If I have a table with three columns like below
CREATE TABLE QUOTATIONS
(
ID INT NOT NULL,
VERSION INT NOT NULL,
PRICE FLOAT NOT NULL
);
In addition to this, lets say that the table consists of the follow records:
ID | VERSION | PRICE
----+-----------+--------
1 | 1 | 50
1 | 2 | 40
1 | 3 | 30
2 | 1 | 100
2 | 2 | 80
3 | 1 | 50
Is there any single SQL query that can be run and return the rows of all quotations with the highest version only?
The results should be like follow:
ID | VERSION | PRICE
----+-----------+--------
1 | 3 | 30
2 | 2 | 80
3 | 1 | 50
I like this method which uses no subqueries:
select top (1) with ties q.*
from quotations q
order by row_number() over (partition by id order by version desc);
Basically, the row_number() assigns "1" to the highest version for each id. The top (1) with ties returns all the 1s.
SELECT id,version,price
FROM tableName t
JOIN (SELECT id,MAX(version) AS version
FROM tableName
GROUP BY id) AS q1 ON q1.id = t.id AND q1.version = t.version
Related
I have a table:
| id | Number |Address
| -----| ------------|-----------
| 1 | 0 | NULL
| 1 | 1 | NULL
| 1 | 2 | 50
| 1 | 3 | NULL
| 2 | 0 | 10
| 3 | 1 | 30
| 3 | 2 | 20
| 3 | 3 | 20
| 4 | 0 | 75
| 4 | 1 | 22
| 4 | 2 | 30
| 5 | 0 | NULL
I need to get: the NUMBER of the last ADDRESS change for each ID.
I wrote this select:
select dh.id, dh.number from table dh where dh =
(select max(min(t.history)) from table t where t.id = dh.id group by t.address)
But this select not correctly handling the case when the address first changed, and then changed to the previous value. For example id=1: group by return:
| Number |
| -------- |
| NULL |
| 50 |
I have been thinking about this select for several days, and I will be happy to receive any help.
You can do this using row_number() -- twice:
select t.id, min(number)
from (select t.*,
row_number() over (partition by id order by number desc) as seqnum1,
row_number() over (partition by id, address order by number desc) as seqnum2
from t
) t
where seqnum1 = seqnum2
group by id;
What this does is enumerate the rows by number in descending order:
Once per id.
Once per id and address.
These values are the same only when the value is 1, which is the most recent address in the data. Then aggregation pulls back the earliest row in this group.
I answered my question myself, if anyone needs it, my solution:
select * from table dh1 where dh1.number = (
select max(x.number)
from (
select
dh2.id, dh2.number, dh2.address, lag(dh2.address) over(order by dh2.number asc) as prev
from table dh2 where dh1.id=dh2.id
) x
where NVL(x.address, 0) <> NVL(x.prev, 0)
);
Say I have a table with the following results:
How is it possible for me to select such that I only want distinct parent_ids with the min result of object0_behaviour?
Expected output:
parent_id | id | object0_behaviour | type
------------------------------------------
1 | 1 | 5 | IP
2 | 3 | 5 | IP
3 | 5 | 7 | ID
4 | 6 | 7 | ID
5 | 8 | 5 | IP
6 | 18 | 7 | ID
7 | 10 | 7 | ID
8 | 9 | 5 | IP
I have tried:
SELECT parent_id, min(object0_behaviour) FROM table GROUP BY parent_id
It works, however if I wanted the other 2 additional columns, I am required to add into GROUP BY clause and things go back to square one.
I saw examples with R : Select the first row by group
Similar output from what I need, but I can't seem to convert it into SQL
You can try using row_number() window function
select * from
(
select *, row_number() over(partition by parent_id order by object0_behaviour) as rn
from tablename
)A where rn=1
select * from table
join (
SELECT parent_id, min(object0_behaviour) object0_behaviour
FROM table GROUP BY parent_id
) grouped
on grouped.parent_id = table.parent_id
and grouped.object0_behaviour = table.object0_behaviour
I have a table which looks like this:
id | name| fk_something
----------------
0 | 25 | 3
1 | 25 | 2
2 | 23 | 1
and I want to add another column with a number which increments everytime row name occurs, e.g.:
id | name| fk_something| n
--------------------------
0 | 25 | 3 | 1
1 | 25 | 2 | 2
2 | 23 | 1 | 1
I'm not really sure how to achieve this. Using count() I will only get the total number of occurances of name but I want to increment n so that I have a distinct value for each row.
