同一表中记录的MySQL联接类型

这个标题有点别扭，但我会尽我所能解释一下我想要实现的目标

我有一个名为

Users

的表和另一个名为

Friends

的表

用户

表的抽象结构为：

+----+------+----------+----------+
| ID | Name | Username | Password |
+----+------+----------+----------+

Friends

表具有如下抽象结构：

+----+--------+----------+--------+
| ID | UserID | FriendID | Hidden |
+----+--------+----------+--------+

其中，

UserID

是发送好友请求的用户的ID，

FriendID

是请求的接收者。如果收件人选择隐藏请求，则隐藏列的值为1

我想把这一切压缩成一个查询，到目前为止，我有两个单独的查询使用

左连接

或

右连接

要查找已发送的请求和当前好友，请执行以下操作：

SELECT 
    *, 
    CASE 
       WHEN C.ID IS Null THEN "Request Sent" 
       ELSE "Friends" 
    END AS Status 
FROM
    (SELECT DISTINCT
         A.ID, A.Name, E.Hidden
     FROM
         Users A 
     INNER JOIN 
         Friends E ON A.ID = E.UserID
     WHERE
         A.ID in (SELECT UserID 
                  FROM Friends 
                  WHERE FriendID = "1" AND Deleted='No')) C
RIGHT JOIN
    (SELECT DISTINCT
         B.ID, B.Name, F.Hidden
     FROM
         Users B 
     INNER JOIN 
         Friends F ON B.ID = F.FriendID
     WHERE
         B.ID in (SELECT FriendID 
                  FROM Friends 
                  WHERE UserID = "1" AND Deleted = 'No')) D ON C.ID = D.ID

SELECT *, CASE WHEN D.ID IS Null THEN "Wants to be your friend" ELSE "Friends" END AS Status FROM
(SELECT DISTINCT
A.ID, A.Name, E.Hidden
FROM
Users A INNER JOIN Friends E ON A.ID=E.UserID
WHERE
A.ID in (SELECT UserID FROM Friends WHERE FriendID = "1" AND Deleted='No')) C
LEFT JOIN
(SELECT DISTINCT
B.ID, B.Name, F.Hidden
FROM
Users B INNER JOIN Friends F ON B.ID=F.FriendID
WHERE
B.ID in (SELECT FriendID FROM Friends WHERE UserID = "1" AND Deleted='No')) D
ON C.ID=D.ID

对于收到的好友请求和当前好友：

SELECT 
    *, 
    CASE 
       WHEN C.ID IS Null THEN "Request Sent" 
       ELSE "Friends" 
    END AS Status 
FROM
    (SELECT DISTINCT
         A.ID, A.Name, E.Hidden
     FROM
         Users A 
     INNER JOIN 
         Friends E ON A.ID = E.UserID
     WHERE
         A.ID in (SELECT UserID 
                  FROM Friends 
                  WHERE FriendID = "1" AND Deleted='No')) C
RIGHT JOIN
    (SELECT DISTINCT
         B.ID, B.Name, F.Hidden
     FROM
         Users B 
     INNER JOIN 
         Friends F ON B.ID = F.FriendID
     WHERE
         B.ID in (SELECT FriendID 
                  FROM Friends 
                  WHERE UserID = "1" AND Deleted = 'No')) D ON C.ID = D.ID

SELECT *, CASE WHEN D.ID IS Null THEN "Wants to be your friend" ELSE "Friends" END AS Status FROM
(SELECT DISTINCT
A.ID, A.Name, E.Hidden
FROM
Users A INNER JOIN Friends E ON A.ID=E.UserID
WHERE
A.ID in (SELECT UserID FROM Friends WHERE FriendID = "1" AND Deleted='No')) C
LEFT JOIN
(SELECT DISTINCT
B.ID, B.Name, F.Hidden
FROM
Users B INNER JOIN Friends F ON B.ID=F.FriendID
WHERE
B.ID in (SELECT FriendID FROM Friends WHERE UserID = "1" AND Deleted='No')) D
ON C.ID=D.ID

如果我没有解释我需要什么，下面的维恩图可能是一个重要的说明

---

请注意，为了构造查询，我已选择了所有列，但我只需要结尾处的朋友姓名和ID。

在两个查询之间执行

union all

或

union distinct

，这将在一个表中返回所有结果

SELECT *, D.ID, CASE WHEN D.ID IS Null THEN "Wants to be your friend" ELSE "Friends" END AS Status FROM
(SELECT DISTINCT
A.ID, A.Name, E.Hidden
FROM
Users A INNER JOIN Friends E ON A.ID=E.UserID
WHERE
A.ID in (SELECT UserID FROM Friends WHERE FriendID = "1" AND Deleted='No')) C
LEFT JOIN
(SELECT DISTINCT
B.ID, B.Name, F.Hidden
FROM
Users B INNER JOIN Friends F ON B.ID=F.FriendID
WHERE
B.ID in (SELECT FriendID FROM Friends WHERE UserID = "1" AND Deleted='No')) D
ON C.ID=D.ID
union all
SELECT *, D.ID, CASE WHEN C.ID IS Null THEN "Request Sent" ELSE "Friends" END AS Status FROM
(SELECT DISTINCT
A.ID, A.Name, E.Hidden
FROM
Users A INNER JOIN Friends E ON A.ID=E.UserID
WHERE
A.ID in (SELECT UserID FROM Friends WHERE FriendID = "1" AND Deleted='No')) C
RIGHT JOIN
(SELECT DISTINCT
B.ID, B.Name, F.Hidden
FROM
Users B INNER JOIN Friends F ON B.ID=F.FriendID
WHERE
B.ID in (SELECT FriendID FROM Friends WHERE UserID = "1" AND Deleted='No')) D
ON C.ID=D.ID

---

在我将

UNION ALL

更改为

UNION DISTINCT

以停止重复行（如果两个用户是朋友）之后，这似乎起到了作用。谢谢你的帮助！再多加一点，是否可以提供用户ID，在这个问题中为1，只提供一次？只是我也有一个会话表，所以我将使用一个内部join语句从会话ID中获取用户ID。如果没有，也没关系！你不是已经有了一个案例结构来决定这是否是1吗？你可以这样做，把它从你的子查询添加到你的主查询中，在两个查询中的案例结构之前。我试试看，你能编辑你的答案吗？只是一点提示？如果你不想，我就不接受