I need to select the data of all my customers with the records displayed in the image. But I need to get the most recent record only, for example I need to get the order # E987 for John and E888 for Adam. As you can see from the example, when I do the select statement, I get all the order records.
You don't mention the specific database, so I'll answer with a generic solution.
You can do:
select *
from (
select t.*,
row_number() over(partition by name order by order_date desc) as rn
from t
) x
where rn = 1
You can use analytical function row_number.
Select * from
(Select t.*,
Row_number() over (partition by customer_id order by order_date desc) as rn
From your_table t) t
Where rn = 1
Or you can use not exists as follows:
Select *
From yoir_table t
Where not exists
(Select 1 from your_table tt
Where t.customer_id = tt.custome_id
And tt.order_date > t.order_date)
You can do it with a subquery that finds the last order date.
SELECT t.*
FROM yoir_table t
JOIN (SELECT tt.custome_id,
MAX(tt.order_date) MaxOrderDate
FROM yoir_table tt
GROUP BY tt.custome_id) AS tt
ON t.custome_id = tt.custome_id
AND t.order_date = tt.MaxOrderDate
I want to get the latest record from my source table based on num and id columns and insert in my target table.
Scenario is explained in the attached screen shot. For latest record date column can be used.
Screenshot
Thanks.
Select num,id, date
FROM
(
Select *, ROW_NUMBER() OVER(partition by num,id Order by date desc) as rnk
FROM source_table
)a
WHERE rnk = 1;
by using corelated Subquery
select * from your_table t
where t.date= (
select max(date) from your_table t1
where t1.num=t.num and t1.id=t.id
)
You can do it using max() function
select num,id,max(date) from your_table t
group by num,id
SELECT NUM,ID,DATE FROM TABLE_TEMP
QUALIFY RANK OVER(PARTITION BY NUM,ID ORDER BY DATE DESC)=1;
You can do this using single line query
SELECT NUM,ID,DATE FROM TABLE_TEMP
QUALIFY RANK OVER(PARTITION BY NUM,ID ORDER BY DATE DESC)=1;
I have a table like this,
Date User
15-06-2018 A
16-06-2018 A
15-06-2018 B
14-06-2018 C
16-06-2018 C
I want to get the output like this,
Date User
16-06-2018 A
15-06-2018 B
16-06-2018 C
I tried Select Max(date),User from Table group by User
Based on your comment, I assume you have duplicated results in those 80 columns when you group by them. Assuming so, here's one option using row_number to always return 1 row per user:
select *
from (
select *, row_number() over (partition by user order by date desc) rn
from yourtable
) t
where rn = 1
You can use correlation subquery :
select t.*
from table t
where date = (select max(t1.date)
from table t1
where t1.user = t.user
);
However, i would also recommend row_number() :
select top (1) with ties *
from table t
order by row_number() over (partition by user order by date desc);
You can also use a ranking function
SELECT User, Date
FROM
(
SELECT User, Date
, Row_id = Row_Number() OVER (Partition by User, ORDER BY User, Date desc)
FROM table
)q
WHERE Row_Id = 1
I would suggest you this
Select * from table t where exist
(Select 1 from
(Select user, max(date) as date from table) A
Where A.user = t.user and A.date = t.date )
I think it's easier to show you an image:
So, for each fld_call_id, go to the next value, if it's identical. When we get to the last value, I need the value in column fld_menu_id.
Or, to put it in another way, eliminate fld_call_id duplicates and save only the last one.
You can use ROW_NUMBER:
WITH CTE AS(
SELECT RN = ROW_NUMBER() OVER (PARTITION BY fld_call_id ORDER BY fld_id DESC),
fld_menu_id
FROM dbo.TableName
)
SELECT fld_menu_id FROM CTE WHERE RN = 1
You can create a Rank column and only select that row, something along the lines of the following:
;WITH cte AS
(
SELECT
*
,RANK() OVER (PARTITION BY fld_call_id ORDER BY fld_id DESC) Rnk
FROM YourTable
)
SELECT
*
FROM cte
WHERE Rnk=1
So you GROUP BY fld_call_id and ORDER BY fld_id in descending order so that the last value comes first. These are the rows where Rnk=1.
Edit after comments of OP.
SELECT Table.*
FROM Table
INNER JOIN
(
SELECT MAX(fldMenuID) AS fldMenuID,
fldCallID
FROM Table
GROUP BY fldCallID
) maxValues
ON (maxValues.fldMenuID = Table.fldMenuID
AND maxValues.fldCallID= Table.fldCallID)
Hope This works
SELECT A.*
FROM table A
JOIN (SELECT fld_id,
ROW_NUMBER() OVER (PARTITION BY Fld_call_id ORDER BY fld_id DESC) [Row]
FROM table) LU ON A.fld_id = LU.fld_id
WHERE LU.[Row] = 1
I have a table that is a collection entries as to when a user was logged on.
username, date, value
--------------------------
brad, 1/2/2010, 1.1
fred, 1/3/2010, 1.0
bob, 8/4/2009, 1.5
brad, 2/2/2010, 1.2
fred, 12/2/2009, 1.3
etc..
How do I create a query that would give me the latest date for each user?
Update: I forgot that I needed to have a value that goes along with the latest date.
This is the simple old school approach that works with almost any db engine, but you have to watch out for duplicates:
select t.username, t.date, t.value
from MyTable t
inner join (
select username, max(date) as MaxDate
from MyTable
group by username
) tm on t.username = tm.username and t.date = tm.MaxDate
Using window functions will avoid any possible issues with duplicate records due to duplicate date values, so if your db engine allows it you can do something like this:
select x.username, x.date, x.value
from (
select username, date, value,
row_number() over (partition by username order by date desc) as _rn
from MyTable
) x
where x._rn = 1
Using window functions (works in Oracle, Postgres 8.4, SQL Server 2005, DB2, Sybase, Firebird 3.0, MariaDB 10.3)
select * from (
select
username,
date,
value,
row_number() over(partition by username order by date desc) as rn
from
yourtable
) t
where t.rn = 1
I see most of the developers use an inline query without considering its impact on huge data.
Simply, you can achieve this by:
SELECT a.username, a.date, a.value
FROM myTable a
LEFT OUTER JOIN myTable b
ON a.username = b.username
AND a.date < b.date
WHERE b.username IS NULL
ORDER BY a.date desc;
From my experience the fastest way is to take each row for which there is no newer row in the table.
Another advantage is that the syntax used is very simple, and that the meaning of the query is rather easy to grasp (take all rows such that no newer row exists for the username being considered).
NOT EXISTS
SELECT username, value
FROM t
WHERE NOT EXISTS (
SELECT *
FROM t AS witness
WHERE witness.username = t.username AND witness.date > t.date
);
ROW_NUMBER
SELECT username, value
FROM (
SELECT username, value, row_number() OVER (PARTITION BY username ORDER BY date DESC) AS rn
FROM t
) t2
WHERE rn = 1
INNER JOIN
SELECT t.username, t.value
FROM t
INNER JOIN (
SELECT username, MAX(date) AS date
FROM t
GROUP BY username
) tm ON t.username = tm.username AND t.date = tm.date;
LEFT OUTER JOIN
SELECT username, value
FROM t
LEFT OUTER JOIN t AS w ON t.username = w.username AND t.date < w.date
WHERE w.username IS NULL
To get the whole row containing the max date for the user:
select username, date, value
from tablename where (username, date) in (
select username, max(date) as date
from tablename
group by username
)
SELECT *
FROM MyTable T1
WHERE date = (
SELECT max(date)
FROM MyTable T2
WHERE T1.username=T2.username
)
This one should give you the correct result for your edited question.
The sub-query makes sure to find only rows of the latest date, and the outer GROUP BY will take care of ties. When there are two entries for the same date for the same user, it will return the one with the highest value.
SELECT t.username, t.date, MAX( t.value ) value
FROM your_table t
JOIN (
SELECT username, MAX( date ) date
FROM your_table
GROUP BY username
) x ON ( x.username = t.username AND x.date = t.date )
GROUP BY t.username, t.date
If your database syntax supports it, then TOP 1 WITH TIES can be a lifesafer in combination with ROWNUMER.
With the example data you provided, use this query:
SELECT TOP 1 WITH TIES
username, date, value
FROM user_log_in_attempts
ORDER BY ROW_NUMBER() OVER (PARTITION BY username ORDER BY date DESC)
It yields:
username | date | value
-----------------------------
bob | 8/4/2009 | 1.5
brad | 2/2/2010 | 1.2
fred | 12/2/2009 | 1.3
Demo
How it works:
ROWNUMBER() OVER (PARTITION BY... ORDER BY...) For each username a list of rows is calculated from the youngest (rownumber=1) to the oldest (rownumber=high)
ORDER BY ROWNUMBER... sorts the youngest rows of each user to the top, followed by the second-youngest rows of each user, and so on
TOP 1 WITH TIES Because each user has a youngest row, those youngest rows are equal in the sense of the sorting criteria (all have rownumber=1). All those youngest rows will be returned.
Tested with SQL-Server.
SELECT DISTINCT Username, Dates,value
FROM TableName
WHERE Dates IN (SELECT MAX(Dates) FROM TableName GROUP BY Username)
Username Dates value
bob 2010-02-02 1.2
brad 2010-01-02 1.1
fred 2010-01-03 1.0
This is similar to one of the answers above, but in my opinion it is a lot simpler and tidier. Also, shows a good use for the cross apply statement. For SQL Server 2005 and above...
select
a.username,
a.date,
a.value,
from yourtable a
cross apply (select max(date) 'maxdate' from yourtable a1 where a.username=a1.username) b
where a.date=b.maxdate
You could also use analytical Rank Function
with temp as
(
select username, date, RANK() over (partition by username order by date desc) as rnk from t
)
select username, rnk from t where rnk = 1
SELECT MAX(DATE) AS dates
FROM assignment
JOIN paper_submission_detail ON assignment.PAPER_SUB_ID =
paper_submission_detail.PAPER_SUB_ID
SELECT Username, date, value
from MyTable mt
inner join (select username, max(date) date
from MyTable
group by username) sub
on sub.username = mt.username
and sub.date = mt.date
Would address the updated problem. It might not work so well on large tables, even with good indexing.
SELECT *
FROM ReportStatus c
inner join ( SELECT
MAX(Date) AS MaxDate
FROM ReportStatus ) m
on c.date = m.maxdate
For Oracle sorts the result set in descending order and takes the first record, so you will get the latest record:
select * from mytable
where rownum = 1
order by date desc
SELECT t1.username, t1.date, value
FROM MyTable as t1
INNER JOIN (SELECT username, MAX(date)
FROM MyTable
GROUP BY username) as t2 ON t2.username = t1.username AND t2.date = t1.date
Select * from table1 where lastest_date=(select Max(latest_date) from table1 where user=yourUserName)
Inner Query will return the latest date for the current user, Outer query will pull all the data according to the inner query result.
I used this way to take the last record for each user that I have on my table.
It was a query to get last location for salesman as per recent time detected on PDA devices.
CREATE FUNCTION dbo.UsersLocation()
RETURNS TABLE
AS
RETURN
Select GS.UserID, MAX(GS.UTCDateTime) 'LastDate'
From USERGPS GS
where year(GS.UTCDateTime) = YEAR(GETDATE())
Group By GS.UserID
GO
select gs.UserID, sl.LastDate, gs.Latitude , gs.Longitude
from USERGPS gs
inner join USER s on gs.SalesManNo = s.SalesmanNo
inner join dbo.UsersLocation() sl on gs.UserID= sl.UserID and gs.UTCDateTime = sl.LastDate
order by LastDate desc
My small compilation
self join better than nested select
but group by doesn't give you primary key which is preferable for join
this key can be given by partition by in conjunction with first_value (docs)
So, here is a query:
select
t.*
from
Table t inner join (
select distinct first_value(ID) over(partition by GroupColumn order by DateColumn desc) as ID
from Table
where FilterColumn = 'value'
) j on t.ID = j.ID
Pros:
Filter data with where statement using any column
select any columns from filtered rows
Cons:
Need MS SQL Server starting with 2012.
I did somewhat for my application as it:
Below is the query:
select distinct i.userId,i.statusCheck, l.userName from internetstatus
as i inner join login as l on i.userID=l.userID
where nowtime in((select max(nowtime) from InternetStatus group by userID));
Here's one way to return only the most recent record for each user in SQL Server:
WITH CTE AS (
SELECT *, ROW_NUMBER() OVER (PARTITION BY user_id ORDER BY date DESC) AS rn
FROM your_table
)
SELECT *
FROM CTE
WHERE rn = 1;
This uses a common table expression (CTE) to assign a unique rn (row number) to each record for each user, based on the user_id and sorted in descending order by date. The final query then selects only the records with rn equal to 1, which represents the most recent record for each user.
SELECT * FROM TABEL1 WHERE DATE= (SELECT MAX(CREATED_DATE) FROM TABEL1)
You would use aggregate function MAX and GROUP BY
SELECT username, MAX(date), value FROM tablename GROUP BY username, value