Question

我正在尝试为以下结果创建SQL查询：

id, name,   Users in group, Messages in groups, Message type 1 in groups, Message type 2 in groups
1   Group1  4               3                   2                         1
2   Group2  3               2                   1                         1
3   Group1  1               1                   1                         0

我的数据库：

create database user_test;

use user_test;


CREATE TABLE `user` (
  `id`             BIGINT      NOT NULL AUTO_INCREMENT,
  `name` VARCHAR(50) NOT NULL,
  `group_id`       BIGINT      NOT NULL,
  PRIMARY KEY (`id`)
);


CREATE TABLE `user_group` (
  `id`      BIGINT           NOT NULL AUTO_INCREMENT,
  `name`    VARCHAR(100)
            CHARACTER SET utf8
            COLLATE utf8_bin NOT NULL,
  UNIQUE KEY `unique` (`name`) USING BTREE,
  PRIMARY KEY (`id`)
);


CREATE TABLE `message` (
  `id`               BIGINT   NOT NULL AUTO_INCREMENT,
  `user_id`           BIGINT   NOT NULL,
  `type`             INT      NOT NULL,
  `date`             DATETIME NOT NULL,
  UNIQUE KEY `unique` (`user_id`, `type`, `date`) USING BTREE,
  PRIMARY KEY (`id`)
);


ALTER TABLE user
    ADD CONSTRAINT user_group_c_fk FOREIGN KEY (group_id) REFERENCES user_group(id);


ALTER TABLE message
    ADD CONSTRAINT user_a_fk FOREIGN KEY (user_id) REFERENCES user(id);

初始值：

INSERT INTO `user_group`
VALUES (1, "Group 1"),
       (2, "Group 2"),
       (3, "Group 3");

INSERT INTO `user`
VALUES (1, "User 1", 1),
       (2, "User 2", 1),
       (3, "User 3", 1),
       (4, "User 4", 1),
       (5, "User 5", 2),
       (6, "User 6", 2),
       (7, "User 7", 2),
       (8, "User 8", 3);

INSERT INTO `message`
VALUES (1, 1, 1, "2019-07-25 00:00:00"),
       (2, 1, 2, "2019-07-25 00:00:01"),
       (3, 1, 2, "2019-07-25 00:00:02"),
       (4, 5, 1, "2019-07-25 00:00:03"),
       (5, 6, 2, "2019-07-25 00:00:04"),
       (6, 8, 1, "2019-07-25 00:00:05");

我想知道是否有可能取得结果。我正在尝试像这样的事情：

use user_test;

select user_group.id, 
       user_group.name,
       count(user.id) As "Users in group",
       count(message.id) AS "Messages in groups",
       count(message.type = 1) AS "Message type1 in groups",
       count(message.type = 2) AS "Message type2 in groups"
FROM user_group
LEFT JOIN user ON user.group_id = user_group.id
LEFT JOIN message ON message.user_id = user.id
WHERE user_group.id = message.user_id
GROUP BY user_group.id;

但是，当然，这完全是错误的查询。我认为在这种情况下，配置数据库无济于事，但它与整个系统相比很小，我无法修改。我想为后端端点创建一个通用查询-最好的解决方案是对所有结果进行一个查询。

Answer 1

我真的很接近获得正确的结果！但是仍然不是正确的方法，但是可能会有所帮助...

use user_test;

 select user_group.id, 
       user_group.name,
        count(user.id) As "Users in group",
        count(message.id) AS "Messages in groups",
        count(CASE message.type WHEN 1 THEN 1 ELSE null end) AS "Message type1 in groups",
        count(CASE message.type WHEN 2 THEN 1 ELSE null end) AS "Message type2 in groups" FROM user_group LEFT JOIN user ON
 user.group_id = user_group.id LEFT JOIN message ON message.user_id =
 user.id WHERE user_group.id = user.group_id GROUP BY user_group.id,
 message.type

Answer 2

希望这会有所帮助。我正在做内部联接，而不是左联接。您可以根据需要进行更改。

select 
ug.id, 
ug.name,
count(u.id) as users_in_group,
count(m.id) as messages_in_group,
count(decode(m.type,'1',count(m.id))) as message_type1_in_group, 
count(decode(m.type,'2',count(m.id))) as message_type1_in_group
from user_group ug
join user u on ug.id=u.group_id
join message m on m.user_id = u.user_id
group by ug.id,ug.name ug.id,ug.name;

Answer 3

您可以使用count(distinct)快速完成此操作：

SELECT ug.id, ug.name,
       count(distinct u.id) As "Users in group",
       count(distinct m.id) AS "Messages in groups",
       count(distinct case when m.type = 1 then m.id end) AS "Message type1 in groups",
       count(distinct case when m.type = 2 then m.id end) AS "Message type2 in groups"
FROM user_group ug  LEFT JOIN
     user u
     ON u.group_id = ug.id LEFT JOIN
     message m
     ON m.user_id = u.id
WHERE ug.id = m.user_id
GROUP BY ug.id;

您还可以通过在进行联接之前汇总来获得准确的计数。另外请注意，我添加了表别名，因此查询更易于编写和阅读。

Answer 4

对Gordon Linoff的答案进行位编辑。

SELECT a.id, a.name,
       COUNT(DISTINCT b.id) AS "Users in group" ,
       COUNT(c.id) AS "Messages in groups",
       COUNT(DISTINCT CASE WHEN c.type = 1 THEN c.id END) AS "Message type 1 in groups",
       COUNT(DISTINCT CASE WHEN c.type = 2 THEN c.id END) AS "Message type 2 in groups"
FROM user_group a LEFT JOIN
     USER b
     ON b.group_id = a.id LEFT JOIN
     message c
     ON c.user_id = b.id GROUP BY a.id;

此处从分组表中获取汇总结果。

SQL-3个表的结果，聚合函数

4 个答案: