I have seen many examples regarding calculating the sum of fields using the fiscal year, but I can not find one that fits my needs. What I am trying to do is get just the current fiscal year totals for a field using SQL Query. The fields I have is userid, startdate, total_hours, and missed_hours. Here is the query I have so far:
SELECT
userid,
SUM(total_hours) - SUM(missed_hours) AS hours
FROM mytable
GROUP BY userid
This works great, but all I need is the total number of hours for the current fiscal year for each of the userid's. Our fiscal year runs from July to June. I only need the current fiscal year and I need it to start over again this coming July.
Assuming this is SQLServer, try:
SELECT userid, SUM(total_hours) - SUM(missed_hours) AS hours
FROM mytable
WHERE startdate >= cast( cast(year(dateadd(MM,-6,getdate())) as varchar(4)) +
'-07-01' as date ) and
startdate < cast( cast(year(dateadd(MM, 6,getdate())) as varchar(4)) +
'-07-01' as date )
GROUP BY userid
Add a where clause:
FROM mytable
WHERE startdate >= '2011-07-01'
GROUP BY userid
Or with the start of the year dynamically:
where startdate >= dateadd(yy, datepart(yy, getdate())-2001, '2000-07-01')
Maybe something like this:
SELECT
userid,
SUM(total_hours) - SUM(missed_hours) AS hours
FROM
mytable
WHERE
MONTH(startdate) BETWEEN 6 AND 7
AND YEAR(startdate) IN (2011,2012)
GROUP BY userid
For the solution, two additional information is needed
the name of the date column
the vendor type of RDBMS you are using
I supposed your date column is date_col and you are using MySQL
SELECT
userid,
SUM(total_hours) - SUM(missed_hours) AS hours
FROM mytable
WHERE date_col between STR_TO_DATE('01,7,2011','%d,%m,%Y') and STR_TO_DATE('01,7,2010','%d,%m,%Y')
GROUP BY userid
Related
I have a situation in which i have to calculate "Total" for each month of a year and i have to show total for each month of the year while maintaining datetime format of the date.i am writing the following query:
SELECT [DATE_FORMATION], sum(CO) CO, sum(Total_Members) Total_Members
from (
select convert(varchar(7),CAST([DATE_FORMATION] AS DATE),126) as [DATE_FORMATION],
count([CO_ID]) as CO,
sum(convert(int,[TOTAL_MEMBERS])) as Total_Members
from COCores
where cast([DATE_FORMATION] as date) >= Dateadd(Month, Datediff(Month, 0, DATEADD(m, -10, current_timestamp)), 0)
group by [DATE_FORMATION]
) tmp
group by [DATE_FORMATION]
It is giving the accurate answer which i required
but the problem is it generates date in "Varchar" format where as i require it in datetime format so that i can use it further.how can i get this result while keeping the date in datetime format and getting the same result.
Dates have no formats, they are binary values, just like int or decimal or binary. DATE_FORMATION should be a date type, eg date, datetime or datetime2. If the column uses the correct type, the query can be simplified to just :
select
datefromparts(Year(DATE_FORMATION),MONTH(DATE_FORMATION),1) as DATE_FORMATION,
count([CO_ID]) as CO,
sum([TOTAL_MEMBERS]) as Total_Members
from COCores
where
[DATE_FORMATION] as date) >= DATEFROMPARTS(Year(current_timestamp),1,1)
group by
YEAR(DATE_FORMATION),
MONTH(DATE_FORMATION)
DATEFROMPARTS(Year(current_timestamp),1,1) returns the first date of the current year. This is used to return only rows in the current year. The results are grouped by the Year and Month of DATE_FORMATION using the YEAR and MONTH functions. Finally, SELECT returns the first day of each month by using DATEFROMPARTS to generate a date value from the group keys
Queries like this become far easier if you use a Calendar table. A calendar table contains rows for each date eg for 50 years and extra columns for year, month, day, week number, names, etc that can be used to make reporting easier. Calendar tables are heavily indexed, allowing easy querying eg by year, quarter, semester etc.
Assuming DATE_FORMATION is a date, and a Calendar table with date, Year, Month and StartOfMonth columns, you could turn the query into :
select
Calendar.StartOfMonth,
count([CO_ID]) as CO,
sum([TOTAL_MEMBERS]) as Total_Members
from COCores inner join Calendar on DATE_FORMATION=Calendar.date
where
Calendar.Year= Year(current_timestamp)
group by
StartOfMonth
This query would be fast too as it can take advantage of any indexes on Year, Date and StartOfMonth
You can convert all dates to first day of the month like so:
DATEADD(DAY, 1, EOMONTH(DATE_FORMATION, -1))
It produces a DATE. You can use this expression in GROUP BY and in SELECT clause:
SELECT DATEADD(DAY, 1, EOMONTH(DATE_FORMATION, -1)), COUNT(CO_ID) AS CO, SUM(CONVERT(INT, TOTAL_MEMBERS)) AS Total_Members
FROM COCores
WHERE DATE_FORMATION >= -- calculate -10 months
GROUP BY DATEADD(DAY, 1, EOMONTH(DATE_FORMATION, -1))
I am trying to write an SQL statement based on the following code.
CREATE TABLE mytable (
year INTEGER,
month INTEGER,
day INTEGER,
hoursWorked INTEGER )
Assuming that each employee works multiple days over each month in a 3 year period.
I need to write an sql statement that returns the total hours worked in each month, grouped by earliest year/month first.
I tried doing this, but I don't think it is correct:
SELECT Sum(hoursWorked) FROM mytable
ORDER BY(year,month)
GROUP BY(month);
I am a little confused about how to operate the sum function in conjunction with thee GROUP BY or ORDER BY function. How does one go about doing this?
Try this:
SELECT year, month, SUM(hoursWorked)
FROM mytable
GROUP BY year, month
ORDER BY year, month
This way you will have for example:
2014 December 30
2015 January 12
2015 February 40
Fields you want to group by always have be present in SELECT part of query. And vice-versa - what you put in SELECT part, need be also in GROUP BY.
SELECT year, month, Sum(hoursWorked)as workedhours
FROM mytable
GROUP BY year,month
ORDER BY year,month;
You have to group by year and month.
Is this what you are trying to do. This will sum by Year/Month and Order by Year/Month.
Select [Year], [Month], Sum(HoursWorked) as WorkedHours
From mytable
Group By [Year], [Month]
Order by [Year], [Month]
You have to group by year and month, otherwise you will have the hours you worked on March 2014 and 2015 in one record :)
SELECT Sum(hoursWorked) as hoursWorked, year, month
FROM mytable
GROUP BY(year, month)
ORDER BY(year,month)
;
I have a Postgres table that I'm trying to analyze based on some date columns.
I'm basically trying to count the number of rows in my table that fulfill this requirement, and then group them by month and year. Instead of my query looking like this:
SELECT * FROM $TABLE WHERE date1::date <= '2012-05-31'
and date2::date > '2012-05-31';
it should be able to display this for the months available in my data so that I don't have to change the months manually every time I add new data, and so I can get everything with one query.
In the case above I'd like it to group the sum of rows which fit the criteria into the year 2012 and month 05. Similarly, if my WHERE clause looked like this:
date1::date <= '2012-06-31' and date2::date > '2012-06-31'
I'd like it to group this sum into the year 2012 and month 06.
This isn't entirely clear to me:
I'd like it to group the sum of rows
I'll interpret it this way: you want to list all rows "per month" matching the criteria:
WITH x AS (
SELECT date_trunc('month', min(date1)) AS start
,date_trunc('month', max(date2)) + interval '1 month' AS stop
FROM tbl
)
SELECT to_char(y.mon, 'YYYY-MM') AS mon, t.*
FROM (
SELECT generate_series(x.start, x.stop, '1 month') AS mon
FROM x
) y
LEFT JOIN tbl t ON t.date1::date <= y.mon
AND t.date2::date > y.mon -- why the explicit cast to date?
