Question

我正在尝试实现“添加好友”页面，但我无法确定为用户显示的所有人可以添加的正确SQL是什么。这是我的表格：

用户：

-id

-name

友谊

-id_user

-id_friend

这是这篇文章的同类数据库：friendship database schema

对于一个用户，我希望所有与他不是朋友的用户，没有让他成为朋友并且没有被要求成为他的朋友

编辑：这是我想要的结果中最接近的，我试过了：

SELECT *
FROM user user
LEFT JOIN friend friend ON user.id=friend.id
WHERE friend.id IS NULL AND friend.id_friend IS NULL

但它只会归还所有没有朋友的人。

提前致谢！

Answer 1

不是他的朋友

这应该这样做：

SELECT *
FROM users U
WHERE NOT id IN (
  SELECT F.id_friend
  FROM friendship F
  WHERE U.id = F.id_friend
    AND F.id_user = 1
)
AND id != 1

或者

SELECT U.*, F.*
FROM users U
LEFT JOIN friendship F
  ON U.id = F.id_friend
  AND F.id_user = 1
WHERE F.id_user IS NULL
  AND U.id != 1

然而，

没有让他成为朋友并且没有被要求与他成为朋友的人

您的架构中没有此信息。

编辑：太早了：/

Answer 2

sqlFiddle demo根据您在其他相关问题中解释的friendship schema。

/* --------------get users who are not friends with 'Tim' */
SELECT u2.id,u2.name
FROM users u1,users u2
WHERE NOT EXISTS(SELECT 1
                 FROM friendship f
                 WHERE f.id_user = u1.id AND
                       f.id_friend = u2.id)
  AND NOT EXISTS(SELECT 1
                 FROM friendship f
                 WHERE f.id_user = u2.id AND
                       f.id_friend = u1.id)
AND u1.id != u2.id
AND u1.id = 1;

/* ---------------get users who have not asked 'Tim' to be a friend*/
SELECT u2.id,u2.name
FROM users u1,users u2
WHERE NOT EXISTS(SELECT 1
                 FROM friendship f
                 WHERE f.id_user = u2.id AND
                       f.id_friend = u1.id)
AND u1.id != u2.id
AND u1.id = 1;

/*  ------------- get users who have not been asked by 'Tim' to be a friend */
SELECT u2.id,u2.name
FROM users u1,users u2
WHERE NOT EXISTS(SELECT 1
                 FROM friendship f
                 WHERE f.id_user = u1.id AND
                       f.id_friend = u2.id)
AND u1.id != u2.id
AND u1.id = 1;

所有三个查询都已设置为根据Tim和u1.id = 1搜索数据，只需将最后一个条件更改为您要查找的任何用户ID。

第一个查询可以合并为OR条件，如下所示

/* get users who are not friends with 'Tim' */
SELECT u2.id,u2.name
FROM users u1,users u2
WHERE NOT EXISTS(SELECT 1
                 FROM friendship f
                 WHERE (f.id_user = u1.id AND
                        f.id_friend = u2.id)
                     OR(f.id_user = u2.id AND
                        f.id_friend = u1.id))
AND u1.id != u2.id
AND u1.id = 1;

左连接解决方案sqlFiddle（您必须替换1的ID与您选择的用户ID的比较，它们在第一个查询中出现3次，在第二和第三个查询）

/* get users who are not friends with 'Tim' */
SELECT u.id,u.name
FROM users u
LEFT JOIN friendship f ON
  (u.id=f.id_user AND f.id_friend = 1)
OR(u.id=f.id_friend AND f.id_user = 1)
WHERE
    f.id_friend IS NULL
AND u.id != 1;

/* get users who have not asked 'Tim' to be a friend */
SELECT u.id,u.name
FROM users u
LEFT JOIN friendship f ON
  (u.id=f.id_user AND f.id_friend = 1)
WHERE
    f.id_friend IS NULL
AND u.id != 1;

/* get users who have not been asked by 'Tim' to be a friend */
SELECT u.id,u.name
FROM users u
LEFT JOIN friendship f ON
  (u.id=f.id_friend AND f.id_user = 1)
WHERE
    f.id_friend IS NULL
AND u.id != 1;

SQL请求：显示不是朋友的人

2 个答案: