I'm struggling with creation of select which shows me penultimate date for each record in my DB.
For example:
id date
1 01.01.2018
1 05.01.2018
1 06.02.2018
2 01.06.2018
2 03.06.2018
3 12.12.2017
Out of this record I need to write select, which shows me following:
ID max_date penultimate
1 06.02.2018 05.01.2018
2 03.06.2018 01.06.2018
3 12.12.2017 NULL
Any idea how to do it? many thanks in advance
Use conditional aggregation and the ANSI-standard row_number() or dense_rank() functions:
select id,
max(date) as max_date,
max(case when seqnum = 2 then date end) as penultimate_date
from (select t.*,
dense_rank() over (partition by id order by date desc) as seqnum
from t
) t
where seqnum in (1, 2)
group by id;
Use row_number() if the dates can be the same in the event of ties.
Use GROUP BY to get the MAX and a correlated subquery with another MAX but this time lower than the former.
SELECT
T.id,
MAX(T.date) max_date,
(
SELECT
MAX(N.date)
FROM
YourTable N
WHERE
N.id = T.id AND
N.date < MAX(T.date)
) penultimate
FROM
YourTable T
GROUP BY
T.id
Just an opitimized query:
;WITH cte AS
(
SELECT id AS ID
,[date] AS max_date
,LEAD ([date], 1, 0) OVER (PARTITION BY id ORDER BY [date] DESC) AS penultimate
,ROW_NUMBER() OVER(PARTITION BY id ORDER BY [date] DESC) AS RN
FROM Table3
)
SELECT ID,max_date,penultimate
FROM cte
WHERE RN=1
SQL Fiddle
I wrote in this way,
SELECT ID
,max(StartDate) MaxDate
,(
SELECT StartDate
FROM YourTable t2
WHERE t2.id = t1.id
ORDER BY StartDate DESC OFFSET 1 ROWS FETCH NEXT 1 ROW ONLY
) penultimate
FROM YourTable t1
GROUP BY id
Related
I have the following dataset:
id
id_rev
time
1
1
08.01.2022
1
0
31.02.2021
2
2
28.01.2017
2
1
25.07.2021
2
0
25.07.2021
I am looking for a SQL query that can return an entry per id but only the one where the id_rev is maximum. So in this case it should return these two rows:
(id=1, id_rev=1,time)
(id=2, id_rev=2, time)
One canonical approach uses ROW_NUMBER:
WITH cte AS (
SELECT t.*, ROW_NUMBER() OVER (PARTITION BY id ORDER BY id_rev DESC) rn
FROM yourTable t
)
SELECT id, id_rev, time
FROM cte
WHERE rn = 1
ORDER BY id;
Another approach would be to use exists logic:
SELECT id, id_rev, time
FROM yourTable t1
WHERE NOT EXISTS (
SELECT 1
FROM yourTable t2
WHERE t2.id = t1.id AND t2.id_rev > t1.id_rev
);
#result =
SELECT
*,
RANK()
OVER (PARTITION BY id ORDER BY id_rev DESC) AS Rank
FROM dataset ORDER BY Rank;
#result =
SELECT *
FROM #result
WHERE Rank = 1;
I have a Oracle table with the following columns
Table Structure
In a query I need to return all the records with CPER>=40 which is trivial. However, apart from CPER>=40 I need to list 5 random records for each CPID.
I have attached a sample list of records. However, in my table I have around 50,000 records.
Appreciate if you can help.
Oracle solution:
with CTE as
(
select t1.*,
row_number() over(order by DBMS_RANDOM.VALUE) as rn -- random order assigned
from MyTable t1
where CPID <40
)
select *
from CTE
where rn <=5 -- pick 5 at random
union all
select t2.*, null
from my_table t2
where CPID >= 40
SQL Server:
with CTE as
(
select t1.*,
row_number() over(order by newid()) as rn -- random order assigned
from MyTable t1
where CPID <40
)
select *
from CTE
where rn <=5 -- pick 5 at random
union all
select t2.*, null
from my_table t2
where CPID >= 40
How about something like this...
SELECT *
FROM (SELECT CID,
CVAL,
CPID,
CPER,
Row_number() OVER (partition BY CPID ORDER BY CPID ASC ) AS RN
FROM Table) tmp
WHERE CPER>=40 OR pids <= 5
However, this is not random.
Assuming that you want five additional random records, you can do:
select t.*
from (select t.*,
row_number() over (partition by cpid,
(case when cper >= 40 then 1 else 2 end)
order by dbms_random.value
) as seqnum
from t
) t
where seqnum <= 5 or cper >= 40;
The row_number() is enumerating the rows for each cpid in two groups -- based on the cper value. The outer where is taking all cper values in the range you want as well as five from the other group.
I need to find the latest 3 records for each user that has a particular status on 'Fail'. At first it seems easy but I just can't seem to get it right.
So in a table of:
ID Date Status
1 2017-01-01 Fail
1 2017-01-02 Fail
1 2017-02-04 Fail
1 2015-03-21 Pass
1 2014-02-19 Fail
1 2016-10-23 Pass
2 2017-01-01 Fail
2 2017-01-02 Pass
2 2017-02-04 Fail
2 2016-10-23 Fail
I would expect ID 1 to be returned as the most recent 3 records are fails, but not ID 2, as they have a pass within their three fails. Each user may have any number of Pass and Fail records. There are thousands of different IDs
So far I've tried a CTE with ROW_NUMBER() to order the attempts but can't think of a way to ensure that the latest three results all have the same status of Fail.
Expected Results
ID Latest Fail Date Count
1 2017-02-04 3
Maybe try something like this:
WITH cte
AS
(
SELECT id,
date,
status,
ROW_NUMBER () OVER (PARTITION BY id ORDER BY date DESC) row
FROM #table
),cte2
AS
(
SELECT id, max(date) as date, count(*) AS count
FROM cte
WHERE status = 'fail'
AND row <= 3
GROUP BY id
)
SELECT id,
date AS latest_fail,
count
FROM cte2
WHERE count = 3
Check This.
Demo : Here
with CTE as
(
select *,ROW_NUMBER () over( partition by id order by date desc) rnk
from temp
where Status ='Fail'
)
select top 1 ID,max(DATE) as Latest_Fail_Date ,COUNT(rnk) as count
from CTE where rnk <=3
group by ID
Ouptut :
I think you can do this using cross apply:
select i.id
from (select distinct id from t) i cross apply
(select sum(case when t.status = 'Fail' then 1 else 0 end) as numFails
from (select top 3 t.*
from t
where t.id = i.id
order by date desc
) ti
) ti
where numFails = 3;
Note: You probably have a table with all the ids. If so, you an use that instead of the select distinct subquery.
Or, similarly:
select i.id
from (select distinct id from t) i cross apply
(select top 3 t.*
from t
where t.id = i.id
order by date desc
) ti
group by i.id
having min(ti.status) = 'Fail' and max(ti.status) = 'Fail' and
count(*) = 3;
Here you go:
declare #numOfTries int = 3;
with fails_nums as
(
select *, row_number() over (partition by ID order by [Date] desc) as rn
from #fails
)
select ID, max([Date]) [Date], count(*) as [count]
from fails_nums fn1
where fn1.rn <= #numOftries
group by ID
having count(case when [Status]='Fail' then [Status] end) = #numOfTries
Example here
So I have data like this:
UserID CreateDate
1 10/20/2013 4:05
1 10/20/2013 4:10
1 10/21/2013 5:10
2 10/20/2012 4:03
I need to group by each user get the average time between CreateDates. My desired results would be like this:
UserID AvgTime(minutes)
1 753.5
2 0
How can I find the difference between CreateDates for all records returned for a User grouping?