ORDER BY y.mon, t.date1, t.date2;
Assuming date2 >= date1.
Compute lower and upper border of time period and truncate to month (adding 1 to upper border to include the last row, too.
Use generate_series() to create the set of months in question
LEFT JOIN rows from your table with the declared criteria and sort by month.
You could also GROUP BY at this stage to calculate aggregates ..
Here is the reasoning. First, create a list of all possible dates. Then get the cumulative number of date1 up to a given date. Then get the cumulative number of date2 after the date and subtract the results. The following query does this using correlated subqueries (not my favorite construct, but handy in this case):
select thedate,
(select count(*) from t where date1::date <= d.thedate) -
(select count(*) from t where date2::date > d.thedate)
from (select distinct thedate
from ((select date1::date as thedate from t) union all
(select date2::date as thedate from t)
) d
) d
This is assuming that date2 occurs after date1. My model is start and stop dates of customers. If this isn't the case, the query might not work.
It sounds like you could benefit from the DATEPART T-SQL method. If I understand you correctly, you could do something like this:
SELECT DATEPART(year, date1) Year, DATEPART(month, date1) Month, SUM(value_col)
FROM $Table
-- WHERE CLAUSE ?
GROUP BY DATEPART(year, date1),
DATEPART(month, date1)
I have a table that contains multiple records for each day of the month, over a number of years. Can someone help me out in writing a query that will only return the last day of each month.
SQL Server (other DBMS will work the same or very similarly):
SELECT
*
FROM
YourTable
WHERE
DateField IN (
SELECT MAX(DateField)
FROM YourTable
GROUP BY MONTH(DateField), YEAR(DateField)
)
An index on DateField is helpful here.
PS: If your DateField contains time values, the above will give you the very last record of every month, not the last day's worth of records. In this case use a method to reduce a datetime to its date value before doing the comparison, for example this one.
The easiest way I could find to identify if a date field in the table is the end of the month, is simply adding one day and checking if that day is 1.
where DAY(DATEADD(day, 1, AsOfDate)) = 1
If you use that as your condition (assuming AsOfDate is the date field you are looking for), then it will only returns records where AsOfDate is the last day of the month.
Use the EOMONTH() function if it's available to you (E.g. SQL Server). It returns the last date in a month given a date.
select distinct
Date
from DateTable
Where Date = EOMONTH(Date)
Or, you can use some date math.
select distinct
Date
from DateTable
where Date = DATEADD(MONTH, DATEDIFF(MONTH, -1, Date)-1, -1)
In SQL Server, this is how I usually get to the last day of the month relative to an arbitrary point in time:
select dateadd(day,-day(dateadd(month,1,current_timestamp)) , dateadd(month,1,current_timestamp) )
In a nutshell:
From your reference point-in-time,
Add 1 month,
Then, from the resulting value, subtract its day-of-the-month in days.
Voila! You've the the last day of the month containing your reference point in time.
Getting the 1st day of the month is simpler:
select dateadd(day,-(day(current_timestamp)-1),current_timestamp)
From your reference point-in-time,
subtract (in days), 1 less than the current day-of-the-month component.
Stripping off/normalizing the extraneous time component is left as an exercise for the reader.
A simple way to get the last day of month is to get the first day of the next month and subtract 1.