You want row_number() :
select t.*, row_number() over (partition by name order by id) as n
from table t;
You may try using COUNT as an analytic function:
SELECT
id,
name,
fk_something,
COUNT(*) OVER (PARTITION BY name ORDER BY id) n
FROM yourTable
ORDER BY
id;
Demo
I have some queries in SQL Server. I have two tables
keyword_text
Keyword_relate
Columns in keyword_text:
key_id
keywords
Columns in keyword_relate:
key_id
product_id
score
status
Sample data for keyword_text:
----|----------
1 | Pencil
2 | Pen
3 | Books
Sample data for keyword_relate:
----------------------------
Sno| Product | SCore|status
---------------------------
1 | 124 | 2 | 1
1 | 125 | 3 | 1
2 | 124 | 3 | 1
2 | 125 | 2 | 1
From this I want to get the product_id, grouped by keywords and which have maximum score
Presuming that key_id of first table is Sno in second table. You can use ROW_NUMBER:
WITH CTE AS
(
SELECT Product AS ProductID, Score As MaxScore,
RN = ROW_NUMBER() OVER (PARTITION BY kt.key_id ORDER BY Score DESC)
FROM keyword_text kt INNER JOIN keyword_relate kr
ON kt.key_id = kr.Sno
)
SELECT ProductID, MaxScore
FROM CTE
WHERE RN = 1
I have this table :
+------+-------+------------------------------------+
| id | rev | class |
+------+-------+------------------------------------+
| 1 | 10 | 2 |
| 1 | 10 | 5 |
| 2 | 40 | 6 |
| 2 | 50 | 6 |
| 2 | 52 | 1 |
| 3 | 33 | 3 |
| 3 | 63 | 5 |
+------+-------+------------------------------------+
I only need the rows where rev AND then class columns have max value.
+------+-------+------------------------------------+
| id | rev | class |
+------+-------+------------------------------------+
| 1 | 10 | 5 |
| 2 | 52 | 1 |
| 3 | 63 | 5 |
+------+-------+------------------------------------+
Query cost is important for me.
Just the rows that satisfy the condition that it has both max values?
Here's an SQL Fiddle;
SELECT h.id, h.rev, h.class
FROM ( SELECT id,
MAX( rev ) rev,
MAX( class ) class
FROM Herp
GROUP BY id ) derp
INNER JOIN Herp h
ON h.rev = derp.rev
AND h.class = derp.class;
The fastest way might be to have an index on t(id, rev) and t(id, class) and then do:
select t.*
from table t
where not exists (select 1
from table t2
where t2.id = t.id and t2.rev > t.rev
) and
not exists (select 1
from table t2
where t2.id = t.id and t2.class > t.class
);
SQL Server is pretty smart in terms of optimization, so the aggregation approach might be just as good. However, in terms of performance, this is just a bunch of index lookups.
Here is a SQL 2012 example. Very straight forward with the implied table and the PARTITION function.
Basically, with each ID as a partition/group, sort the values of the other fields in a descending order assigning each one an incrementing RowId, then only take the first one.
select id, rev, [class]
from
(
SELECT id, rev, [class],
ROW_NUMBER() OVER(PARTITION BY id ORDER BY rev DESC, [class] desc) AS RowId
FROM sample
) t
where RowId = 1
Here is the SQL Fiddle
Keep in mind, this works with the criteria in the example dataset, and not the MAX of two fields as stated in the question's title.
I guess you mean: the max of rev and the max of class. If not, please clarify what to do when there is no row where both fields have the highest value.
select id
, max(rev)
, max(class)
from table
group
by id
If you mean total value of rev and class use this:
select id
, max
, rev
from table
where id in
( select id
, max(rev + class)
from table
group
by id
)