Consider a table with multiple rows with dates:
Date
------
5/9/2022
5/27/2022
4/18/2022
6/2/2022
7/1/2022
6/6/2022
7/8/2022
7/6/2022
7/22/2022
7/19/2022
7/11/2022
current query is returning all rows DATE <= GETDATE()+1
so it would be returning all dates <=7/9/2022 when it actually needs to be <=7/11/22. This being this column never has a date that falls on a weekend or holiday. So on days like Fridays, the query should be trying to pull all rows <= to that coming Monday (the next date in the column or work day). This query has multiple where conditions as well.
WHERE (_Header.CODE IN ('10', '15')) AND (_Header.LOCATION = '89') AND (_Header.OR_NO > '0') AND (_Header.DATE <= GETDATE()+1)
This would give you the next day in a date field after today. It is a simple common table expression you could use in your query. You would just request all dates be <= to the date that is retrieved from the cte.
with cte(nextDay)
AS (SELECT TOP 1 cal_date FROM calendar
WHERE cal_date > GETDATE()
ORDER BY cal_date)
SELECT * FROM cte
If you have table like this:
Name
Data type
UserID
INT
StartDate
DATETIME
EndDate
DATETIME
With data like this:
UserID
StartDate
EndDate
21
2021-01-02 00:00:00
2021-01-02 23:59:59
21
2021-01-03 00:00:00
2021-01-04 15:42:00
24
2021-01-02 00:00:00
2021-01-06 23:59:59
And you want to calculate number of users that is represented on each day in a week with a result like this:
Year
Week
NumberOfTimes
2021
1
8
2021
2
10
2021
3
4
Basically I want to to a Select like this:
SELECT YEAR(dateColumn) AS yearname, WEEK(dateColumn)as week name, COUNT(somecolumen)
GROUP BY YEAR(dateColumn) WEEK(dateColumn)
The problem I have is the start and end date if the date goes over several days I want it to counted each day. Preferably I don't want the same user counted twice each day. There are millions of rows that are constantly being deleted and added so speed is key.
The database is MS-SQL 2019
I would suggest a recursive CTE:
with cte as (
select userid, startdate, enddate
from t
union all
select userid, startdate,
enddate
from cte
where startdate < enddate and
week(startdate) <> week(enddate)
)
select year(startdate), week(startdate), count(*)
from cte
group by year(startdate), week(startdate)
option (maxrecursion 0);
The CTE expands the data by adding 7 days to each row. This should be one day per week.
There is a little logic in the second part to handle the situation where the enddate ends in the same week as the last start date. The above solution assumes that the dates are all in the same year -- which seems quite reasonable given the sample data. There are other ways to prevent this problem.
You need to cross-join each row with the relevant dates.
Create a calendar table with columns of years and weeks, include a start and end date of the week. See here for an example of how to create one, and make sure you index those columns.
Then you can cross-join like this
SELECT
YEAR(dateColumn) AS yearname,
WEEK(dateColumn)as weekname,
COUNT(somecolumen)
FROM Table t
JOIN CalendarWeek c ON c.StartDate >= t.StartDate AND c.EndDate <= t.EndDate
GROUP BY YEAR(dateColumn), WEEK(dateColumn)
Using SQL I need to return a smooth set of results (i.e. one per day) from a dataset that contains 0-N records per day.
The result per day should be the most recent previous value even if that is not from the same day. For example:
Starting data:
Date: Time: Value
19/3/2014 10:01 5
19/3/2014 11:08 3
19/3/2014 17:19 6
20/3/2014 09:11 4
22/3/2014 14:01 5
Required output:
Date: Value
19/3/2014 6
20/3/2014 4
21/3/2014 4
22/3/2014 5
First you need to complete the date range and fill in the missing dates (21/3/2014 in you example). This can be done by either joining a calendar table if you have one, or by using a recursive common table expression to generate the complete sequence on the fly.
When you have the complete sequence of dates finding the max value for the date, or from the latest previous non-null row becomes easy. In this query I use a correlated subquery to do it.
with cte as (
select min(date) date, max(date) max_date from your_table
union all
select dateadd(day, 1, date) date, max_date
from cte
where date < max_date
)
select
c.date,
(
select top 1 max(value) from your_table
where date <= c.date group by date order by date desc
) value
from cte c
order by c.date;
May be this works but try and let me know
select date, value from test where (time,date) in (select max(time),date from test group by date);
I have exchange rate table in which there are multiple date wise records with exchange rate.
Date Rate
17/05/2012 5
23/05/2012 6
27/05/2012 7
Now I want rate while passing any date like if, I pass 20/05/2012 then rate 5 should return because 20/05/2012 elapse in date range 17 and 23 may 2012.
Assuming you have correct datatypes (that is, not varchar to store date values...)
SELECT TOP 1
Rate
FROM
MyTable
WHERE
DateColumn <= '20120520'
ORDER BY
DateColumn DESC
Something like this will work:
select Rate from tablename where Date in (
select max(Date) as Date
from tablename
where Date <= convert(datetime, '20/05/2012', 103)
)
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