This should work on Oracle DB
select distinct last_day(trunc(sysdate - rownum)) dt
from dual
connect by rownum < 430
order by 1
I did the following and it worked out great. I also wanted the Maximum Date for the Current Month. Here is what I my output is. Notice the last date for July which is 24th. I pulled it on 7/24/2017, hence the result
Year Month KPI_Date
2017 4 2017-04-28
2017 5 2017-05-31
2017 6 2017-06-30
2017 7 2017-07-24
SELECT B.Year ,
B.Month ,
MAX(DateField) KPI_Date
FROM Table A
INNER JOIN ( SELECT DISTINCT
YEAR(EOMONTH(DateField)) year ,
MONTH(EOMONTH(DateField)) month
FROM Table
) B ON YEAR(A.DateField) = B.year
AND MONTH(A.DateField) = B.Month
GROUP BY B.Year ,
B.Month
SELECT * FROM YourTableName WHERE anyfilter
AND "DATE" IN (SELECT MAX(NameofDATE_Column) FROM YourTableName WHERE
anyfilter GROUP BY
TO_CHAR(NameofDATE_Column,'MONTH'),TO_CHAR(NameofDATE_Column,'YYYY'));
Note: this answer does apply for Oracle DB
Here's how I just solved this. day_date is the date field, calendar is the table that holds the dates.
SELECT cast(datepart(year, day_date) AS VARCHAR)
+ '-'
+ cast(datepart(month, day_date) AS VARCHAR)
+ '-'
+ cast(max(DATEPART(day, day_date)) AS VARCHAR) 'DATE'
FROM calendar
GROUP BY datepart(year, day_date)
,datepart(month, day_date)
ORDER BY 1
How do I get a maximium daily value of a numerical field over a year in MS-SQL
This would query the daily maximum of value over 2008:
select
datepart(dayofyear,datecolumn)
, max(value)
from yourtable
where '2008-01-01' <= datecolumn and datecolumn < '2009-01-01'
group by datepart(dayofyear,datecolumn)
Or the daily maximum over each year:
select
datepart(year,datecolumn),
, datepart(dayofyear,datecolumn)
, max(value)
from yourtable
group by datepart(year,datecolumn), datepart(dayofyear,datecolumn)
Or the day(s) with the highest value in a year:
select
Year = datepart(year,datecolumn),
, DayOfYear = datepart(dayofyear,datecolumn)
, MaxValue = max(MaxValue)
from yourtable
inner join (
select
Year = datepart(year,datecolumn),
, MaxValue = max(value)
from yourtable
group by datepart(year,datecolumn)
) sub on
sub.Year = yourtable.datepart(year,datecolumn)
and sub.MaxValue = yourtable.value
group by
datepart(year,datecolumn),
datepart(dayofyear,datecolumn)
You didn't mention which RDBMS or SQL dialect you're using. The following will work with T-SQL (MS SQL Server). It may require some modifications for other dialects since date functions tend to change a lot between them.
SELECT
DATEPART(dy, my_date),
MAX(my_number)
FROM
My_Table
WHERE
my_date >= '2008-01-01' AND
my_date < '2009-01-01'
GROUP BY
DATEPART(dy, my_date)
The DAY function could be any function or combination of functions which gives you the days in the format that you're looking to get.
Also, if there are days with no rows at all then they will not be returned. If you need those days as well with a NULL or the highest value from the previous day then the query would need to be altered a bit.
Something like
SELECT dateadd(dd,0, datediff(dd,0,datetime)) as day, MAX(value)
FROM table GROUP BY dateadd(dd,0, datediff(dd,0,datetime)) WHERE
datetime < '2009-01-01' AND datetime > '2007-12-31'
Assuming datetime is your date column, dateadd(dd,0, datediff(dd,0,datetime)) will extract only the date part, and then you can group by that value to get a maximum daily value. There might be a prettier way to get only the date part though.
You can also use the between construct to avoid the less than and greater than.
Group on the date, use the max delegate to get the highest value for each date, sort on the value, and get the first record.
Example:
select top 1 theDate, max(theValue)
from TheTable
group by theDate
order by max(theValue) desc
(The date field needs to only contain a date for this grouping to work, i.e. the time component has to be zero.)
If you need to limit the query for a specific year, use a starting and ending date in a where claues:
select top 1 theDate, max(theValue)
from TheTable
where theDate between '2008-01-01' and '2008-12-13'
group by theDate
order by max(theValue) desc