I have this Table with one row Transaction Date the first row is the checkIn and the second one is the Checkout if we organized the result by date asc.
I need to pass the second row value to another column named Checkout. This table has at least 1000 records
Try using the query below
With CteCheckOut as(
Select
ROW_NUMBER() over (Partition by studentID Order by studentID,transactionDate) as Rownumber,
*
From student
)
Select studentID, transactionDate as CheckOutDate
From CteCheckOut
Where Rownumber%2 = 0
You can use row_number():
select t.*
from (select t.*,
row_number() over (partition by studentid order by transactiondate) as seqnum
from t
) t
where seqnum = 2;
This assumes that only studentid is used to identify the "first" and "second" rows. If more columns are needed, add them to the partition by clause.
If the checkout date you need is the second (max) date grouped by reason and you want to single out this date then try:
select
studentID,
reason,
max(transactionDate) as checkoutDate
from student
group by
reason
Related
I have a table that contains three columns: ACCOUNT_ID, STATUS, CREATE_DATE.
I want to grab only the LAST status for each account_id based on the latest create_date.
In the example above, I should only see three records and the last STATUS per that account_2.
Do you know a way to do this?
create table TBL 1 (
account_id int,
status string,
create_date date)
select account_id, max(create_date) from table group by account_id;
will give you the account_id and create_date at the closest past date to today (assuming create_date can never be in the future, which makes sense).
Now you can join with that data to get what you want, something along the lines for example:
select account_id, status, create_date from table where (account_id, create_date) in (<the select expression from above>);
If you use that frequently (account with the latest create date), then consider defining a view for that.
If you have many columns and want keep the row that is the last line, you can use QUALIFY to run the ranking logic, and keep the best, like so:
SELECT *
FROM tbl
QUALIFY row_number() over (partition by account_id order by create_date desc) = 1;
The long form is the same pattern the Ely shows in the second answer. But with the MAX(CREATE_DATE) solution, if you have two rows on the same last day, the IN solution with give you both. you can also get via QUALIFY if you use RANK
So the SQL is the same as:
SELECT account_id, status, create_date
FROM (
SELECT *,
row_number() over (partition by account_id order by create_date desc) as rn
FROM tbl
)
WHERE rn = 1;
So the RANK for, which will show all equal rows is:
SELECT *
FROM tbl
QUALIFY rank() over (partition by account_id order by create_date desc) = 1;
I have a table with multiple rows of the same member id. I need only distinct rows based on 2 unique columns
Ex: there are 100 different customers, the table has 1000 rows because every customer has multiple cities and segments assigned to him.
I need 100 distinct rows for these customers depending on a unique segment and city combination. There is no specific requirement for this combination, just the first from the table is fine.
So, currently the table is somewhat like this,
Hope this helps.
use row_number()
select * from (select *,row_number() over(partition by memberid order by sales) rn
from table_name
) a where a.rn=1
Handy sql-server top(1) with ties syntax for that
select top(1) with ties t.*
from table_name t
order by row_number() over(partition by memberid order by sales)
As you have no paticular requirement for which exactly row to select, any column will do at order by, it can be null as well
select top(1) with ties t.*
from table_name t
order by row_number() over(partition by memberid order by (select null))
The simplest way to do this is to use the ROW_NUMBER() OVER(GROUP BY...) syntax. You have no need to use an order by, since you want an arbitrary row, but only one, for each member.
Since you need only the expected data, and not the Row_Number value, make sure that you detail the fields returned, like below:
SELECT
MemberId,
city,
segment,
sales
FROM (
SELECT *
ROW_NUMBER() OVER (GROUP BY MemberId) as Seq
FROM [Status]
) src
WHERE Seq = 1
I've a situation where there is one ticket history table. it saves all the actions done against a ticket. how to write a query which will return the first record and the last record against specific ticket.
for example in the above table I've one ticket with id 78580. I want to get the first row and last row based on date column.
Just use row_number():
select t.*
from (select t.*,
row_number() over (partition by ticket_id order by action_when asc) as seqnum_a,
row_number() over (partition by ticket_id order by action_when desc) as seqnum_d
from tickets t
) t
where seqnum_a = 1 or seqnum_d = 1;
Use min and max to get first and last date, grouped by ticket id.
SELECT ticket_id, min(action_when), max(action_when)
FROM table_name
GROUP BY ticket_id;
Using R, I want to grab the two most recently dated entries for each UserID, assuming there is 1 or more entries per UserID.
The key elements of my data would be an identifier (UserID), and a date, that is of type date.
Thank you.
In SQL Server, which has the ROW_NUMBER() analytic function, you can try this query:
SELECT t.UserID, t.date, ...other columns
FROM
(
SELECT UserID, date, ...other columns,
ROW_NUMBER() OVER (PARTITION BY UserID ORDER BY date DESC) rn
FROM yourTable
) t
WHERE t.rn <= 2
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;