MySQL的：用戶在線狀態[求助]

Question

我已經做了朋友的唯一

SELECT * FROM users 
WHERE 
    (id_user IN 
     (SELECT CASE 
      WHEN friend_one = '{$id}' 
       THEN friend_two 
      WHEN friend_two = '{$id}' 
       THEN friend_one 
      END 
     FROM friends 
     WHERE status = '1') 
    ) AND (status = '1' OR ($date - last_login) <= 300) 

而且它的工作原理用戶的在線狀態，這MySQL只顯示好友登錄狀態列表。

但問題是，當我嘗試使另一個用戶登錄狀態，只顯示另一個人登錄，但沒有在我們的朋友列表中的列表。

SELECT * FROM users 
WHERE 
    (id_user NOT IN 
     (SELECT CASE 
      WHEN friend_one = '{$id}' 
       THEN friend_two 
      WHEN friend_two = '{$id}' 
       THEN friend_one 
      END 
     FROM friends 
     WHERE status = '1') 
    ) AND (status = '1' OR ($date - last_login) <= 300) 

所以這是結果，我想

Logged as: khrisna 
Friend list: Ben, Alpha, Ralph 
All User: Ben, Alpha, John, Mark, Zayn 

Online friend list: Ben, Alpha 

Another online user: John, Mark, Zayn 

有人可以幫助我達到我想要的結果呢？

---

UPDATE

users表

friends表

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我想要像這樣的結果

在線用戶：Khrisna Gunanasurya // logined佔

好友列表在線：阿爾法，約翰logined帳戶

的//在線好友

其他用戶在線：馬克，拉爾夫//其他用戶登錄並且不是登錄帳戶的朋友

Answer 1

一種方法：

獲取通過在線

SELECT * 
    FROM 
(
    SELECT CASE WHEN friend_one = ? THEN friend_two ELSE friend_one END friend 
    FROM friends 
    WHERE ? IN(friend_one, friend_two) 
    AND status = 1 
) f JOIN users u 
    ON f.friend = u.id_user 
WHERE status = 1 OR last_login >= UNIX_TIMESTAMP() - 300; 

獲取所有非朋友是網上

SELECT * 
    FROM users u 
WHERE id_user <> ? 
    AND NOT EXISTS 
(
    SELECT * 
    FROM friends 
    WHERE ? IN(friend_one, friend_two) 
    AND status = 1 
    AND u.id_user = CASE WHEN friend_one = ? THEN friend_two ELSE friend_one END 
) 
    AND (status = 1 OR last_login >= UNIX_TIMESTAMP() - 300); 

問號代表的ID的值的登錄用戶的所有好友（例如Khrisna Gunanasurya）

這裏是SQLFiddle演示

---

你的PHP代碼可能看起來像

$sql = "SELECT * 
    FROM 
(
    SELECT CASE WHEN friend_one = ? THEN friend_two ELSE friend_one END friend 
    FROM friends 
    WHERE ? IN(friend_one, friend_two) 
    AND status = 1 
) f JOIN users u 
    ON f.friend = u.id_user 
WHERE status = 1 OR last_login >= UNIX_TIMESTAMP() - 300"; 

$stmt = $pdo->prepare($sql); 
$stmt->execute(array_fill(0, 2, $id));