How would I construct a query to receive the MAX TurnTime per ID of the first 2 rounds? Rounds being defined as minimum Beginning_Date to mininmum End_Date of an ID. Without reusing either of the dates for the second round Turn Time calculation.
You can use row_number() . . . twice:
select d.*
from (select d.*,
row_number() over (partition by id order by turn_time desc) as seqnum_turntime
from (select d.*,
row_number() over (partition by id order by beginning_end desc) as seqnum_round
from data d
) d
where seqnum_round <= 2
) d
where seqnum_turntime = 1;
The innermost subquery gets the first two rounds. The outer subquery gets the maximum.
You could express this without window functions as well:
select top (1) with ties d.*
from data d
where d.beginning_date <= (select d2.beginning_date
from data d2
where d2.id = d.id
offset 1 fetch first 1 row only
)
order by row_number() over (partition by id order by turntime desc);
SELECT
ID
,turn_time
,beginning_date
,end_date
FROM
(
SELECT
ID
,MAX(turn_time) OVER (PARTITION BY Id ORDER BY BeginningDate ROWS BETWEEN 1 PRECEDING AND CURRENT ROW) AS turn_time --Maximum turn time of the current row and preceding row
,MIN(BeginningDate) OVER (PARTITION BY Id ORDER BY BeginningDate ROWS BETWEEN 1 PRECEDING AND CURRENT ROW) AS beginning_date --Minimum begin date over current row and preceding row (could also use LAG)
,end_date
,ROW_NUMBER() OVER (PARTITION BY Id ORDER BY BeginningDate) AS Turn_Number
FROM
<whatever your table is>
) turn_summary
WHERE
Turn_Number = 2
Related
I need to get first and last record (ordered by Date column) from table for certain SSID. It is not a problem if there is more records with same max or min date. All I need is union all.
I am getting last record having max(date) with:
with c as (
select *, rnk = rank() over (partition by Date order by Date ASC)
from table
where SSID = '00921834800'
)
select top 1 Date, City, Title
from c
order by Date desc
How to I get first record (min(Date)) as well (same thing only with order by Date asc) with single select and without using ranking again?
I'm using MSSQL 2017.
; with c as (
select *,
rnk = rank() over (partition by Date order by Date ASC),
rnk2 = rank() over (partition by Date order by Date desc)
from table
where SSID= '00921834800'
)
select Date,
City,
Title
from c
where rnk = 1 or rnk2 = 1
order by Date desc
I would use the following query:
select * from (select top 1 with ties * from t where ssid = '00921834800' order by date) as a
union all
select * from (select top 1 with ties * from t where ssid = '00921834800' order by date desc) as b
One other solution is :
with
c as
(
select *,
rank() over (partition by Date order by Date ASC) AS RNK,
count() OVER (partition by Date) AS CNT
from table
where SSID= '00921834800')
select Date, City, Title
from c
WHERE RNK = 1
OR CNT = RNK
order by Date desc
I have a peculiar problem at hand. I need to rank in the following manner:
Each ID gets a new rank.
rank #1 is assigned to the ID with the lowest date. However, the subsequent dates for that particular ID can be higher but they will get the incremental rank w.r.t other IDs.
(E.g. ADF32 series will be considered to be ranked first as it had the lowest date, although it ends with dates 09-Nov, and RT659 starts with 13-Aug it will be ranked subsequently)
For a particular ID, if the days are consecutive then ranks are same, else they add by 1.
For a particular ID, ranks are given in date ASC.
How to formulate a query?
You need two steps:
select
id_col
,dt_col
,dense_rank()
over (order by min_dt, id_col, dt_col - rnk) as part_col
from
(
select
id_col
,dt_col
,min(dt_col)
over (partition by id_col) as min_dt
,rank()
over (partition by id_col
order by dt_col) as rnk
from tab
) as dt
dt_col - rnk caluclates the same result for consecutives dates -> same rank
Try datediff on lead/lag and then perform partitioned ranking
select t.ID_COL,t.dt_col,
rank() over(partition by t.ID_COL, t.date_diff order by t.dt_col desc) as rankk
from ( SELECT ID_COL,dt_col,
DATEDIFF(day, Lag(dt_col, 1) OVER(ORDER BY dt_col),dt_col) as date_diff FROM table1 ) t
One way to think about this problem is "when to add 1 to the rank". Well, that occurs when the previous value on a row with the same id_col differs by more than one day. Or when the row is the earliest day for an id.
