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
I hope i can explain the issue i'm having and hopefully so can point me in the same direction.
I'm trying to do a group by (Email Address) on a subset of data, then i'm using a max() on a date field but because of different values in other fields its bring back more rows then require.
I would just like to return the max record per email address and return the fields that are on the same row that are on the max record.
Not sure how i can write this query?
This is a task for ROW_NUMBER:
select *
from
(
select t.*,
-- assign sequential number starting with 1 for the maximum date
row_number() over (partiton by email_address order by datecol desc) as rn
from tab
) as dt
where rn = 1 -- only return the latest row
You can write this query using row_number():
select t.*
from (select t.*,
row_number() over (partition by emailaddress order by date desc) as seqnum
from t
) t
where seqnum = 1;
How about something like this?
select a.*
from baseTable as a
inner join
(select Email,
Max(EmailDate) as EmailDate
from baseTable
group by Email) as b
on a.Email = b.Email
and a.EmailDate = b.EmailDate
This is the sample table
What I need to achieve is to get or display only the record of tenant with the highest month value. If ever month is equal, I need to base on the latest date value. Here is the sample desired output
With this, I started by this code using max function and incorporated temp table, but unable to get the desired result.
select tenant, name, date, month
into #sample
from tenant
select *
from #sample
where months = (select max(months)from #sample)
and output to something like this. As I believe, the code is getting the max value in the whole list not considering per tenant filtering.
Any help will be greatly appreciated :)
This can be done with the row_number window function:
select tenant, name, date, months
from (select t.*,
row_number() over (partition by t.tenant, t.name order by t.months desc, t.date desc) as rn
from TableName t) x
where rn = 1
You can use a row_number function.
Query
;with cte as
(
select rn = row_number() over
(
partition by tenant
order by months desc,[date] desc
),*
from table_name
)
select tenant,name,[date],months from cte
where rn = 1;
I have table of duplicate records like
Now I want only one record from duplicate records which has latest created date as How can I do it ?
use row_number():
select EnquiryId, Name, . . .
from (select t.*,
row_number() over (partition by enquiryID order by CreatedDate desc) as seqnum
from table t
) t
where seqnum = 1;
Use ROW_NUMBER function to tag the duplicate records ordered by CreatedDate, like this:
;with CTE AS (
select *, row_NUMBER() over(
partition by EnquiryID -- add columns on which you want to identify duplicates
ORDER BY CreatedDate DESC) as rn
FROM TABLE
)
select * from CTE
where rn = 1
I'm working with Sql server 2008.i have a table contains following columns,
Id,
Name,
Date
this table contains more than one record for same id.i want to get distinct id having maximum date.how can i write sql query for this?
Use the ROW_NUMBER() function and PARTITION BY clause. Something like this:
SELECT Id, Name, Date FROM (
SELECT *, ROW_NUMBER() OVER (PARTITION BY Id ORDER BY Date desc) AS ROWNUM
FROM [MyTable]
) x WHERE ROWNUM = 1
If you need only ID column and other columns are NOT required, then you don't need to go with ROW_NUMBER or MAX or anything else. You just do a Group By over ID column, because whatever the maximum date is you will get same ID.
SELECT ID FROM table GROUP BY ID
--OR
SELECT DISTINCT ID FROM table
If you need ID and Date columns with maximum date, then simply do a Group By on ID column and select the Max Date.
SELECT ID, Max(Date) AS Date
FROM table
GROUP BY ID
If you need all the columns but 1 line having Max. date then you can go with ROW_NUMBER or MAX as mentioned in other answers.
SELECT *
FROM table AS M
WHERE Exists(
SELECT 1
FROM table
WHERE ID = M.ID
HAVING M.Date = Max(Date)
)
One way, using ROW_NUMBER:
With CTE As
(
SELECT Id, Name, Date, Rn = Row_Number() Over (Partition By Id
Order By Date DESC)
FROM dbo.TableName
)
SELECT Id --, Name, Date
FROM CTE
WHERE Rn = 1
If multiple max-dates are possible and you want all you could use DENSE_RANK instead.
Here's an overview of sql-server's ranking function: http://technet.microsoft.com/en-us/library/ms189798.aspx
By the way, CTE is a common-table-expression which is similar to a named sub-query. I'm using it to be able to filter by the row_number. This approach allows to select all columns if you want.
select Max(Date) as "Max Date"
from table
group by Id
order by Id
Try with Max(Date) and GROUP BY the other two columns (the ones with repeating data)..
SELECT ID, Max(Date) as date, Name
FROM YourTable
GROUP BY ID, Name
You may try with this
DECLARE #T TABLE(ID INT, NAME VARCHAR(50),DATE DATETIME)
INSERT INTO #T VALUES(1,'A','2014-04-20'),(1,'A','2014-04-28')
,(2,'A2','2014-04-22'),(2,'A2','2014-04-24')
,(3,'A3','2014-04-20'),(3,'A3','2014-04-28')
,(4,'A4','2014-04-28'),(4,'A4','2014-04-28')
,(5,'A5','2014-04-28'),(5,'A5','2014-04-28')
SELECT T.ID FROM #T T
WHERE T.DATE=(SELECT MAX(A.DATE)
FROM #T A
WHERE A.ID=T.ID
GROUP BY A.ID )
GROUP BY T.ID
select id, max(date) from NameOfYourTable group by id;