我創建了一個加友系統,這樣人們可以通過他們的個人資料相互關注。我還沒有完成樣式,它是純粹的功能。我的問題如下:「已發送邀請」按鈕不斷顯示,儘管邀請已發送。老實說,我不知道爲什麼。我在phpmyadmin中測試了這個查詢,它工作正常。有2個用戶的入口。添加朋友系統故障
這裏有疑問:
$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 ");
這是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 $my_username ?></b>
<?php endif; ?>
Bye.
</body>
所以基本上 '添加好友' 按鈕不斷顯示沉綿有一個條目,其中invite_sent被設置爲'1'。這意味着它應該顯示第二個查詢,即'INVITE SENT'按鈕。
我無法弄清楚什麼是錯這裏爲')
編輯#1 - SAFE()函數
function safe($mysqli,$value) {
return mysqli_real_escape_string($mysqli,$value);
}
你的'safe()'方法做什麼? –
我編輯了我的帖子,關於這個:) – user3512502
嘗試在您的HTML中替換'==='以找到'==',以捕獲任何失敗的查詢,這將返回false而不是零行。 – Oli