This turns the problem into a cumulative sum:
select t.*,
sum(case when prev_dt_col = dt_col - 1 then 0 else 1
end) over
(order by min_dt_col, id_col, dt_col) as ranking
from (select t.*,
lag(dt_col) over (partition by id_col order by dt_col) as prev_dt_col,
min(dt_col) over (partition by id_col) as min_dt_col
from t
) t;
I have table that looks like the following
I have to select every second record per PatientID that would give the following result (my last query returns this result)
I then have to select the record with the oldest date which would be the following (this is the end result I want)
What I have done so far: I have a CTE that gets all the data I need
WITH cte
AS
(
SELECT visit.PatientTreatmentVisitID, mat.PatientMatchID,pat.PatientID,visit.RegimenDate AS VisitDate,
ROW_NUMBER() OVER(PARTITION BY mat.PatientMatchID, pat.PatientID ORDER BY visit.VisitDate ASC) AS RowNumber
FROM tblPatient pat INNER JOIN tblPatientMatch mat ON mat.PatientID = pat.PatientID
LEFT JOIN tblPatientTreatmentVisit visit ON visit.PatientID = pat.PatientID
)
I then write a query against the CTE but so far I can only return the second row for each patientID
SELECT *
FROM
(
SELECT PatientTreatmentVisitID,PatientMatchID,PatientID, VisitDate, RowNumber FROM cte
) as X
WHERE RowNumber = 2
How do I return the record with the oldest date only? Is there perhaps a MIN() function that I could be including somewhere?
If I follow you correctly, you can just order your existing resultset and retain the top row only.
In standard SQL, you would write this using a FETCH clause:
SELECT *
FROM (
SELECT
visit.PatientTreatmentVisitID,
mat.PatientMatchID,
pat.PatientID,
visit.RegimenDate AS VisitDate,
ROW_NUMBER() OVER(PARTITION BY mat.PatientMatchID, pat.PatientID ORDER BY visit.VisitDate ASC) AS rn
FROM tblPatient pat
INNER JOIN tblPatientMatch mat ON mat.PatientID = pat.PatientID
LEFT JOIN tblPatientTreatmentVisit visit ON visit.PatientID = pat.PatientID
) t
WHERE rn = 2
ORDER BY VisitDate
OFFSET 0 ROWS FETCH FIRST 1 ROW ONLY
This syntax is supported in Postgres, Oracle, SQL Server (and possibly other databases).
If you need to get oldest date from all selected dates (every second row for each patient ID) then you can try window function Min:
SELECT * FROM
(
SELECT *, MIN(VisitDate) OVER (Order By VisitDate) MinDate
FROM
(
SELECT PatientTreatmentVisitID,PatientMatchID,PatientID, VisitDate,
RowNumber FROM cte
) as X
WHERE RowNumber = 2
) Y
WHERE VisitDate=MinDate
Or you can use SELECT TOP statement. The SELECT TOP clause allows you to limit the number of rows returned in a query result set:
SELECT TOP 1 PatientTreatmentVisitID,PatientMatchID,PatientID, VisitDate FROM
(
SELECT *
FROM
(
SELECT PatientTreatmentVisitID,PatientMatchID,PatientID, VisitDate,
RowNumber FROM cte
) as X
WHERE RowNumber = 2
) Y
ORDER BY VisitDate
For simplicity add order desc on date column and use TOP to get the first row only
SELECT TOP 1 *
FROM
(
SELECT PatientTreatmentVisitID,PatientMatchID,PatientID, VisitDate, RowNumber FROM cte
) as X
WHERE RowNumber = 2
order by VisitDate desc
I am using SQL Server and trying to get the order of transaction in a month, for example if I have 10 transactions in a month I want to know when the selected transaction occurred, if it was the 1st, 2nd, 3rd etc. transaction in that month. I tried using ROW_NUMBER() like
SELECT
transactionid,
ROW_NUMBER() OVER(ORDER BY transactionexecutiontime ASC) rowOrderNumber
FROM
dbo.exportebop
WHERE
MONTH(transactionexecutiontime) = 1
ORDER BY
transactionexecutiontime ASC
This works if I do not specify the TransactionID because I get multiple rows, the moment I use TransactionID = idNumber It returns a single row and row_number() is always 1
I know ROW_NUMBER() counts the result set but I need a way to calculate when order number in a month using the TransactionExecutionTime instead.
Does anyone know how I can achieve this?
