This question already has answers here:
Retrieving last record in each group from database - SQL Server 2005/2008
(2 answers)
Closed 2 years ago.
I have a table that contains items with three values Id, RevisionId and Data like this.
| Id | RevisionId | Data |
| 0 | 0 | Value1 |
| 0 | 1 | Value2 |
| 1 | 0 | Value1 |
| 2 | 0 | Value1 |
| 2 | 1 | Value2 |
| 2 | 3 | Value3 |
If I want only the Id with the highest RevisionId for each I can make an SQL statement like this.
SELECT Id, MAX(RevisionId) AS RevisionId FROM RevisionTable GROUP BY Id
But I don't know how to get the Data value connected to that max RevisionId for each Id.
You can use row_number():
select *
from (
select t.*, row_number() over(partition by id order by revisionid desc) rn
from mytable t
) t
where rn = 1
top(1) with ties also comes to mind:
select top (1) with ties t.*
from mytable t
order by row_number() over(partition by id order by revisionid desc) rn
Related
Given the following example table:
+-----------+
| Id | Name |
+----+------+
| 1 | A |
| 2 | B |
| 3 | B |
| 4 | C |
| 5 | A |
| 6 | B |
| 7 | B |
| 8 | B |
| 9 | B |
| 10 | X |
+----+------+
I would like a query to get the following result:
+----+------+
| 6 | B |
| 7 | B |
| 8 | B |
| 9 | B |
+----+------+
The best query I could do was:
SELECT * FROM
(SELECT id, name, LEAD(id) OVER (ORDER BY id) t
FROM test WHERE name = 'B' ORDER BY id)
WHERE ID <> t-1;
sqlfiddle here
If you want the length and where it starts:
select min(id), max(id)
from (select t.*,
row_number() over (order by id) as seqnum,
row_number() over (partition by name order by id) as seqnum_1
from test t
) t
where name = 'B'
group by (seqnum - seqnum_1)
order by min(id) desc
fetch first 1 row only;
You can join back to the table to get the original rows.
Another method using window functions to count the number of non-Bs after a given row . . . and then choose the first:
select t.*
from (select t.*,
dense_rank() over (order by nonbs_after asc) as grp
from (select t.*,
sum(case when name <> 'B' then 1 else 0 end) over (order by id desc) as nonbs_after
from test t
) t
where name = 'B'
) t
where grp = 1;
Here is a db<>fiddle.
i have a table in a postgres DB which has the following structure:
id | date | groupme1 | groupme2 | value
----------------------------------------
1 |
2 |
3 |
Now i want to achieve the following:
Grouping the table after groupme1 and groupme2
Get the value for every group
But only the last entry for each group-compination (odered after date)
Example:
id | date | groupme1 | groupme2 | value
---------------------------------------
| | A | 1 | 4
| | A | 2 | 7
| | A | 3 | 3
| | B | 1 | 9
My current approach looks like this:
SELECT a.*
FROM table AS a
JOIN (SELECT max(id) AS id
FROM table
GROUP BY groupme1, groupme2) AS b
ON a.id = b.id
The Problems of this approach:
it asumes that higher dates have a higher id
it takes long
Is there a faster and better way of doing this? Can windowing function help with this?
I think you just want window functions:
select t.*
from (select t.*,
row_number() over (partition by groupme1, groupme2 order by date desc) as seqnum
from t
) t
where seqnum = 1;
Or, a better way to do this in Postgres uses distinct on:
select distinct on (groupme1, groupme2) t.*
from t
order by groupme1, groupme2, date desc;
This question already has answers here:
Select first row in each GROUP BY group?
(20 answers)
Closed 8 years ago.
I have a table with fields:
id
category_id
property_id
Weight
How can I get list with {category_id, property_id} where property_id from the highest weight?
Example:
id | category_id | property_id | weight |
1 | 1 | 1 | 20 |
2 | 1 | 2 | 10 |
3 | 2 | 2 | 30 |
4 | 2 | 3 | 40 |
Right results after query:
category_id | property_id
1 | 1 (because 20 > 10)
2 | 3 (because 40 > 30)
It is simple issue, but I am searching easiest and right way, how I can do it with postgresql instruments without subqueries and temporary tables.
Use distinct on:
select distinct on (category_id) t.*
from tablewithfields t
order by category_id, weight desc;
EDIT:
You can do this with window functions, but the above is probably more efficient:
select t.*
from (select t.*, row_number() over (partition by category_id order by weight desc) as seqnum
from tablewithfields t
) t
where seqnum = 1;
I have this table shown below...How do I select only the latest data of the id based on changeno?
+----+--------------+------------+--------+
| id | data | changeno | |
+----+--------------+------------+--------+
| 1 | Yes | 1 | |
| 2 | Yes | 2 | |
| 2 | Maybe | 3 | |
| 3 | Yes | 4 | |
| 3 | Yes | 5 | |
| 3 | No | 6 | |
| 4 | No | 7 | |
| 5 | Maybe | 8 | |
| 5 | Yes | 9 | |
+----+---------+------------+-------------+
I would want this result...
+----+--------------+------------+--------+
| id | data | changeno | |
+----+--------------+------------+--------+
| 1 | Yes | 1 | |
| 2 | Maybe | 3 | |
| 3 | No | 6 | |
| 4 | No | 7 | |
| 5 | Yes | 9 | |
+----+---------+------------+-------------+
I currently have this SQL statement...
SELECT id, data, MAX(changeno) as changeno FROM Table1 GROUP BY id;
and clearly it doesn't return what I want. This should return an error because of the aggrerate function. If I added fields under the GROUP BY clause it works but it doesn't return what I want. The SQL statement is by far the closest I could think of. I'd appreciate it if anybody could help me on this. Thank you in advance :)
This is typically referred to as the "greatest-n-per-group" problem. One way to solve this in SQL Server 2005 and higher is to use a CTE with a calculated ROW_NUMBER() based on the grouping of the id column, and sorting those by largest changeno first:
;WITH cte AS
(
SELECT id, data, changeno,
rn = ROW_NUMBER() OVER (PARTITION BY id ORDER BY changeno DESC)
FROM dbo.Table1
)
SELECT id, data, changeno
FROM cte
WHERE rn = 1
ORDER BY id;
You want to use row_number() for this:
select id, data, changeno
from (SELECT t.*,
row_number() over (partition by id order by changeno desc) as seqnum
FROM Table1 t
) t
where seqnum = 1;
Not a well formed or performance optimized query but for small tasks it works fine.
SELECT * FROM TEST
WHERE changeno IN (SELECT MAX(changeno)
FROM TEST
GROUP BY id)
for other alternatives :
DECLARE #Table1 TABLE
(
id INT, data VARCHAR(5), changeno INT
);
INSERT INTO #Table1
SELECT 1,'Yes',1
UNION ALL
SELECT 2,'Yes',2
UNION ALL
SELECT 2,'Maybe',3
UNION ALL
SELECT 3,'Yes',4
UNION ALL
SELECT 3,'Yes',5
UNION ALL
SELECT 3,'No',6
UNION ALL
SELECT 4,'No',7
UNION ALL
SELECT 5,'Maybe',8
UNION ALL
SELECT 5,'Yes',9
SELECT Y.id, Y.data, Y.changeno
FROM #Table1 Y
INNER JOIN (
SELECT id, changeno = MAX(changeno)
FROM #Table1
GROUP BY id
) X ON X.id = Y.id
WHERE X.changeno = Y.changeno
ORDER BY Y.id
I have table in my sql server database which I want to convert to PK column
To do that I want to change value of each row in this column to 1,2,3 ...
Could You write T-Sql query for that task ?
Thanks for help
begin state:
Id | Name |
----------
1 | One |
2 | Two |
2 | Three|
x | xxx |
result:
Id | Name |
----------
1 | One |
2 | Two |
3 | Three|
4 | xxx |
;with cte as
(
SELECT Id, ROW_NUMBER() over (order by Id) as rn
from YourTable
)
UPDATE cte SET Id = rn
you can also do it with name if you dont have the id!
;with cte as
(
SELECT Id, ROW_NUMBER() over (order by name) as rn
from YourTable
)
UPDATE cte SET Id = rn