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?
答案 0 :(得分:3)
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
time_column 是表格中的列, table1 是表格的名称。例如:
create table table1 (
time_column varchar(10)
);
insert into table1 (time_column) values ('12:20'), ('10:40'), ('15:50');
结果: 38:50
答案 1 :(得分:2)
试试这个 -
<强>查询:强>
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
<强>输出:强>
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