Mysql 连接和多重连接条件

我有一张用户表

ID     NAME
1      John
2      Mike
3      Jack

和具有属性和用户ID的表

USER   ATTRIBUTE
1      1
1      2
2      4

我需要选择属性为1和2的所有用户，因此，在本例中，用户为1。属性可以多于两个

我试过了

SELECT * FROM user u LEFT JOIN attributes a ON u.id = a.user 
WHERE a.attribute = 1 AND a.attribute = 2

---

但是它当然不起作用。

having子句可以与sum一起使用

SELECT u.id FROM user u 
INNER JOIN attributes a ON u.id = a.user 
group by u.id
having ( sum(case when attribute in (1,2) then 1 else 0 end) ) =2

---

您正在查找属性1和属性2存在的所有用户。解决这个问题的一种方法——顾名思义——是EXISTS子句：

select * 
from users u
where exists
(
  select *
  from user_attributes ua
  where ua.user = u.id
  and ua.attribute = 1
)
and exists
(
  select *
  from user_attributes ua
  where ua.user = u.id
  and ua.attribute = 2
);

另一种方法是：找到具有这两个属性的所有用户ID，然后从users表中选择

select *
from users
where id in
(
  select user
  from user_attributes
  where attribute in (1,2)
  group by user
  having count(*) = 2 
);

如果属性中存在重复项，则必须用countdistinct属性替换count*

---

还有其他方法可以做到这一点。我认为上面提到的这两个问题相当直截了当。

您需要使用IN和GROUP BY的组合。。。必须做到这一点。如果您只需要用户ID，也不需要加入。比如：

SELECT user, COUNT(attribute) AS attribute_count
FROM attributes
WHERE attribute IN(...) /* include your set of attributes here */
GROUP BY user
HAVING attribute_count = ? /* include number equal to number of attribute ID's in IN() above */

如果需要用户id和名称，只需将从上述查询派生的此记录集作为筛选器连接到users表：

SELECT user.id, user.name
FROM user
INNER JOIN
  (
    SELECT user, COUNT(attribute) AS attribute_count
    FROM attributes
    WHERE attribute IN(...) /* include your set of attributes here */
    GROUP BY user
    HAVING attribute_count = ? /* include number equal to number of attribute ID's in IN() above */
  ) AS filter
  ON user.id = filter.user

---

你需要一个由……组成的小组。。。。拥有

SELECT u.name
FROM
    users u
JOIN
    attributes a
    ON u.id = a.user
WHERE a.id IN (1,2)
GROUP BY u.name
    HAVING COUNT(*) = 2

---

它是一个大查询，用户表是核心。如果我已经习惯了在别的地方吃饭呢？以X公里为单位显示所选用户area@poh实际上，当您键入注释时，我正在演示如何通过联接将该筛选器应用于主表。