I have a SQL table where one field can have the same value in multiple records. For example:
I would like to select the EndTime and UserId of each UserID once. If there are multiple records with the same UserID I want to select the most recent record and override the past ones.
I'm working with Sqlite in python.
You can use a window function such as ROW_NUMBER()
SELECT EndTime, userID
FROM
(
SELECT ROW_NUMBER() OVER (PARTITION BY userID ORDER BY StartTime DESC) AS rn,
t.*
FROM t
) AS tt
WHERE rn = 1
where grouping by userID is provided by PARTITION BY userID, and picking the recent values is provided through sorting descendingly by StartTime
Demo
All you need is aggregation:
SELECT UserID, MAX(StartTime) StartTime, EndTime
FROM tablename
GROUP BY UserID;
because SQLite will return for each UserID the row with the max StartTime.
See the demo.
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;
These are the columns in my table
id (autogenerated)
created_user
created_date
post_text
This table has lot of values. I wanted to take latest 3 posts of every created_user
I am new to SQL and need help. I ran the below query in my Postgres database and it is not helpful
SELECT * FROM posts WHERE created_date IN
(SELECT MAX(created_date) FROM posts GROUP BY created_date)
You could use the row_number() window function to create an ordered row count per user. After that you can easily filter by this value
demo:db<>fiddle
SELECT
*
FROM (
SELECT
*,
row_number() OVER (PARTITION BY created_user ORDER BY created_date DESC)
FROM
posts
) s
WHERE row_number <= 3
PARTITION BY groups the users
ORDER BY date DESC orders the posts of each user into a descending order to get the most recent as row_count == 1, ...
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
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;