I'm trying to display a result set based on a min date value and today's date but can't seem to make it work. It's essentially a date sensitive price list.
Example Data
ID Title Value ExpireDate
1 Fred 10 2019-03-01
2 Barney 15 2019-03-01
3 Fred2 20 2019-06-01
4 Barney2 25 2019-06-01
5 Fred3 30 2019-07-01
6 Barney3 55 2019-07-01
Required Results:
Display records based on minimum date > GetDate()
3 Fred2 20 2019-06-01
4 Barney2 25 2019-06-01
Any assistance would be great - thank you.
Use where clause to filter all future rows and row_number() to find the first row per group:
SELECT *
FROM (
SELECT *, ROW_NUMBER() OVER (PARTITION BY Title ORDER BY ExpireDate) AS rn
FROM t
WHERE ExpireDate >= CAST(CURRENT_TIMESTAMP AS DATE)
) AS x
WHERE rn = 1
Based on your revised question, you can simply do this:
SELECT TOP 1 WITH TIES *
FROM t
WHERE ExpireDate >= CAST(CURRENT_TIMESTAMP AS DATE)
ORDER BY ExpireDate
Related
I have a table with the following entries
CustomeID
TransDate
WorkID
1
2012-12-01
12
1
2012-12-03
45
1
2013-01-21
3
2
2012-12-23
11
3
2013-01-04
13
3
2013-12-24
16
4
2014-01-02
2
I am trying get the data between two dates and the required date values are minimum and maximum values of the column. I am able to get the desired output when I hard code the values.
SELECT *
FROM dbo.MyTable
WHERE TransDate >= '2012-12-01' AND TransDate <= '2014-01-02'
I am aware the if I remove the where clause it will solve all the issues, But my actual query is much complex and has other conditions. The only way is to get maximum date values and minimum date value from the table and pass that reference to it.
I tried the below step but that does not work and throws the below error.
SELECT *
FROM dbo.MyTable
WHERE TransDate >= '2012-12-01' AND TransDate <= MAX(TransDate)
Error
An aggregate may not appear in the WHERE clause unless it is in a subquery contained in a HAVING clause or a select list, and the column being aggregated is an outer reference.
Expected Output:
CustomeID
TransDate
WorkID
1
2012-12-01
12
1
2012-12-03
45
1
2013-01-21
3
2
2012-12-23
11
3
2013-01-04
13
3
2013-12-24
16
4
2014-01-02
2
Use a scalar subquery to find the maximum date across the whole table:
SELECT *
FROM dbo.MyTable
WHERE TransDate >= '2012-12-01' AND
TransDate < (SELECT DATEADD(DAY, 1, MAX(TransDate)) FROM dbo.MyTable);
Note that I am using a strict inequality (less than) in the WHERE clause against one day later than the max date. This will include all days which fall on or earlier than the maximum date.
You can also declare variables and then use them as given below:
DECLARE #minDate DATE = (SELECT MIN(TransDate) FROM Customer);
DECLARE #maxDate DATE = (SELECT MAX(TransDate) FROM Customer);
SELECT * FROM dbo.MyTable WHERE TransDate >= #minDate AND
TransDate <= #maxDate
I have a table for which I have to perform a rather complex filter: first a filter by date is applied, but then records from the previous and next days should be included if their time difference does not exceed 8 hours compared to its prev or next record (depending if the date is less or greater than filter date).
For those adjacent days the selection should stop at the first record that does not satisfy this condition.
This is how my raw data looks like:
Id
Desc
EntryDate
1
Event type 1
2021-03-12 21:55:00.000
2
Event type 1
2021-03-12 01:10:00.000
3
Event type 1
2021-03-11 20:17:00.000
4
Event type 1
2021-03-11 05:04:00.000
5
Event type 1
2021-03-10 23:58:00.000
6
Event type 1
2021-03-10 11:01:00.000
7
Event type 1
2021-03-10 10:00:00.000
In this example set, if my filter date is '2021-03-11', my expected result set should be all records from that day plus adjacent records from 03-12 and 03-10 that satisfy the 8 hours condition. Note how record with Id 7 is not be included because record with Id 6 does not comply:
Id
EntryDate
2
2021-03-12 01:10:00.000
3
2021-03-11 20:17:00.000
4
2021-03-11 05:04:00.000
5
2021-03-10 23:58:00.000
Need advice how to write this complex query
This is a variant of gaps-and-islands. Define the difference . . . and then groups based on the differences:
with e as (
select t.*
from (select t.*,
sum(case when prev_entrydate > dateadd(hour, -8, entrydate) then 0 else 1 end) over (order by entrydate) as grp
from (select t.*,
lag(entrydate) over (order by entrydate) as prev_entrydate
from t
) t
)
select e.*
from e.*
where e.grp in (select e2.grp
from t e2
where date(e2.entrydate) = #filterdate
);
Note: I'm not sure exactly how filter date is applied. This assumes that it is any events on the entire day, which means that there might be multiple groups. If there is only one group (say the first group on the day), the query can be simplified a bit from a performance perspective.
declare #DateTime datetime = '2021-03-11'
select *
from t
where t.EntryDate between DATEADD(hour , -8 , #DateTime) and DATEADD(hour , 32 , #DateTime)
This question already has answers here:
Calculate a Running Total in SQL Server
(15 answers)
Closed 3 years ago.
I have the following table:
________________________
date | amount
________________________
01-01-2019 | 10
01-01-2019 | 10
01-01-2019 | 10
01-01-2019 | 10
02-01-2019 | 5
02-01-2019 | 5
02-01-2019 | 5
02-01-2019 | 5
03-01-2019 | 20
03-01-2019 | 20
These are mutation values by date. I would like my query to return the summed amount by date. So for 02-01-2019 I need 40 ( 4 times 10) + 20 ( 4 times 5). For 03-01-2019 I would need ( 4 times 10) + 20 ( 4 times 5) + 40 ( 2 times 20) and so on. Is this possible in one query? How do I achieve this?
My current query to get the individual mutations:
Select s.date,
Sum(s.amount) As Sum_amount
From dbo.Financieel As s
Group By s.date
You can try below -
DEMO
select dateval,
SUM(amt) OVER(ORDER BY dateval ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) as amt
from
(
SELECT
dateval,
SUM(amount) amt
FROM t2 group by dateval
)A
OUTPUT:
dateval amt
01/01/2019 00:00:00 40
01/02/2019 00:00:00 60
01/03/2019 00:00:00 100
Try this below script to get your desired output-
SELECT A.date,
(SELECT SUM(amount) FROM <your_table> WHERE Date <= A.Date) C_Total
FROM <your_table> A
GROUP BY date
ORDER BY date
Output is-
date C_Total
01-01-2019 40
02-01-2019 60
03-01-2019 100
I suggest to use a window function, like this:
select date, sum(amount) over( order by date)
from table
I have the following table, I am using SQL Server 2008
BayNo FixDateTime FixType
1 04/05/2015 16:15:00 tyre change
1 12/05/2015 00:15:00 oil change
1 12/05/2015 08:15:00 engine tuning
1 04/05/2016 08:11:00 car tuning
2 13/05/2015 19:30:00 puncture
2 14/05/2015 08:00:00 light repair
2 15/05/2015 10:30:00 super op
2 20/05/2015 12:30:00 wiper change
2 12/05/2016 09:30:00 denting
2 12/05/2016 10:30:00 wiper repair
2 12/06/2016 10:30:00 exhaust repair
4 12/05/2016 05:30:00 stereo unlock
4 17/05/2016 15:05:00 door handle repair
on any given day need do find the highest number of fixes made on a given bay number, and if that calculated number is repeated then it should also appear in the resultset
so would like to see the result set as follows
BayNo FixDateTime noOfFixes
1 12/05/2015 00:15:00 2
2 12/05/2016 09:30:00 2
4 12/05/2016 05:30:00 1
4 17/05/2016 15:05:00 1
I manage to get the counts of each but struggling to get the max and keep the highest calculated repeated value. can someone help please
Use window functions.
Get the count for each day by bayno and also find the min fixdatetime for each day per bayno.
Then use dense_rank to compute the highest ranked row for each bayno based on the number of fixes.
Finally get the highest ranked rows.
select distinct bayno,minfixdatetime,no_of_fixes
from (
select bayno,minfixdatetime,no_of_fixes
,dense_rank() over(partition by bayno order by no_of_fixes desc) rnk
from (
select t.*,
count(*) over(partition by bayno,cast(fixdatetime as date)) no_of_fixes,
min(fixdatetime) over(partition by bayno,cast(fixdatetime as date)) minfixdatetime
from tablename t
) x
) y
where rnk = 1
Sample Demo
You are looking for rank() or dense_rank(). I would right the query like this:
select bayno, thedate, numFixes
from (select bayno, cast(fixdatetime) as date) as thedate,
count(*) as numFixes,
rank() over (partition by cast(fixdatetime as date) order by count(*) desc) as seqnum
from t
group by bayno, cast(fixdatetime as date)
) b
where seqnum = 1;
Note that this returns the date in question. The date does not have a time component.
i have table and it has following data:
USERID NAME DATEFROM DATETO
1 xxx 2014-05-10 2014-05-15
1 xxx 2014-05-20 2014-05-25
4 yyy 2014-04-20 2014-04-21
now i have sql query like :
select * from leave where datefrom>='2014-05-01' and dateto<='2014-05-31'
so now i want output :
userid name total_leave_days
1 xxx 12
4 yyy 2
(2014-05-10 - 2014-05-15 )=6 days
(2014-05-20 - 2014-05-25 )=6 days
total = 12 days for useid 1
(2014-04-20 - 2014-04-21)= 2 days for userid 4
how can i calculate this total days .?
Please try:
select
USERID,
NAME,
SUM(DATEDIFF(day, DATEFROM, DATETO)+1) total_leave_days
From leave
group by USERID, NAME
SQL Fiddle Demo
It's important to note that you need "+1" to emulate the expected calculations because there is an inherent assumption of ""start of day" for the Start date and "end of day" for end date - but dbms's don't think that way. a date is always stored as "start of day".
select
USERID
, name
, sum( datediff(day,DATEFROM,DATETO) + 1 ) as leave_days
from leavetable
group by
USERID
, name
produces this:
| USERID | NAME | LEAVE_DAYS |
|--------|------|------------|
| 1 | xxx | 12 |
| 4 | yyy | 2 |
see: http://sqlfiddle.com/#!3/ebe5d/1
You can use DateDiff.
SELECT UserID, Name, SUM(DATEDIFF(DAY, DateFrom, DateTo) + 1) AS total_leave_days
FROM leave
WHERE datefrom >= '2014-05-01' AND dateto <= '2014-05-31'
GROUP BY UserID, Name
The + 1 ,of course, is because DATEDIFF will return the exclusive count, where it sounds like you want the inclusive number of days.
Try this:
select userid, name, sum (1 + datediff(day,datefrom,dateto)) as total_leave_days
from leaves
where datefrom>='2014-05-01' and dateto<='2014-05-31'
group by userid, name
This will sum the total leaves per userid. Note that datediff will give you 5 days difference for the range 2014-05-10 to 2014-05-15, so we need to add 1 to the result to get 6 days i.e. range inclusive of both ends.
Demo