SUM分钟SQL服务器

时间:2010-04-12 10:00:09

标签: sql sql-server time sum

嗨,伙计们, 我的问题是这样的: 我有这样的列:PHONE_NR,TIME(时间字段),意思是呼叫电话号码和呼叫持续时间。我需要分组电话和总和分钟。 Filds看起来像这样:

 nr          time 
726028xxx 00:07:07 
735560css 00:07:37 
726028xxx 00:07:55 

3 个答案:

答案 0 :(得分:5)

SELECT PHONE_NR, SUM(DATEPART(minute, TIME)) FROM [table] GROUP BY PHONE_NR;

据我所知,这应该适用于SQL Server DATETIME和2008 TIME数据类型。

示例:

DECLARE @tbl TABLE
(
    phone_nr varchar(10),
    call_time TIME
);

INSERT INTO @tbl VALUES ('726028xxx', '00:07:07');
INSERT INTO @tbl VALUES ('735560css', '00:07:37');
INSERT INTO @tbl VALUES ('726028xxx', '00:07:55');

SELECT phone_nr, SUM(DATEPART(minute, call_time)) AS total_call_time FROM @tbl GROUP BY phone_nr;

结果:

phone_nr  | minutes
726028xxx | 14
735560css | 7

修改

此版本与上述版本相同,不同之处在于它还考虑了秒数和小时数(例如1小时7分07秒= 67.117分钟),因此它更准确。

SELECT 
    phone_nr, 
    SUM(CAST(DATEPART(second, call_time) + (DATEPART(minute, call_time) * 60) + (DATEPART(hour, call_time) * 3600) AS decimal) / 60) AS total_call_time
FROM 
    @tbl
GROUP BY
    phone_nr;

如果第一条记录为01:07:07,则会产生以下结果。

phone_nr  | minutes
726028xxx | 75.033332
735560css | 7.616666

答案 1 :(得分:1)

您可以使用DATEDIFF获取自“零”时间以来已经过的总分钟数

SELECT
 [PHONE_NR],
 SUM(DATEDIFF(minute, '00:00:00', [TIME]))
FROM
 [YourTable]
GROUP BY
 [PHONE_NR]

答案 2 :(得分:0)

declare @v1 varchar(50)
declare @v2 varchar(50)
set @v1='03:10'
set @v2='01:50'

--select cast(left(v1,2) as int) + cast(left(v2,2)as int)

select 
cast(
cast(left(@v1,2) as int) + cast(left(@v2,2)as int)
+(cast(right(@v1,2) as int) + cast(right(@v2,2) as int))/60 as varchar)

+':'+cast(( cast(right(@v1,2) as int) + cast(right(@v2,2) as int))%60 as varchar)