Question

我现在正在加入user和address表。现在，我需要对in_time和out_time填充的日志表进行计数

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

id  name    address  total_count proper_count
1   Alpha   Chennai  4           3

Answer 1

根据需要使用以下架构设计。 User_id必须为INTEGER，并在输入和输出时间列中使用datetime数据类型。

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

Answer 2

您可以这样做吗？

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;

Answer 3

您仅比预期的查询退了一步。只需要另一个与日志表的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

SQL连接并在表中计数

3 个答案: