【sql练习】计算时间差的秒数(考虑非工作日)

例如：审批开始时间是2020-02-07 20：09：57，结束时间是2020-02-10 22：30：30，请问审批共耗时多少秒？(扣除所有的法定节假日和周六日)

在mysql中进行测试

数据准备

节假日表数据

DROP TABLE IF EXISTS holiday_table;
CREATE TABLE holiday_table (
  DATE_ID date,
  DAY_WORK int,
  isWork varchar(50)
) ;

插入数据

INSERT INTO holiday_table VALUES ('2018-03-30 00:00:00',1,'工作日');
INSERT INTO holiday_table VALUES ('2018-03-31 00:00:00',2,'放假'  );
INSERT INTO holiday_table VALUES ('2018-04-01 00:00:00',2,'放假'  );
INSERT INTO holiday_table VALUES ('2018-04-02 00:00:00',1,'工作日');
INSERT INTO holiday_table VALUES ('2018-04-03 00:00:00',1,'工作日');
INSERT INTO holiday_table VALUES ('2018-04-04 00:00:00',1,'工作日');
INSERT INTO holiday_table VALUES ('2018-04-05 00:00:00',2,'放假'  );
INSERT INTO holiday_table VALUES ('2018-04-06 00:00:00',2,'放假'  );
INSERT INTO holiday_table VALUES ('2018-04-07 00:00:00',2,'放假'  );
INSERT INTO holiday_table VALUES ('2018-04-08 00:00:00',1,'工作日');
INSERT INTO holiday_table VALUES ('2018-04-09 00:00:00',1,'工作日');
INSERT INTO holiday_table VALUES ('2018-04-10 00:00:00',1,'工作日');
INSERT INTO holiday_table VALUES ('2018-04-11 00:00:00',1,'工作日');

DATE_ID	DAY_WORK	isWork
2018-03-30 00:00:00	1	工作日
2018-03-31 00:00:00	2	放假
2018-04-01 00:00:00	2	放假
2018-04-02 00:00:00	1	工作日
2018-04-03 00:00:00	1	工作日
2018-04-04 00:00:00	1	工作日
2018-04-05 00:00:00	2	放假
2018-04-06 00:00:00	2	放假
2018-04-07 00:00:00	2	放假
2018-04-08 00:00:00	1	工作日
2018-04-09 00:00:00	1	工作日
2018-04-10 00:00:00	1	工作日

审批时间表数据

app_info 表存放审批信息的数据

DROP TABLE IF EXISTS app_info ;
CREATE TABLE app_info  (
  ID int,
  start_date TIMESTAMP,
  end_date TIMESTAMP
) ;

插入三条数据

INSERT INTO app_info VALUES (1,'2018-03-30 12:10:00','2018-04-02 17:10:00');
INSERT INTO app_info VALUES (2,'2018-04-02 11:00:05','2018-04-03 10:40:05');
INSERT INTO app_info VALUES (3,'2018-04-03 17:20:00','2018-04-11 11:10:05');

方法1

将日期表中的数据去关联假日表的数据，把非工作日的去除，剩下的条数就是工作日的天数。

1. 关联表取时间

SELECT
	A.id,
	A.start_date,
	A.end_date,
	B.DATE_ID AS Interval_date
from app_info A
join holiday_table B
on B.DATE_ID BETWEEN date(A.start_date) AND date(A.end_date)
	and B.DAY_WORK=1

id	start_date	end_date	Interval_date
1	2018/3/30 12:10	2018/4/2 17:10	2018/3/30
1	2018/3/30 12:10	2018/4/2 17:10	2018/4/2
2	2018/4/2 11:00	2018/4/3 10:40	2018/4/2
2	2018/4/2 11:00	2018/4/3 10:40	2018/4/3
3	2018/4/3 17:20	2018/4/11 11:10	2018/4/3
3	2018/4/3 17:20	2018/4/11 11:10	2018/4/4
3	2018/4/3 17:20	2018/4/11 11:10	2018/4/8
3	2018/4/3 17:20	2018/4/11 11:10	2018/4/9
3	2018/4/3 17:20	2018/4/11 11:10	2018/4/10
3	2018/4/3 17:20	2018/4/11 11:10	2018/4/11

2. 统计天数

SELECT
	A.id,
	A.start_date,
	A.end_date,
	count(1) AS Interval_num
from app_info A
join holiday_table B
on B.DATE_ID BETWEEN date(A.start_date) AND date(A.end_date)
	and B.DAY_WORK=1
group by 
		A.id,
	A.start_date,
	A.end_date

id	start_date	end_date	Interval_num
1	2018/3/30 12:10	2018/4/2 17:10	2
2	2018/4/2 11:00	2018/4/3 10:40	2
3	2018/4/3 17:20	2018/4/11 11:10	5

3.时间间隔的时间

所得的相差秒数应为时间相差的秒数减去多算的工作日的天数对应的描述（即天数*24*60*60）

select
	id,
	start_date,
	end_date,
	TIMESTAMPDIFF(SECOND,start_date,end_date)-(DATEDIFF(end_date,start_date)+1-Interval_num)*24*60*60 AS Interval_sec
from
	(SELECT
		A.id,
		A.start_date,
		A.end_date,
		count(1) as Interval_num
	from app_info A
	join holiday_table B
	on B.DATE_ID BETWEEN date(A.start_date) AND date(A.end_date)
		and B.DAY_WORK=1
	group by 
			A.id,
		A.start_date,
		A.end_date) t

最终结果

id	start_date	end_date	Interval_sec
1	2018-03-30 12:10:00	2018-04-02 17:10:00	104400
2	2018-04-02 11:00:05	2018-04-03 10:40:05	85200
3	2018-04-03 17:20:00	2018-04-11 11:10:05	409805

4. 如果提交审批是休息日

上述代码对于审批日是休息日，计算会有差错

插入例子

INSERT INTO app_info VALUES (4,'2018-04-01 09:20:00','2018-04-02 11:10:05');
INSERT INTO app_info VALUES (5,'2018-04-04 20:20:00','2018-04-05 11:10:05');

id为4不考虑休息日应为 39600+600+5=40205秒，用之前代码的得出的结果为6605秒结果存在误差

id为5不考虑休息日应为10800+2400=13200秒

对代码进行进一步的处理

SELECT
	id,
	start_date,
	end_date,
	TIMESTAMPDIFF(SECOND,start_date_new,end_date_new)-(DATEDIFF(end_date_new,start_date_new)+1-Interval_num_new)*24*60*60 AS Interval_sec	
FROM
	(SELECT
		id,
		start_date,
		end_date,
		case when date(start_date) = first_workday then start_date else date_format(first_workday,'%Y-%m-%d %H:%i:%s') end AS start_date_new ,
		case when date(end_date) = last_workday then end_date else  date_format(last_workday,'%Y-%m-%d %H:%i:%s') end AS end_date_new,
		case when date(end_date) <> last_workday then Interval_num + 1 else Interval_num end AS Interval_num_new
	FROM
		(SELECT
			A.id,
			A.start_date,
			A.end_date,
			count(1) AS Interval_num,
			MIN(DATE_ID) AS first_workday,
			MAX(DATE_ID) AS last_workday
		FROM app_info A
		JOIN holiday_table B
		ON B.DATE_ID BETWEEN date(A.start_date) AND date(A.end_date) 
			AND B.DAY_WORK=1
		GROUP BY 
			A.id,
			A.start_date,
			A.end_date
		) t1
	) t2

id为4的结果为40205

id	start_date	end_date	Interval_sec
1	2018-03-30 12:10:00	2018-04-02 17:10:00	104400
2	2018-04-02 11:00:05	2018-04-03 10:40:05	85200
3	2018-04-03 17:20:00	2018-04-11 11:10:05	409805
4	2018-04-01 09:20:00	2018-04-02 11:10:05	40205
5	2018/4/4 20:20	2018/4/5 11:10	13200

方法2

自定义函数

create  function getWorkDay
( 
    beginday date, 
    endday date
) 
returns int
begin
	declare Interval_num int;
    SELECT COUNT(1) from holiday_table WHERE DATE_ID BETWEEN beginday AND endday AND DAY_WORK=1 into Interval_num;
	return Interval_num;
end;

测试

select getWorkDay('2018-03-30 12:10:00','2018-04-02 17:10:00');

结果为2天，还需将天换算成对应的时分秒再进行其他的转换，暂时未写完此方法。

Sql Boy

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