I think you need to apply Row_Number first then after that you can apply filtering. I will use CTE to do that -
;WITH CTE AS (
SELECT
TransactionID,
ROW_NUMBER() OVER(ORDER BY TransactionExecutionTime ASC) rowOrderNumber
FROM dbo.EXPORTEBOP
WHERE MONTH(TransactionExecutionTime) = 1
)
SELECT *
FROM CTE
WHERE TransactionID = 1
ORDER BY rowOrderNumber ASC
As a sub-query-
DECLARE #Month INT = 1
SELECT *
FROM (
SELECT
TransactionID,
ROW_NUMBER() OVER(ORDER BY TransactionExecutionTime ASC) rowOrderNumber
FROM dbo.EXPORTEBOP
WHERE MONTH(TransactionExecutionTime) = #Month
) X
WHERE TransactionID = 1
ORDER BY rowOrderNumber ASC
If you want all months listed, you can partition the row number by month:
SELECT
transactionid,
Month = MONTH(transactionexecutiontime),
rowOrderNumber = ROW_NUMBER() OVER(
PARTITION BY
MONTH(transactionexecutiontime) -- Will start different "row numbers" for each month
ORDER BY
transactionexecutiontime ASC)
FROM
dbo.exportebop
ORDER BY
MONTH(transactionexecutiontime),
transactionexecutiontime ASC
If you want a particular transaction ID, you can't filter it on the same result set as the ROW_NUMBER() because it will only generate a row number for that result set, which will only contain 1 row (your filtered transaction). You will have to filter it on a another query that references the first (with a CTE or a subquery):
;WITH RankedTransactionsByMonth AS
(
SELECT
transactionid,
Month = MONTH(transactionexecutiontime),
rowOrderNumber = ROW_NUMBER() OVER(
PARTITION BY
MONTH(transactionexecutiontime) -- Will start different "row numbers" for each month
ORDER BY
transactionexecutiontime ASC)
FROM
dbo.exportebop
-- No Order by in subqueries without TOP!!
)
SELECT
R.*
FROM
RankedTransactionsByMonth AS R
WHERE
R.transactionid = 1
how do i get unique name with the latest date and lowest value.
Name date value
brad 1/2/10 1.1
brad 1/2/10 2.3
bob 1/6/10 1.0
brad 2/4/09 13.2
this query does not seem to work
SELECT distinct
A.[ViralLoadMemberID]
,B.LastName
,B.FirstName
,A.[Date]
,A.[vaule]
FROM [t].[dbo].[tblViralLoad] A
left join [dbo].[tblEnrollees] B on A.ViralLoadMemberID = B.MemberID
where
A.Date =
(
select MAX(Date)
from dbo.tblViralLoad
where ViralLoadMemberID = A.ViralLoadMemberID
and
( Date >= '07/01/2014'
and Date <= '12/3/2014' ) )
The idea is to use order by and fetch only one row. If you want the lowest value on the latest date, the standard SQL would be:
select t.*
from table t
order by desc desc, value asc
fetch first 1 row only;
For older versions of SQL Server, you would omit the last line and do select top 1 * . . .. For MySQL, the last line would be limit 1.
Fun with rank()
declare #t as table (name varchar(50),dte date,val decimal(18,10));
insert into #t(name,dte,val) values
('Dave','1/1/2015',1.0),
('Dave','1/3/2015',1.2),
('Dave','1/4/2015',1.5),
('Dave','1/10/2015',1.3),
('Dave','1/15/2015',1.2),
('Steve','1/11/2015',1.6),
('Steve','1/12/2015',1.1),
('Steve','1/15/2015',1.2),
('Bill','1/21/2015',1.9),
('Ted','1/1/2015',1.8),
('Ted','1/10/2015',1.0),
('Ted','1/12/2015',1.7)
-- This will show the lowest prices by each person
select name,dte,val from (select name,dte,val, rank() over (partition by name order by val) as r from #t) as data where r = 1
-- This will be users lowest price and the last day they sublitted a prices regurdless if it is the lowest
select name,max(dte) as [last Date] ,min(val) as [Lowest Value] from #t group by name
-- Who had the lowest price last regurdless if they have raised there price later.
select top(1) name,dte [last lowest quote],val from (select name,dte,val, rank() over (order by val) as r from #t) as data where r = 1 order by dte desc
-- what is the lowest price cueently quoted reguarless who quoted it
select top(1) name,dte [best active quote],val from (select name,dte,val, rank() over (partition by name order by dte desc) as r from #t) as data where r = 1 order by val