sql - 如何以特定格式对一组周期时间求和

Question

如果我有这样的结果集：

Work_hour(hh:mm)
10:24
12:59
06:28

---

where Work_hour is of type varchar

---

如何以相同的格式汇总这些小时和分钟？

Answer 1

SELECT CAST(FLOOR(TMP1.MINS/60) AS VARCHAR) + ':' + CAST((TMP1.MINS % 60) AS VARCHAR) FROM (
    SELECT SUM (CAST(LEFT(time_column, 2) AS INT) * 60 + CAST(RIGHT(time_column, 2) AS INT)) as MINS FROM table1
) AS TMP1

其中time_column是表中的列，table1是表的名称。例子：

create table table1 (
    time_column varchar(10)
);
insert into table1 (time_column) values ('12:20'), ('10:40'), ('15:50');

结果：38 :50

Answer 2

试试这个——

询问：

DECLARE @temp TABLE
(
      work_hour CHAR(5)
)

INSERT INTO @temp (work_hour)
VALUES 
    ('10:24'),
    ('12:59'),
    ('06:28')

;WITH cte AS
(
    SELECT mn = SUM(DATEDIFF(MINUTE, '19000101', CAST('19000101 ' + work_hour AS DATETIME)))
    FROM @temp
)
SELECT CAST(FLOOR(mn / 60) AS VARCHAR(5)) + ':' + CAST(mn % 60 AS VARCHAR(2))
FROM cte

输出：

hm
--------
29:51

更新 2：

DECLARE @temp TABLE
(
      transtime_out DATETIME
    , transtime_in DATETIME
)

INSERT INTO @temp (transtime_out, transtime_in)
VALUES 
    ('2013-05-19 16:40:53.000', '2013-05-19 08:58:07.000'), 
    ('2013-05-19 16:40:53.000', '2013-05-19 08:58:07.000')

SELECT diff = LEFT(CONVERT(VARCHAR(10), CAST(SUM(CAST(a.transtime_out - a.transtime_in AS FLOAT)) AS DATETIME), 108), 5)
FROM @temp a