EDIT:
Using SQL Server 2012
Try this:
SELECT A.UserID,
AVG(CAST(DATEDIFF(MINUTE,B.CreateDate,A.CreateDate) AS FLOAT)) AvgTime
FROM #YourTable A
OUTER APPLY (SELECT TOP 1 *
FROM #YourTable
WHERE UserID = A.UserID
AND CreateDate < A.CreateDate
ORDER BY CreateDate DESC) B
GROUP BY A.UserID
This approach should aslo work.
Fiddle demo here:
;WITH CTE AS (
Select userId, createDate,
row_number() over (partition by userid order by createdate) rn
from Table1
)
select t1.userid,
isnull(avg(datediff(second, t1.createdate, t2.createdate)*1.0/60),0) AvgTime
from CTE t1 left join CTE t2 on t1.UserID = t2.UserID and t1.rn +1 = t2.rn
group by t1.UserID;
Updated: Thanks to #Lemark for pointing out number of diff = recordCount - 1
since you're using 2012 you can use lead() to do this
with cte as
(select
userid,
(datediff(second, createdate,
lead(CreateDate) over (Partition by userid order by createdate)
)/60) datdiff
From table1
)
select
userid,
avg(datdiff)
from cte
group by userid
Demo
Something like this:
;WITH CTE AS
(
SELECT
ROW_NUMBER() OVER (PARTITION BY UserID ORDER BY CreateDate) RN,
UserID,
CreateDate
FROM Tbl
)
SELECT
T1.UserID,
AVG(DATEDIFF(mi, ISNULL(T2.CreateDate, T1.CreateDate), T1.CreateDate)) AvgTime
FROM CTE T1
LEFT JOIN CTE T2
ON T1.UserID = T2.UserID
AND T1.RN = T2.RN - 1
GROUP BY T1.UserID
With SQL 2012 you can use the ROW_NUMBER function and self-join to find the "previous" row in each group:
WITH Base AS
(
SELECT
ROW_NUMBER() OVER (PARTITION BY UserID ORDER BY CreateDate) RowNum,
UserId,
CreateDate
FROM Users
)
SELECT
B1.UserID,
ISNULL(
AVG(
DATEDIFF(mi,B2.CreateDate,B1.CreateDate) * 1.0
)
,0) [Average]
FROM Base B1
LEFT JOIN Base B2
ON B1.UserID = B2.UserID
AND B1.RowNum = B2.RowNum + 1
GROUP BY B1.UserId
Although I get a different answer for UserID 1 - I get an average of (5 + 1500) / 2 = 752.
This only works in 2012. You can use the LEAD analytic function:
CREATE TABLE dates (
id integer,
created datetime not null
);
INSERT INTO dates (id, created)
SELECT 1 AS id, '10/20/2013 4:05' AS created
UNION ALL SELECT 1, '10/20/2013 4:10'
UNION ALL SELECT 1, '10/21/2013 5:10'
UNION ALL SELECT 2, '10/20/2012 4:03';
SELECT id, isnull(avg(diff), 0)
FROM (
SELECT id,
datediff(MINUTE,
created,
LEAD(created, 1, NULL) OVER(partition BY id ORDER BY created)
) AS diff
FROM dates
) as diffs
GROUP BY id;
http://sqlfiddle.com/#!6/4ce89/22
For example I have the following database entries:
timestamp | value1 | value 2
----------
1452|5|7
1452|1|6
1452|2|7
1623|1|2
1623|5|6
1623|4|5
1623|4|7
1855|1|2
Now I want to have a sql query which returns me value1 only for the timestamp which is availble the most. Therefore it should return only the timestamp 1623 and it's values.
I was first thinking of count, but that will return only the number of the availability and not the entries.
select *
from T
inner join (select timestamp
from T
group by timestamp
order by count(*) desc
limit 1) t2
on T.timestamp = t2.timestamp
see it's working live in a sqlfiddle
WITH CTE AS (
SELECT *, COUNT(timestamps) OVER (PARTITION BY value1, timestamps) AS cnt
FROM mytable
), cte2 as (select *, row_number() over (partition by value1 order by cnt DESC, timestamps) as Rn FROM cte)
SELECT value1, timestamps , cnt FROM CTE2 WHERE Rn = 1;