SQL加入并在表中获取计数

问题描述 投票:0回答:3

这是我的SQL Fiddle

我现在正加入useraddress表。现在我需要记录in_timeout_time填充的日志表

这是我到目前为止的SQL查询

select u.id, u.name, a.address from user u 
left join address a on u.id = a.user_id
where u.id = 1

即,输出应该是这样的

id  name    address  total_count proper_count
1   Alpha   Chennai  4           3
mysql sql join left-join
3个回答
0
投票

根据您的要求使用以下架构设计。 User_id必须为INTEGER,并使用datetime数据类型为in&out时间列。

CREATE TABLE log (
    id BIGINT,
    user_id BIGINT,
    in_time datetime,
    out_time datetime
);

INSERT INTO log (id,user_id, in_time, out_time) VALUES (1,1,'2018-07-21 06:50:41','2018-07-21 10:50:41');
INSERT INTO log (id,user_id, in_time, out_time) VALUES (2,1,'2018-07-22 06:50:41','2018-07-22 10:50:41');
INSERT INTO log (id,user_id, in_time) VALUES (3,1,'2018-07-23 06:50:41');
INSERT INTO log (id,user_id, in_time, out_time) VALUES (4,1,'2018-07-24 06:50:41','2018-07-22 10:50:41');

select u.id as user_id, u.name, a.address, COUNT(in_time) AS total_count, COUNT(out_time) as proper_count
from log l
INNER JOIN user u on u.id = l.user_id
INNER JOIN address a on a.user_id = u.id
GROUP BY u.id , u.name, a.address

0
投票

你可以这样:

SELECT u.id, u.name, a.address, 
    COUNT(*) AS total_count,
    SUM(CASE WHEN l.in_time = 0 OR l.out_time = 0 THEN 0 ELSE 1 END) AS proper_count
FROM USER u 
LEFT JOIN address a 
ON u.id = a.user_id
LEFT JOIN log l 
ON u.id = l.user_id
WHERE u.id = 1
GROUP BY u.id, u.name, a.address; 

0
投票

您距离预期的查询只有一步之遥。只需要与日志表的另一个连接并使用聚合函数

select u.id, u.name, a.address,
sum(case when in_time is not null and out_time is not null 
    then 1 else 0 end ) as total_count ,
    SUM(CASE WHEN l.in_time = 0 OR l.out_time = 0 THEN 0 ELSE 1 END) AS proper_count
from user u 
left join address a on u.id = a.user_id
left join log l on u.id=l.user_id
where u.id = 1
group by u.id, u.name, a.address

http://sqlfiddle.com/#!9/b2efe0/6

id  name    address total_count proper_count
1   Alpha   Chennai   4            3
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