I have a database table which looks like this.
id account_id action time_point
3 234 delete 100
1 656 create 600
1 4435 update 900
3 645 create 50
I need to group this table by id and select particular row where time_point has a largest value.
Result table should look like this:
id account_id action time_point
3 234 delete 100
1 4435 update 900
Thanks for help,
qwew
In Postgres, I would recommend distinct on to solve this top 1 per group problem:
select distinct on (id) *
from mytable
order by id, time_point desc
However, this does not allow possible to ties. If so, rank() is a better solution:
select *
from (
select t.*, rank() over(partition by id order by time_point desc) rn
from mytable t
) t
where rn = 1
Or, if you are running Postgres 13:
select *
from mytable t
order by rank() over(partition by id order by time_point desc)
fetch first row with ties
check this.
select * from x
where exists (
select 1 from x xin
where xin.id = x.id
having max(time_point) = time_point
);
Related
I want to exclude records where id > 5 then select the top 1 of it order by date. How can I achieve this? Each record has audit_line which is unique field for each record. Recent SQL script is on below:
SELECT *
FROM db.table
HAVING COUNT(id) > 5
If you want id > 5 then you want where:
select top (1) t.*
from db.table t
where id > 5
order by date;
You can use row-numbering for this.
Note that if you have no other column to order by, you can do ORDER BY (SELECT NULL), but then you may get different results on each run.
SELECT *
FROM (
SELECT *, ROW_NUMBER() OVER (PARTITION BY id ORDER BY some_other_column) rn
FROM table
) t
WHERE rn = 5;
I have a following tables:
TABLE A:
ID ID NAME PRICE CODE
00001 B 1000 1
00002 A 2000 1
00003 C 3000 1
Here is the SQL I use:
Select Min (ID),
Min (ID NAME),
Sum(PRICE)
From A
GROUP BY CODE
Here is what I get:
ID ID NAME PRICE
00001 A 6000
As you can see, ID NAME don't match up with the min row value. I need them to match up.
I would like the query to return the following
ID ID NAME PRICE
00001 B 6000
What SQL can I use to get that result?
If you want one row, use limit or fetch first 1 row only:
select a.*
from a
order by a.price asc
fetch first 1 row only;
If, for some reason, you want the sum() of all prices, then you can use window functions:
select a.*, sum(a.price) over () as sum_prices
from a
order by a.price asc
fetch first 1 row only;
You can use row_number() function :
select min(id), max(case when seq = 1 then id_name end) as id_name, sum(price) as price, code
from (select t.*, row_number() over (partition by code order by id) seq
from table t
) t
group by code;
you can also use sub-query
select t1.*,t2.* from
(select ID,Name from t where ID= (select min(ID) from t)
) as t1
cross join (select sum(Price) as total from t) as t2
https://dbfiddle.uk/?rdbms=postgres_10&fiddle=a496232b552390a641c0e5c0fae791d1
id name total
1 B 6000
I have table with some data, for example
ID Specified TIN Value
----------------------
1 0 tin1 45
2 1 tin1 34
3 0 tin2 23
4 3 tin2 47
5 3 tin2 12
I need to get rows with all fields by MAX(Specified) column. And if I have few row with MAX column (in example ID 4 and 5) i must take last one (with ID 5)
finally the result must be
ID Specified TIN Value
-----------------------
2 1 tin1 34
5 3 tin2 12
This will give the desired result with using window function:
;with cte as(select *, row_number(partition by tin order by specified desc, id desc) as rn
from tablename)
select * from cte where rn = 1
Edit: Updated query after question edit.
Here is the fiddle
http://sqlfiddle.com/#!9/20e1b/1/0
SELECT * FROM TBL WHERE ID IN (
SELECT max(id) FROM
TBL WHERE SPECIFIED IN
(SELECT MAX(SPECIFIED) FROM TBL
GROUP BY TIN)
group by specified)
I am sure we can simplify it further, but this will work.
select * from tbl where id =(
SELECT MAX(ID) FROM
tbl where specified =(SELECT MAX(SPECIFIED) FROM tbl))
One method is to use window functions, row_number():
select t.*
from (select t.*, row_number() over (partition by tim
order by specified desc, id desc
) as seqnum
from t
) t
where seqnum = 1;
However, if you have an index on tin, specified id and on id, the most efficient method is:
select t.*
from t
where t.id = (select top 1 t2.id
from t t2
where t2.tin = t.tin
order by t2.specified desc, id desc
);
The reason this is better is that the index will be used for the subquery. Then the index will be used for the outer query as well. This is highly efficient. Although the index will be used for the window functions; the resulting execution plan probably requires scanning the entire table.
suppose i have this table:
group_id | image | image_id |
-----------------------------
23 blob 1
23 blob 2
23 blob 3
21 blob 4
21 blob 5
25 blob 6
25 blob 7
how to get results of only 1 of each group id? in this case,there may be multiple images for one group id, i just want one result of each group_id
i tried distinct but i will only get group_id. max for image also would not work.
There are no standard aggregate functions in Oracle that would work with BLOBs, so GROUP BY solutions won't work.
Try this one based on ROW_NUMBER() in a sub-query.
SELECT inn.group_id, inn.image, inn.image_id
FROM
(
SELECT t.group_id, t.image, t.image_id,
ROW_NUMBER() OVER (PARTITION BY t.group_id ORDER BY t.image_id) num
FROM theTable t
) inn
WHERE inn.num = 1;
The above should return the first (based on image_id) row for each group.
SQL Fiddle
SELECT group_id, image, image_id
FROM a_table
WHERE (group_id, image_id) IN
(
SELECT group_id, MIN(image_id)
FROM a_table
GROUP BY
group_id
)
;
select * from
(select t1.*,
ROW_NUMBER() OVER (PARTITION BY group_id ORDER BY group_id desc) as seqnum
from tablename t1)
where seqnum=1;
I tried this with solutions avaialble online, but none worked for me.
Table :
Id rank
1 100
1 100
2 75
2 45
3 50
3 50
I want Ids 1 and 3 returned, beacuse they have duplicates.
I tried something like
select * from A where rank in (
select rank from A group by rank having count(rank) > 1
This also returned ids without any duplicates. Please help.
Try this:
select id from table
group by id, rank
having count(*) > 1
select id, rank
from
(
select id, rank, count(*) cnt
from rank_tab
group by id, rank
having count(*) > 1
) t
This general idea should work:
SELECT id
FROM your_table
GROUP BY id
HAVING COUNT(*) > 1 AND COUNT(DISTINCT rank) = 1
In plain English: get every id that exists in multiple rows, but all these rows have the same value in rank.
If you want ids that have some duplicated ranks (but not necessarily all), something like this should work:
SELECT id
FROM your_table
GROUP BY id
HAVING COUNT(*) > COUNT(DISTINCT rank)