Question

我正在创建一个添加好友系统，以便人们可以通过他们的个人资料互相关注。我还没有完成造型，它是纯粹的功能。我的问题如下：'发送邀请'按钮继续显示，尽管邀请已经发送。老实说，我无法弄清楚原因。我在phpmyadmin中测试了查询，它运行正常。这2个用户有一个条目。

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>

所以基本上“添加朋友”按钮会继续显示，尽管有一个条目，其中invite_sent设置为“1”。这意味着它应该显示第二个查询，也就是'INVITE SENT'按钮。

我无法弄清楚这里的错误：'）

编辑＃1 - 安全（）功能

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

Answer 1

我自己解决了。我阅读了代码的每一小部分，并注意到我使用了$ myqsli而不是$ mysqli。因此，它没有显示任何内容。

添加朋友系统故障

1 个答案: