I am trying to group the number of hours that employees worked for the last 4 weeks but I want to group them on a weekly basis. For example:
WEEK HOURS
Feb 24 to March 2 55
March 3 to March 9 40
March 10 to March 16 48
March 17 to March 23 37
This is what I have so far, please help. thanks
SET DATEFIRST 1
SELECT CAST(MIN( [DT]) AS VARCHAR(20))+' TO '+CAST (MAX([DT]) AS VARCHAR(20)) AS DATE,
SUM(HOURS) AS NUM_HRS
FROM MyTable
GROUP BY DATEPART(WEEK,[DT])
HAVING COUNT(DISTINCT[DT])=7
Create a Calendar auxilliary table, with Year, Month, Week, Date columns (you can also add holidays and other interesting stuff to it, it has many potential uses) and populate it for the period of interest.
After that, it's as easy as this:
SELECT sum(hours), cast(min(date) as varchar), cast(max(date) as varchar)
FROM Calendar c
LEFT OUTER JOIN MyTable h on h.Date = c.date
GROUP BY year, week
ORDER BY year, week
SET DATEFIRST 1
SELECT DATEPART(WEEK,DT) AS WEEK,
SUM(HOURS) AS NUM_HRS
FROM MyTable
WHERE DT >= DATEADD(WEEK, -4, GetDate()),
GROUP BY DATEPART(WEEK,[DT])
Try something like
SELECT
DATEADD(DD,
CONVERT(INT, (DATEDIFF(DD, '1/1/1900', t.DT)/7)) * 7,
'1/1/1900') [WeekBeginDate],
DATEADD(DD,
(CONVERT(INT, (DATEDIFF(DD, '1/1/1900', t.DT)/7)) * 7) + 6,
'1/1/1900') [WeekEndDate],
SUM(HOURS) AS NUM_HRS
FROM MyTable t
GROUP BY CONVERT(INT, DATEDIFF(DD, '1/1/1900', t.DT)/7)
Though this is the brute force trick, I think in your case it will work.
EDIT : Modified the query a little bit, the error was caused because of the order in which DATEDIFF calculates the difference.
Also here is a SQL FIDDLE with a working example.
EDIT 2 : Updated the Fiddle with the Date Format. To customize the date format, this article would help.
Related
I am trying to update certain fields for employees whose date of joining falls in between 10 Jun and 31 Dec, irrespective of the year. I am trying using 'Between' Operator but it requires year to be included in the dates. Is there a way to generalise it in order to consider Day and Month excluding the Year?
Use the DatePart function - replace thedate with your column, and thetable with the column.
Something like this:
select datepart(MONTH, thedate), datepart(DAY, thedate),*
from thetable
where datepart(MONTH, thedate) between 6 and 12
and datepart(DAY, thedate) between 10 and 31
You may try this:
WITH Emp AS (
SELECT *, DATEPART(MONTH, JoinDate) AS MonthJoin, DATEPART(DAY, JoinDate) AS DayJoin
FROM Employees)
SELECT *
FROM Emp
WHERE (MonthJoin > 1 AND MonthJoin < 12)
OR (MonthJoin = 1 AND DayJoin >= 10)
OR (MonthJoin = 12 AND DayJoin <= 31)
Where Employees is your table and JoinDate is your date of joining in this table
I would like to create an SQL query that gets the count of employees that were employed as of the 15th of each month of the last 5 years.
This query gets me a single month:
SELECT
SUM(X.CountOfEmployees)
FROM (SELECT
COUNT(CNCEmployeeID) AS CountOfEmployees
FROM dbo.CNCEmployees
GROUP BY CNCEmployeeStartDate,
CNCEmployeeDateLeft
HAVING (CNCEmployeeStartDate < CONVERT(datetime, '2016-07-15 00:00:00', 102))
AND ((CNCEmployeeDateLeft > CONVERT(datetime, '2016-07-15 00:00:00', 102))
OR (CNCEmployeeDateLeft IS NULL))) AS X
What I am looking for would output:
Jan 2016 - 32
Feb 2016 - 33
Mar 2016 - 33
etc. for each month that we have data.
I know how to create a Query and at least make it quick to change the dates by hand by adding a variable and changing that over and over (in fact I will probably do that to get the report done today for the last 12 months). I believe that there is a better way to do this in one step without the need to manually go through each month.
One method generates for the 60 months and use that in the join:
with dates as (
select cast(dateadd(day, 16 - day(getdate()), getdate()) as date) as thedate, 1 as num
union all
select dateadd(month, -1, thedate), num + 1
from dates
where num <= 60
)
select d.thedate, count(e.CNCEmployeeStartDate)
from dates d left join
dbo.CNCEmployees e
on e.CNCEmployeeStartDate <= d.thedate and
(e.CNCEmployeeDateLeft >= d.thedate or e.CNCEmployeeDateLeft is null)
group by d.thedate;
This is not the most efficient method, but if you have a few hundred or thousand employees it should be fine in terms of performance.
I'm not really good when it comes to database...
I'm wondering if it is possible to get the weeks of a certain month and year..
For example: 1 (January) = month and 2016 = year
Desired result will be:
week 1
week 2
week 3
week 4
week 5
This is what I have tried so far...
declare #date datetime = '01/01/2016'
select datepart(day, datediff(day, 0, #date) / 7 * 7) / 7 + 1
This only returns the total of the weeks which is 5.
declare #MonthStart datetime
-- Find first day of current month
set #MonthStart = dateadd(mm,datediff(mm,0,getdate()),0)
select
Week,
WeekStart = dateadd(dd,(Week-1)*7,#MonthStart)
from
( -- Week numbers
select Week = 1 union all select 2 union all
select 3 union all select 4 union all select 5
) a
where
-- Necessary to limit to 4 weeks for Feb in non-leap year
datepart(mm,dateadd(dd,(Week-1)*7,#MonthStart)) =
datepart(mm,#MonthStart)
Got the answer in the link: http://www.sqlservercentral.com/Forums/Topic1328013-391-1.aspx
Here is one way to approach this:
A month has a minimum of 29 or more days, and a max of 31 or less. Meaning there are almost always 5 weeks a month, with the exception of a non-leap year's feburary, and in those cases, 4 weeks a month.
You can refer to this to find out which years are "leap".
Check for leap year
Hope this helps!
The following code will allow you to select a start and end date, and output one row per week, with numbered weeks, between those dates:
declare #start date = '1/1/2016'
declare #end date = '5/1/2016'
;with cte as (select #start date
, datename(month, #start) as month
union all
select dateadd(dd, 7, date)
, datename(month, dateadd(dd, 7, date))
from CTE
where date <= #end
)
select *, 'week '
+ cast(row_number() over (partition by month order by date) as varchar(1))
from CTE
order by date
I am looking for the number of Mon,Tues, Wed, Thur, Fri, Sat, Sun in the past 30 days. Can I select the last 30 days date and day of week without an actual database table? Something like
SELECT --everything between
convert(date,GETDATE()), DATENAME(DW, GETDATE())
--and
convert(date,GETDATE() - 30), DATENAME(DW, GETDATE())
You can use a recursive CTE:
;WITH CTE AS
(
SELECT convert(date,GETDATE()) sDate, DATENAME(DW, GETDATE()) sDayofWeek
UNION ALL
SELECT DATEADD(DAY,-1,sDate), DATENAME(DW, DATEADD(DAY,-1,sDate))
FROM CTE
WHERE sDate > GETDATE()-29
)
SELECT * FROM CTE
WITH cteCount AS (
SELECT DATENAME(dw, GETDATE()) dw, 1 ix
UNION ALL
SELECT DATENAME(dw, DATEADD(d, -ix, GETDATE())), ix+1 FROM cteCount WHERE ix<30
)
SELECT dw, COUNT(1) cnt FROM cteCount GROUP BY dw
A couple solutions:
SELECT ... From ... WHERE date > DATEADD(year, -1, GETDATE())
Also, I think this statement will work with MySQL:
select date_sub(now(),interval 30 day)as Datebefore30days;
Well, there are a couple of ways to do it.
You could fill a temp table, using a loop and INSERT statements, and then select the contents of the table. You could create a table-valued UDF to do this, in fact.
You could also create 30 SELECT statements, and UNION them all together. But, frankly, I think you're better off with option 1.
ETA: Thinking about it, if all you want is the number of each day of the week in the past 30 days, you can probably do that just with some math, without returning 30 records.
There are 4 instances of each day of the week in any 30 day period, plus 2 extra days. So all you really need is to know what day of the week the first day in your period is, and the second day. Those days of the week have 5 instances.
I'm pretty lazy and just load a temp table and then do a group by select on that temp table
DECLARE #tmpDates TABLE (calDate DATETIME)
DECLARE #beginDate DATETIME
SET #beginDate = DATEADD(day,-30,GETDATE())
WHILE #beginDate < GETDATE()
BEGIN
INSERT INTO #tmpDates ([calDate]) VALUES (#beginDate)
SET #beginDate = DATEADD(DAY,1,#beginDate)
END
SELECT DATEPART(dw,[calDate]) AS [weekDay], COUNT(1) AS [dayCount]
FROM #tmpDates
GROUP BY DATEPART(dw,[calDate])
Number of times each day of the week got hit in the last 30 days:
SELECT DATENAME(dw,GETDATE())+' 5 times' as results
UNION ALL
SELECT DATENAME(dw,DATEADD(day,-1,GETDATE()))+' 5 times'
UNION ALL
SELECT DATENAME(dw,DATEADD(day,-2,GETDATE()))+' 4 times'
UNION ALL
SELECT DATENAME(dw,DATEADD(day,-3,GETDATE()))+' 4 times'
UNION ALL
SELECT DATENAME(dw,DATEADD(day,-4,GETDATE()))+' 4 times'
UNION ALL
SELECT DATENAME(dw,DATEADD(day,-5,GETDATE()))+' 4 times'
UNION ALL
SELECT DATENAME(dw,DATEADD(day,-6,GETDATE()))+' 4 times'
This really is about dividing 30 by 7
This gives me
results
Thursday 5 times
Wednesday 5 times
Tuesday 4 times
Monday 4 times
Sunday 4 times
Saturday 4 times
Friday 4 times
For example,
the start date = '20100530' and
the end date = '20100602'
How can I write a SQL to display the below result?
month: may, 2 days
month: June, 2 days
Use a recursive CTE to generate all of the dates between the start and end, and then do a simple group and count (caution, not tested, but should be close if not exactly right):
with dates (the_date) as (
select #start_date
UNION ALL
select dateadd(dd, 1, the_date) from dates where the_date <= #end_date
)
select
datepart(mm, the_date) month,
count(*) num_days
from
dates
group by
datepart(mm, the_date)
TBH, you really need to provide more schema and information about the source data. However, given what little we know, you should be able to write:
Select DateName('month', [Date]) As Month
, Cast(DateDiff(d, #StartDate, #EndDate) As varchar(10) - 1) + ' days'
From Table
Where [Date] Between #StartDate And #EndDate
What we'd need to know to refine our solutions is exactly how 2 days is supposed to be calculated. Is it the days between the start and end date? Is it the day of the second parameter?