How to calculate sum of 24 hours data of a month

Question

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The table structure has a timestamp(datetime), and a value, similar to this..

SQL

-----timestamp-----             --value--
2009-08-13 08:12:00              200.40
2009-08-13 07:47:00               70.00
2009-08-13 07:33:00               84.00
2009-08-13 07:12:00               350.00
2009-08-12 06:12:00                10.70
2009-08-12 05:12:00                50.86

I want output as following format for a month:

SQL

------------Time-------------                 -----Sum_Value------

28-AUG-2014 06 AM - 29-AUG-2014 06 AM	       553.212
29-AUG-2014 06 AM - 30-AUG-2014 06 AM	       553.212
30-AUG-2014 06 AM - 31-AUG-2014 06 AM	      1053.212

please help me to write a sql query

Posted 12-Sep-14 1:22am

KManishS

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Gihan Liyanage 12-Sep-14 8:24am

This query returns the sum for the day. I have tested for you. I dont have time now(got to go), Please try and see whether you can improve the query.

select CAST([timestamp] AS DATE), SUM(value) as 'Sum FOr The Day'
from tblExample
group by CAST([timestamp] AS DATE)

2 solutions

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Mike Meinz · Answer 1 · 2014-09-12T04:11:00

I don't have access to an Oracle database. I did this solution using SQL Server 2012. Maybe you can use it to help you with your Oracle solution.

The requirement that a "day" for your query is from 6:00AM on one day until 6:00AM on the next day presents a bit of a challenge. To solve that, I first used a Common Table Expression to sum the values by a date. I use the CASE statement to compute the date. If the time is less than 6AM, then the date is set to the day before. After creating the CTE, I select the results and format the "Date" column to your specification including the dashes in the date.

Create the table

SQL

SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
CREATE TABLE [dbo].[tblLog](
    [TimeStamp] [datetime] not NULL,
    [Value] [decimal](18, 3) not NULL
) ON [PRIMARY]
GO

Put some data into the table

SQL

insert into tblLog values('2014-09-01 07:00:00',100.1);
insert into tblLog values('2014-09-01 08:00:00',100.1);
insert into tblLog values('2014-09-02 07:00:00',100.1);
insert into tblLog values('2014-09-02 08:00:00',100.1);
insert into tblLog values('2014-09-03 07:00:00',100.1);
insert into tblLog values('2014-09-03 08:00:00',100.1);
insert into tblLog values('2014-09-01 05:00:00',100.1);
insert into tblLog values('2014-09-01 06:00:00',100.1);
insert into tblLog values('2014-09-02 05:00:00',100.1);
insert into tblLog values('2014-09-02 06:00:00',100.1);
insert into tblLog values('2014-09-03 05:00:00',100.1);
insert into tblLog values('2014-09-03 06:00:00',100.1);

Execute the query

SQL

With Sums (LogDate,Sum)
AS
(
SELECT
 CASE
 WHEN CAST(timestamp AS time)<'06:00 am' THEN DATEADD(day,-1,CAST(timestamp AS date)) ELSE CAST(timestamp AS date)
 END AS LogDate,
 SUM(value) AS Sum
 FROM tblLog
 GROUP BY
 CASE
 WHEN CAST(timestamp AS time)<'06:00 am' THEN DATEADD(day,-1,CAST(timestamp AS date)) ELSE CAST(timestamp AS date)
 END
 )
 SELECT CONCAT(REPLACE(CONVERT(NVARCHAR,LogDate,113),' ','-'),' 06 AM - ',REPLACE(CONVERT(NVARCHAR,DATEADD(day,1,LogDate),113),' ','-'),' 06 AM') AS LogDate,Sum from Sums
 ORDER BY CAST(LogDate AS DATE)

The results

LogDate                                 Sum
31-Aug-2014 06 AM - 01-Sep-2014 06 AM	100.100
01-Sep-2014 06 AM - 02-Sep-2014 06 AM	400.400
02-Sep-2014 06 AM - 03-Sep-2014 06 AM	400.400
03-Sep-2014 06 AM - 04-Sep-2014 06 AM	300.300

pradiprenushe · Answer 2 · 2014-09-12T02:15:00

Solution 1

Please chechk this

SQL

select trunc(datecolumn) - 1 || 'AM - ' || trunc(datecolumn) || 'AM' as RecordDate,sum(valuecolumn) as AverageValue  from Yourtable group by trunc(datecolumn);

Posted 12-Sep-14 2:15am

pradiprenushe

Updated 12-Sep-14 2:41am

v2