Question

我正在尝试抓取一条消息，然后在我的通知标签下显示它，但它没有显示消息我正在尝试从两个表成员表和通知表中进行INNER JOIN这是我的表格结构已经建立。

会员表

MEMBERID，

用户名，

ownerIP，

以及更多

现在在通知表中，这就是我所拥有的。

通知

ID

MEMBERID

消息

状态

这是我的PHP代码。

                            <li class="dropdown hidden-xs">
                                <a href="#" data-target="#" class="dropdown-toggle waves-effect waves-light" data-toggle="dropdown" aria-expanded="true">
                                    <i class="icon-bell"></i> <span class="badge badge-xs badge-danger">3</span>
                                </a>
                                <ul class="dropdown-menu dropdown-menu-lg">
                                <?php 

                                // Load Notfications
                                $loadNotfications = $db->fetchAll("SELECT * FROM notfications INNER JOIN members.memberID ON notfications.memberID = members.memberID WHERE memberID = '".$memberID."'");

                                if($loadNotfications) {
                                    foreach($loadNotfications as $loadNotfication) {
                                        $id = $loadNotfication['id'];
                                        $memberID = $loadNotfication['memberID'];
                                        $message = $loadNotfication['message'];
                                        $status = $loadNotfication['status'];

                                ?>                                  
                                    <li class="notifi-title"><span class="label label-default pull-right">New 3</span>Notification</li>
                                    <li class="list-group nicescroll notification-list">

                                       <!-- list item-->
                                        <a href="javascript:void(0);" class="list-group-item">
                                            <div class="media">
                                                <div class="pull-left p-r-10">
                                                 <em class="fa fa-cog fa-2x text-custom"></em>
                                                </div>
                                                <div class="media-body">
                                                  <h5 class="media-heading">New settings</h5>
                                                  <p class="m-0">
                                                    <small><?php echo $message; ?></small>
                                                </p>
                                                </div>
                                          </div>
                                       </a>
                                    </li>
                                    <li>
                                        <a href="javascript:void(0);" class="list-group-item text-right">
                                            <small class="font-600">See all notifications</small>
                                        </a>
                                    </li>
                                <?php
                                    }
                                }
                                ?>                                  
                                </ul>
                            </li>

我需要帮助，任何人都可以帮助我。

Answer 1

我想你可以查一下这句话：

$loadNotfications = $db->fetchAll("SELECT * FROM notfications INNER JOIN members.memberID ON notfications.memberID = members.memberID WHERE members.memberID = '".$memberID."'");

我已将members.memberId放在WHERE子句之后。我认为它可行。

Answer 2

所以我解决了我的问题，这就是我要解决的问题

                                // Load Notfications
                                $loadNotifications = $db->fetchAll("SELECT * FROM notifications WHERE memberID='{$myID}' AND status = 1 ORDER BY id DESC LIMIT 25");

                                if($loadNotifications) {
                                    foreach($loadNotifications as $loadNotification) {
                                        $id = $loadNotification['id'];
                                        $memberID = $loadNotification['memberID'];
                                        $message = $loadNotification['message'];
                                        $status = $loadNotification['status'];
                                        $subject = $loadNotification['subject'];
                                ?>                                  

                                       <!-- list item-->
                                        <a href="javascript:void(0);" class="list-group-item">
                                            <div class="media">
                                                <div class="pull-left p-r-10">
                                                 <em class="fa fa-bell fa-2x text-custom"></em>
                                                </div>
                                                <div class="media-body">
                                                  <h5 class="media-heading"><?php echo $subject; ?></h5>
                                                  <p class="m-0" style="white-space: normal;">
                                                    <small><?php echo $message; ?></small>
                                                </p>
                                                </div>
                                          </div>
                                       </a>

                                <?php
                                    }
                                } else {
                                    echo '<span class="text-center">There are currently no unread notifications.</span>';
                                }
                                ?>

                                </li>

PHP MySql INNER JOIN用户ID错误

2 个答案: