak=tbl_users.userID 
          AND follow.followrec = $d1 AND follow.status = 1
     ORDER BY 
          follow.time DESC";



         $q1 = $pdo->prepare($sql1);
        $q1->execute([$usid]);
        $q1->setFetchMode(PDO::FETCH_ASSOC);



    } catch (PDOException $e) {
        die("Could not connect to the database $dbname :" . $e->getMessage());
    }

    ?>

单击下拉列表

2 个答案: