如何編寫mysql查詢來實現以下類型的連接（基於列的值加入）

Question

表用戶：

id----email 
1-----s@s.com 
2-----p@p.com 

表user_buddies

id-----user_id----buddy_id----status enum('accepted','pending') 
1------1----------2-----------pending 

假設我正在查詢用戶2然后我想獲得用戶2的電子郵件。 如果我正在查詢用戶1，那么我想獲得用戶1的電子郵件。

以下查詢始終返回空行。

SELECT
    users.email,
    ub1.*, ub2.*
FROM
    users
JOIN user_buddies ub1 ON users.id = ub1.user_id
JOIN user_buddies ub2 ON users.id = ub2.buddy_id
WHERE
    users.id = 1;

有人可以幫我嗎？ 謝謝。

預期結果：

email-----user_id-----buddy_id-----status
s@s.com --1-----------2------------pending (If queryed on user 2)
p@p.com---1-----------2------------pending (If queried on user 1)

語境：

當user1向user2發送請求時， 只有一行被添加到user_buddies表， status=pending 。 因此，對於user1和user2伙伴列表都使用相同的行創建。 因此，如果查詢user2所需的電子郵件是user2's 。

Answer 1

您可以檢查以下查詢：

SELECT
    users.email,
    ub1.user_id,
    ub1.buddy_id,
    ub1.`status`
FROM
    users
JOIN user_buddies ub1 ON users.id = ub1.user_id OR users.id = ub1.buddy_id
WHERE   users.id = 2;

SQL FIDDLE DEMO

結果：

email      user_id     buddy_id     status
p@p.com       1           2         pending (if queried on users.id = 2)
s@s.com       1           2         pending (if queried on users.id = 1)

編輯：

SELECT
    (
        SELECT
            u.email
        FROM users u
        WHERE   u.id =
        IF (
            users.id = ub1.user_id,
            ub1.buddy_id,
            ub1.user_id
        )
    ) AS buddyEmail,
    ub1.user_id,
    ub1.buddy_id,
    ub1.`status`
FROM
    users
INNER JOIN user_buddies ub1 ON users.id = ub1.user_id
OR users.id = ub1.buddy_id
WHERE   users.id = 2;

檢查這個新查詢的小提琴

Answer 2

SELECT 'Userid' querysource, u.id, u.email, ub.*, u1.EMAIL AS Buddyemail
FROM @users u 
JOIN @user_buddies ub ON u.id = ub.user_id
JOIN @users u1 ON u1.id = ub.BUDDY_ID
WHERE u.id = 1

UNION

SELECT 'Buddyidid', u.id, u.email, ub.*, u1.email AS Buddyemail
FROM @users u 
JOIN @user_buddies ub ON u.id = ub.buddy_id 
JOIN @users u1 ON u1.id = ub.user_ID
WHERE u.id = 1