I have a log table that records dates.
I want to find out the date of the Second and Third call.
I get the first call by using MIN(DateCreated)
I get the last date using MAX(DateCreated). This could also be the second call.
What is the best way to find out the date of the second and third call?
I would just get all 3 rows together in one query
SELECT TOP 3 *
FROM LogTable
ORDER BY LogDate ASC -- DESC if you need the last 3
If you do need them one by one, you can use:
WITH LogByDate AS
(
SELECT LogDate,
ROW_NUMBER() OVER (ORDER BY LogDate) AS 'RowNumber'
FROM LogTable
)
SELECT *
FROM LogByDate
WHERE RowNumber = 2; -- RowNumber=3 for the third line
Try using analytical functions:
;WITH CTE AS
(
SELECT *, ROW_NUMBER() OVER(PARTITION BY CallId ORDER BY DateCreated) Corr
FROM dbo.YourTable
)
SELECT *
FROM CTE
WHERE Corr IN (2,3)
Related
Hopefully, someone can help me...
I'm trying to get the last two values from a row_number() window function. Let's say my results contain row numbers up to 6, for example. How would it be possible to get the rows where the row number is 5 and 6?
Let me know if it can be done with another window function or in another way.
Kind regards,
Using QUALIFY:
SELECT *
FROM tab
QUALIFY ROW_NUMBER() OVER(ORDER BY ... DESC) <= 2;
This approach could be further extended to get two rows per each partition:
SELECT *
FROM tab
QUALIFY ROW_NUMBER() OVER(PARTITION BY ... ORDER BY ... DESC) <= 2;
You can use top with order by desc like:
select top 2 row_number() over([partition by] [order by]) as rn
from table
order by rn desc
I'd say #Shmiel is the formal and elegant way, just in case, would be the same as :
WITH CTE AS
(SELECT product,
user_id,
ROW_NUMBER() OVER (PARTITION BY user_id order by product desc)
as RN
FROM Mytable)
SELECT product, user_id
FROM CTE
WHERE RN < 3;
You will use order by [order_condition] with "desc". And then you will use RN(row number) to select as many rows as you want
I have table like this
And I want to have only earliest time from column time for each day from column date. Rest of table has to be unaffected.
So result would be that for example that I have only time 9:25 for 2018-07-13 and rest of rows with later times for 2018-07-13 are deleted
To delete you can use a CTE with ROW_NUMBER window function
;WITH cteDups
AS(
SELECT *, RN=ROW_NUMBER()OVER (PARTITION BY M.Date ORDER BY M.Time ASC)
FROM dbo.yourtable M
)
--SELECT *
DELETE
FROM cteDups D WHERE D.RN > 1
You can use a window function to return all rows
select
*,
min([time]) over (partition by [date] order by [time])
from YourTable
Or just the aggregate to remove them
select *
from YourTable
inner join
(select whatever, min(FullDate) dt
from yourtable
group by whatever) x on x.whatever = YourTable.whatever and x.dt = YourTable.FullDate
If the whatever column doesn't matter, and you only want the date and time:
Select
[date],
min([time])
from YourTable
group by [Date]
The simplest way to keep certain records and remove the rest would be by using a CTE with a windowing function to rank (or add rownumbers). Check this out:
;WITH EarliestHourEveryDay AS (
SELECT
whatever
,FullDate
,[date]
,[time]
,rn = ROW_NUMBER() OVER (PARTITION BY [date] ORDER BY [time])
FROM TableName
)
SELECT *
FROM EarliestHourEveryDay
WHERE rn = 1
/*
DELETE FROM EarliestHourEveryDay
WHERE rn > 1
*/
I have commented out the delete statement so that you can test this first. Run the CTE as-is, and if the result set contains the exact rows which you want, remove the SELECT statement from the CTE and uncomment the DELETE statement and you'll be good to go.
Group by day and select the MIN time.
Use the MIN function and a GROUP BY clause.
Something like:
SELECT date, MIN(time) AS EarliestTime
FROM MyTable
GROUP BY date
ORDER BY date ASC
Here is an example of this working: SQL Fiddle
I think this is a simple problem but I cannot find an easy answer. I need to retrieve the last 2 entries by date. I have used max() to get the latest date; but do not know how to retrieve the next most recent.
The stored procedure code for latest date is:
SELECT *
FROM Table
WHERE Date=(SELECT MAX(Date) FROM Table);
So using a separate procedure how do I get the next most recent?
You can use order by and top:
select top 2 t.*
from t
order by date desc;
Or just to get the next most recent only as you stated...thus returning only one row...
select top 1 t.*
from t
where t.date != (select max(date) from table)
order by date desc;
or...
with cte as(
select
t.*
,row_number() over (order by t.date desc) as RN
from table t)
select *
from cte
where RN = 2
Let us say that I have a database table with the following two records:
CACHE_ID BUSINESS_DATE CREATED_DATE
1183 13-09-06 13-09-19 16:38:59.336000000
1169 13-09-06 13-09-24 17:19:05.762000000
1152 13-09-06 13-09-17 14:18:59.336000000
1173 13-09-05 13-09-19 15:48:59.136000000
1139 13-09-05 13-09-24 12:59:05.263000000
1152 13-09-05 13-09-27 13:28:59.332000000
I need to write a query that will return the CACHE_ID for the record which has the most recent CREATED_DATE.
I am having trouble crafting such a query. I can do a GROUP BY based on BUSINESS_DATE and get the MAX(CREATED_DATE)...of course, I won't have the CACHE_ID of the record.
Could someone help with this?
Not positive on oracle syntax, but use the ROW_NUMBER() function:
SELECT BUSINESS_DATE, CACHE_ID
FROM (SELECT t.*,
ROW_NUMBER() OVER(PARTITION BY BUSINESS_DATE ORDER BY CREATED_DATE DESC) RN
FROM YourTable t
)sub
WHERE RN = 1
The ROW_NUMBER() function assigns a number to each row. PARTITION BY is optional, but used to start the numbering over for each value in that group, ie: if you PARTITION BY BUSINESS_DATE then for each unique BUSINESS_DATE value the numbering would start over at 1. ORDER BY of course is used to define how the counting should go, and is required in the ROW_NUMBER() function.
You want to group on business date, and get the CACHE_ID with the most current created date? Use something like this:
select yt.CACHE_ID, yt.BUSINESS_DATE, yt.CREATED_DATE
from YourTable yt
where yt.CREATED_DATE = (select max(yt1.CREATED_DATE)
from YourTable yt1
where yt1.BUSINESS_DATE = yt.BUSINESS_DATE)
Not sure of the exact syntax, but conceptually, can't you just sort by CREATED_DATE descending and take the first one?
Across all records -
select top 1 CACHE_ID from YourTable order by CREATED_DATE desc
For each BUSINESS_DATE -
select distinct
a.BUSINESS_DATE,
(
select top 1 b.CACHE_ID
from YourTable b where a.BUSINESS_DATE = b.BUSINESS_DATE
order by b.CREATED_DATE desc
) as Last_CREATED_DATE
from YourTable a
I have a table, myTable that has two fields in it ID and patientID. The same patientID can be in the table more than once with a different ID. How can I make sure that I get only ONE instance of every patientID.?
EDIT: I know this isn't perfect design, but I need to get some info out of the database and today and then fix it later.
You could use a CTE with ROW_NUMBER function:
WITH CTE AS(
SELECT myTable.*
, RN = ROW_NUMBER()OVER(PARTITION BY patientID ORDER BY ID)
FROM myTable
)
SELECT * FROM CTE
WHERE RN = 1
It sounds like you're looking for DISTINCT:
SELECT DISTINCT patientID FROM myTable
you can get the same "effect" with GROUP BY:
SELECT patientID FROM myTable GROUP BY patientID
The simple way would be to add LIMIT 1 to the end of your query. This will ensure only a single row is returned in the result set.
WITH CTE AS
(
SELECT tableName.*,ROW_NUMBER() OVER(PARTITION BY patientID ORDER BY patientID) As 'Position' FROM tableName
)
SELECT * FROM CTE
WHERE
Position = 1