繁体   English   中英

添加好友系统失败

[英]Add friend system failure

我正在创建一个添加好友系统,以便人们可以通过自己的个人资料互相关注。 我还没有做样式,这仅仅是功能。 我的问题如下:即使已发送邀请,“已发送邀请”按钮仍持续显示。 老实说,我不知道为什么。 我在phpmyadmin中测试了查询,并且工作正常。 有两个用户的条目。

以下是查询:

$user_id = $_SESSION['user_id'];
//SEARCH THE USERNAME OF THE LOGGED IN USER
$stmt = $mysqli->prepare("SELECT username, email FROM members WHERE user_id = ? ");
$stmt->bind_param('s', $user_id); 
$stmt->execute();  
$stmt->store_result();
$stmt->bind_result($my_username, $my_email);
$stmt->fetch();
$stmt->close();

$username = safe($mysqli,$_GET["username"]);
//LOOK UP THE DETAILS OF THE USERNAME
$stmt = $mysqli->prepare("SELECT user_id, email FROM members WHERE username = ? ");
$stmt->bind_param('s', $username); 
$stmt->execute();  
$stmt->store_result();
$stmt->bind_result($userid, $email);
$stmt->fetch();
$stmt->close();

//Sent invite <---IT KEEPS SAYING THIS EVENTHOUGH THERE'S AN ACTIVE ENTRY
$friendQuery1 = mysqli_query($myqsli,"SELECT * FROM friends WHERE friend_one = '$my_username' AND friend_two = '$username' AND invite_sent = 0 ");

//Invite sent, awaiting acceptance
$friendQuery2 = mysqli_query($myqsli,"SELECT * FROM friends WHERE friend_one = '$my_username' AND friend_two = '$username' AND invite_sent = 1 ");

//Invite accepted. User is friends.
$friendQuery3 = mysqli_query($myqsli,"SELECT * FROM friends WHERE friend_one = '$my_username' AND friend_two = '$username' AND invite_accepted = 1 ");

这是html部分

<body>
DEBUG:
This is the profile page from <b><?php echo $username ?></b><br />
Emailadres: <b><?php echo $email ?></b><br />
<br />

<?php if (login_check($mysqli) == true) : ?> 

<?php if  (mysqli_num_rows($friendQuery1) == 0) {
    echo '<div style="width: 150px; text-align:center; border:1px solid #cecece;">Add friend</div>';
}

if (mysqli_num_rows($friendQuery2) == 1) {
    echo '<div style="width: 200px; text-align:center; border:1px solid #cecece;">Invite sent</div>';
}

if (mysqli_num_rows($friendQuery3) == 1) {
    echo '<div style="width: 200px; text-align:center; border:1px solid #cecece;">Already Friends</div>';
}
?>

<br>
<br>
<br>
<b>CURRENT LOGGED IN USER: <?php echo $user_id ?> AKA USERNAME: <?php echo $my_username ?></b>

<?php endif; ?>

Bye.

</body>

因此,即使有一个将vital_sent设置为“ 1”的条目,“添加朋友”按钮也基本上会一直显示。 这意味着它应该显示第二个查询又名“邀请发送”按钮。

我无法弄清楚这里出了什么问题:')

编辑#1-SAFE()函数

function safe($mysqli,$value) { 
   return mysqli_real_escape_string($mysqli,$value); 
}

我自己解决了。 我阅读了代码的每个小部分,并注意到我使用$ myqsli而不是$ mysqli。 因此,它什么也没显示。

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM