I have table name m_option:
m_option_id m_student_id value
1 1 5
2 1 5
3 1 6
4 1 7
5 2 1
6 2 2
7 2 3
8 2 3
9 2 4
I want to get the 2 rows with min value for each m_student_id:
m_option_id m_student_id value
1 1 5
2 1 5
5 2 1
6 2 2
You can use the row_number window function for that:
SELECT m_option_id, m_student_id, value
FROM (
SELECT
m_option_id, m_student_id, value,
row_number() OVER (PARTITION BY m_student_id ORDER BY value)
FROM m_option
) t
WHERE
row_number <= 2;
row_number will calculate the number of each row within its group. We then use that number to filter the the top 2 rows (i.e. lowest value) from each group.
Alternatively, you could use a LATERAL subquery:
SELECT m_option_id, m_student_id, value
FROM (SELECT DISTINCT m_student_id FROM m_option) s,
LATERAL (
SELECT m_option_id, value
FROM m_option
WHERE s.m_student_id=m_student_id
ORDER BY value
LIMIT 2
) t;
This will go through all distinct values of m_student_id and for each one of them will find the top 2 rows using a LATERAL subquery.
Assuming there can be many rows per student in table m_option, the key to performance is index usage. And that's most efficient if you have a separate student table listing all students uniquely (which you would typically have). Then:
SELECT m.m_option_id, s.student_id AS m_student_id, m.value
FROM student s
, LATERAL (
SELECT m_option_id, value
FROM m_option
WHERE m_student_id = s.student_id -- PK of table student
ORDER BY value
LIMIT 2
) m;
A multicolumn index on m_option makes this fast:
CREATE INDEX m_option_combo_idx ON m_option (m_student_id, value);
If you can get index-only scans out of it, append the column m_option_id as last index item:
CREATE INDEX m_option_combo_idx ON m_option (m_student_id, value, m_option_id)
Index columns in this order.
Is a composite index also good for queries on the first field?
Distilling a unique list of student_id from m_option would incur an expensive sequential scan over m_option and void any performance benefit.
This excludes students without any related rows in m_option. Use LEFT JOIN LATERAL () ON true to include such students in the result (extended with NULL values for the missing option):
What is the difference between LATERAL and a subquery in PostgreSQL?
If you do not have a student table, the other fast option is a recursive CTE.
Detailed explanation for either variant:
Optimize GROUP BY query to retrieve latest record per user
Related
How to repeat rows based on column value in snowflake using sql.
I tried a few methods but not working such as dual and connect by.
I have two columns: Id and Quantity.
For each ID, there are different values of Quantity.
So if you have a count, you can use a generator:
with ten_rows as (
select row_number() over (order by null) as rn
from table(generator(ROWCOUNT=>10))
), data(id, count) as (
select * from values
(1,2),
(2,4)
)
SELECT
d.*
,r.rn
from data as d
join ten_rows as r
on d.count >= r.rn
order by 1,3;
ID
COUNT
RN
1
2
1
1
2
2
2
4
1
2
4
2
2
4
3
2
4
4
Ok let's start by generating some data. We will create 10 rows, with a QTY. The QTY will be randomly chosen as 1 or 2.
Next we want to duplicate the rows with a QTY of 2 and leave the QTY =1 as they are.
Obviously you can change all parameters above to suit your needs - this solution works super fast and in my opinion way better than table generation.
Simply stack SPLIT_TO_TABLE(), REPEAT() with a LATERAL() join and voila.
WITH TEN_ROWS AS (SELECT ROW_NUMBER()OVER(ORDER BY NULL)SOME_ID,UNIFORM(1,2,RANDOM())QTY FROM TABLE(GENERATOR(ROWCOUNT=>10)))
SELECT
TEN_ROWS.*
FROM
TEN_ROWS,LATERAL SPLIT_TO_TABLE(REPEAT('hire me $10/hour',QTY-1),'hire me $10/hour')ALTERNATIVE_APPROACH;
I work in healthcare. In a Postgres database, we have a table member IDs and dates. I'm trying to pull the latest two dates for each member ID.
Simplified sample data:
A 1
B 1
B 2
C 1
C 5
C 7
D 1
D 2
D 3
D 4
Desired result:
A 1
B 1
B 2
C 1
C 5
D 1
D 2
I get a strong feeling this is for a homework assignment and would recommend that you look into partitioning and specifically rank() function by yourself first before looking at my solution.
Moreover, you have not specified how you received the initial result you provided, so I'll have to assume you just did select letter_column, number_column from my_table; to achieve the result.
So, what you actually want here is partition the initial query result into groups by the letter_column and select the first two rows in each. rank() function lets you assign each row a number, counting within groups:
select letter_column,
number_column,
rank() over (partition by letter_column order by number_column) as rank
from my_table;
Since it's a function, you can't use it in a predicate in the same query, so you'll have to build another query around this one, this time filtering the results where rank is over 2:
with ranked_results as (select letter_column,
number_column,
rank() over (partition by letter_column order by number_column asc) as rank
from my_table mt)
select letter_column,
number_column
from ranked_results
where rank < 3;
Here's an SQLFiddle to play around: http://sqlfiddle.com/#!15/e90744/1/0
Hope this helps!
I am using oracle 12c database and I have a table with the following structure:
Id NUMBER
SeqNo NUMBER
Val NUMBER
Valid VARCHAR2
A composite primary key is created with the field Id and SeqNo.
I would like to fetch the data with Valid = 'Y' and apply ketset pagination with a page size of 3. Assume I have the following data:
Id SeqNo Val Valid
1 1 10 Y
1 2 20 N
1 3 30 Y
1 4 40 Y
1 5 50 Y
2 1 100 Y
2 2 200 Y
Expected result:
----------------------------
Page 1
----------------------------
Id SeqNo Val Valid
1 1 10 Y
1 3 30 Y
1 4 40 Y
----------------------------
Page 2
----------------------------
Id SeqNo Val Valid
1 5 50 Y
2 1 100 Y
2 2 200 Y
Offset pagination can be done like this:
SELECT * FROM table ORDER BY Id, SeqNo OFFSET 3 ROWS FETCH NEXT 3 ROWS ONLY;
However, in the actual db it has more than 5 millions of records and using OFFSET is going to slow down the query a lot. Therefore, I am looking for a ketset pagination approach (skip records using some unique fields instead of OFFSET)
Since a composite primary key is used, I need to offset the page with information from more than 1 field.
This is a sample SQL that should work in PostgreSQL (fetch 2nd page):
SELECT * FROM table WHERE (Id, SeqNo) > (1, 4) AND Valid = 'Y' ORDER BY Id, SeqNo LIMIT 3;
How do I achieve the same in oracle?
Use row_number() analytic function with ceil arithmetic fuction. Arithmetic functions don't have a negative impact on performance, and row_number() over (order by ...) expression automatically orders the data without considering the insertion order, and without adding an extra order by clause for the main query. So, consider :
select Id,SeqNo,
ceil(row_number() over (order by Id,SeqNo)/3) as page
from tab
where Valid = 'Y';
P.S. It also works for Oracle 11g, while OFFSET 3 ROWS FETCH NEXT 3 ROWS ONLY works only for Oracle 12c.
Demo
You can use order by and then fetch rows using fetch and offset like following:
Select ID, SEQ, VAL, VALID FROM TABLE
WHERE VALID = 'Y'
ORDER BY ID, SEQ
--FETCH FIRST 3 ROWS ONLY -- first page
--OFFSET 3 ROWS FETCH NEXT 3 ROWS ONLY -- second pages
--OFFSET 6 ROWS FETCH NEXT 3 ROWS ONLY -- third page
--Update--
You can use row_number analytical function as following.
Select id, seqNo, Val, valid from
(Select t.*,
Row_number(order by id, seq) as rn from table t
Where valid = 'Y')
Where ceil(rn/3) = 2 -- for page no. 2
Cheers!!
I have two Tables.
table A
id name Size
===================
1 Apple 7
2 Orange 15
3 Banana 22
4 Kiwi 2
5 Melon 28
6 Peach 9
And Table B
id size
==============
1 14
2 5
3 31
4 9
5 1
6 16
7 7
8 25
My desired result will be (add one column to Table A, which is the number of rows in Table B that have size smaller than Size in Table A)
id name Size Num.smaller.in.B
==============================
1 Apple 7 2
2 Orange 15 5
3 Banana 22 6
4 Kiwi 2 1
5 Melon 28 7
6 Peach 9 3
Both Table A and B are pretty huge. Is there a clever way of doing this. Thanks
Use this query it's helpful
SELECT id,
name,
Size,
(Select count(*) From TableB Where TableB.size<Size)
FROM TableA
The standard way to get your result involves a non-equi-join, which will be a product join in Explain. First duplicating 20,000 rows, followed by 7,000,000 * 20,000 comparisons and a huge intermediate spool before the count.
There's a solution based on OLAP-functions which is usually quite efficient:
SELECT dt.*,
-- Do a cumulative count of the rows of table #2
-- sorted by size, i.e. count number of rows with a size #2 less size #1
Sum(CASE WHEN NAME = '' THEN 1 ELSE 0 end)
Over (ORDER BY SIZE, NAME DESC ROWS Unbounded Preceding)
FROM
( -- mix the rows of both tables, an empty name indicates rows from table #2
SELECT id, name, size
FROM a
UNION ALL
SELECT id, '', size
FROM b
) AS dt
-- only return the rows of table #1
QUALIFY name <> ''
If there are multiple rows with the same size in table #2 you better count before the Union to reduce the size:
SELECT dt.*,
-- Do a cumulative sum of the counts of table #2
-- sorted by size, i.e. count number of rows with a size #2 less size #1
Sum(CASE WHEN NAME ='' THEN id ELSE 0 end)
Over (ORDER BY SIZE, NAME DESC ROWS Unbounded Preceding)
FROM
( -- mix the rows of both tables, an empty name indicates rows from table #2
SELECT id, name, size
FROM a
UNION ALL
SELECT Count(*), '', SIZE
FROM b
GROUP BY SIZE
) AS dt
-- only return the rows of table #1
QUALIFY NAME <> ''
There is no clever way of doing that, you just need to join the tables like this:
select a.*, b.size
from TableA a join TableB b on a.id = b.id
To improve performance you'll need to have indexes on the id columns.
maybe
select
id,
name,
a.Size,
sum(cnt) as sum_cnt
from
a inner join
(select size, count(*) as cnt from b group by size) s on
s.size < a.size
group by id,name,a.size
if you're working with large tables. Indexing table b's size field could help. I'm also assuming the values in table B converge, that there's many duplicates you don't care about, other than you want to count them.
sqlfiddle
#Ritesh solution is perfectly correct, another similar solution is using CROSS JOIN as shown below
use tempdb
create table dbo.A (id int identity, name varchar(30), size int );
create table dbo.B (id int identity, size int);
go
insert into dbo.A (name, size)
values ('Apple', 7)
,('Orange', 15)
,('Banana', 22)
,('Kiwi', 2 )
,('Melon', 28)
,('Peach', 6 )
insert into dbo.B (size)
values (14), (5),(31),(9),(1),(16), (7),(25)
go
-- using cross join
select a.*, t.cnt
from dbo.A
cross apply (select cnt=count(*) from dbo.B where B.size < A.size) T(cnt)
try this query
SELECT
A.id,A.name,A.size,Count(B.size)
from A,B
where A.size>B.size
group by A.size
order by A.id;
I'm attempting to perform a GROUP BY on a join table table. The join table essentially looks like:
CREATE TABLE user_foos (
id SERIAL PRIMARY KEY,
user_id INT NOT NULL,
foo_id INT NOT NULL,
effective_at DATETIME NOT NULL
);
ALTER TABLE user_foos
ADD CONSTRAINT user_foos_uniqueness
UNIQUE (user_id, foo_id, effective_at);
I'd like to query this table to find all records where the effective_at is the max value for any pair of user_id, foo_id given. I've tried the following:
SELECT "user_foos"."id",
"user_foos"."user_id",
"user_foos"."foo_id",
max("user_foos"."effective_at")
FROM "user_foos"
GROUP BY "user_foos"."user_id", "user_foos"."foo_id";
Unfortunately, this results in the error:
column "user_foos.id" must appear in the GROUP BY clause or be used in an aggregate function
I understand that the problem relates to "id" not being used in an aggregate function and that the DB doesn't know what to do if it finds multiple records with differing ID's, but I know this could never happen due to my trinary primary key across those columns (user_id, foo_id, and effective_at).
To work around this, I also tried a number of other variants such as using the first_value window function on the id:
SELECT first_value("user_foos"."id"),
"user_foos"."user_id",
"user_foos"."foo_id",
max("user_foos"."effective_at")
FROM "user_foos"
GROUP BY "user_foos"."user_id", "user_foos"."foo_id";
and:
SELECT first_value("user_foos"."id")
FROM "user_foos"
GROUP BY "user_foos"."user_id", "user_foos"."foo_id"
HAVING "user_foos"."effective_at" = max("user_foos"."effective_at")
Unfortunately, these both result in a different error:
window function call requires an OVER clause
Ideally, my goal is to fetch ALL matching id's so that I can use it in a subquery to fetch the legitimate full row data from this table for matching records. Can anyone provide insight on how I can get this working?
Postgres has a very nice feature called distinct on, which can be used in this case:
SELECT DISTINCT ON (uf."user_id", uf."foo_id") uf.*
FROM "user_foos" uf
ORDER BY uf."user_id", uf."foo_id", uf."effective_at" DESC;
It returns the first row in a group, based on the values in parentheses. The order by clause needs to include these values as well as a third column for determining which is the first row in the group.
Try:
SELECT *
FROM (
SELECT t.*,
row_number() OVER( partition by user_id, foo_id ORDER BY effective_at DESC ) x
FROM user_foos t
)
WHERE x = 1
If you don't want to use a sub query based on a composite of all three keys then you need to create a "dense rank" window function field that orders subsets of id, user_id and foo_id by effective date with the rank order field. Then subquery that and take the records where rank_order=1. Since the rank ordering was by effective date you are getting all fields of the record with the highest effective date for each foo and user.
DATSET
1 1 1 01/01/2001
2 1 1 01/01/2002
3 1 1 01/01/2003
4 1 2 01/01/2001
5 2 1 01/01/2001
DATSET WITH RANK ORDER PARTITIONED BY FOO_ID, USER_ID ORDERED BY DATE DESC
1 3 1 1 01/01/2001
2 2 1 1 01/01/2002
3 1 1 1 01/01/2003
4 1 1 2 01/01/2001
5 1 2 1 01/01/2001
SELECT * FROM QUERY ABOVE WHERE RANK_ORDER=1
3 1 1 1 01/01/2003
4 1 1 2 01/01/2001
5 1 2 1 01/01/2001