I have a table as follows:
CREATE TABLE results (
id uuid primary key UNIQUE,
score integer NOT NULL
)
I need to select a record with particular UUID and what's around it (say, 5 before and after) ordered by score
SELECT * FROM results
WHERE id = <SOME_UUID>
ORDERED BY score
OFFSET -5 LIMIT 10; -- apparently this is wrong
How can I effectively do that?
Its not 'effective', but you could try this:
select a.* from (SELECT * FROM results
WHERE id <> <SOME_UUID> and score <= (select score from results WHERE id = <SOME_UUID>)
ORDERED BY score,id desc
LIMIT 5) as a
UNION ALL
SELECT * FROM results
WHERE id = <SOME_UUID>
UNION ALL
select b.* from (SELECT * FROM results
WHERE id <> <SOME_UUID> and score >= (select score from results WHERE id = <SOME_UUID>)
ORDERED BY score, id asc
LIMIT 5) as b
I tried this an SQL-Server, which needded the 'ALL' to compute.
So you may get records with equal score as duplicates. To avoid this make it again to a subquery and use select distinct.
One way of solving this is with a rank for each row assigned using a window function and then finding out which ranks you are interested in:
WITH ranked AS (
SELECT id, score, rank() OVER (ORDER BY score) AS rnk
FROM results),
this_rank AS (
SELECT rnk - 5 AS low_rnk FROM ranked
WHERE id = <some uuid>::uuid)
SELECT id, score
FROM ranked, this_rank
WHERE rnk >= low_rnk
ORDER BY rnk
LIMIT 11;
For very low or high scores you get fewer than 11 rows, rather than rows with NULLs.
SQLFiddle
One further detail: A PRIMARY KEY already implies uniqueness so you do not have to use the UNIQUE clause in your table definition.
Related
I have this query who create pagination system, I want to SELECT only A* , I dont want to show row_number value even if I need it .
SELECT *
FROM (SELECT A.*, rownum row_number
FROM (select * from dual
) A
WHERE rownum <= 10)
WHERE row_number >= 1
The result :
D ROW_NUMBER
- ----------
X 1
Whats I want
D
-
X
Thanks for help
If your table has a primary key, you may perform the pagination filter only on this key and in the second step select the data based on the PK.
This will allow you to use SELECT *
select * from tab
where id in (
SELECT id
FROM (SELECT id, rownum row_number
FROM (select id from tab
) A
WHERE rownum <= 10)
WHERE row_number >= 1)
You'll pay a little performance penalty, as each selected row must be additionaly accessed by the primary key index (but this will be not visible for 10 rows or so).
An other point with pagination is that you typically need to present the data in some order and not randonly as in your example.
In that case the innermost subquery will be
select id from tab order by <some column>
Here you can profit, as you need to sort only the PK and the sort key and not the whole row (but again it will be not visible for 10 rows).
To describe my query problem, the following data is helpful:
A single table contains the columns ID (int), VAL (varchar) and ORD (int)
The values of VAL may change over time by which older items identified by ID won't get updated but appended. The last valid item for ID is identified by the highest ORD value (increases over time).
T0, T1 and T2 are points in time where data got entered.
How do I get in an efficient manner to the Result set?
A solution must not involve materialized views etc. but should be expressible in a single SQL-query. Using Postgresql 9.3.
The correct way to select groupwise maximum in postgres is using DISTINCT ON
SELECT DISTINCT ON (id) sysid, id, val, ord
FROM my_table
ORDER BY id,ord DESC;
Fiddle
You want all records for which no newer record exists:
select *
from mytable
where not exists
(
select *
from mytable newer
where newer.id = mytable.id
and newer.ord > mytable.ord
)
order by id;
You can do the same with row numbers. Give the latest entry per ID the number 1 and keep these:
select sysid, id, val, ord
from
(
select
sysid, id, val, ord,
row_number() over (partition by id order by ord desc) as rn
from mytable
)
where rn = 1
order by id;
Left join the table (A) against itself (B) on the condition that B is more recent than A. Pick only the rows where B does not exist (i.e. A is the most recent row).
SELECT last_value.*
FROM my_table AS last_value
LEFT JOIN my_table
ON my_table.id = last_value.id
AND my_table.ord > last_value.ord
WHERE my_table.id IS NULL;
SQL Fiddle
Currently i am trying to output the top row for 2 condition. One is max and one is min.
Current code
Select *
from (MY SELECT STATEMENT order by A desc)
where ROWNUM <= 1
UPDATE
I am now able to do for both condition. But i need the A to be the highest, if same then check for the B lowest.
E.g Lets say there is 2 rows, Both A is 100 and B is 50 for one and 60 for other.
In this case the 100:50 shld be choose because A is same then B is lowest.
E.g
Lets say there is 2 rows, A is 100 for one and 90 for other, since one is higher no need to check for B.
I tried using max and min but this method seems to work better, any suggestions
Well, after your clarification, you are looking for one record. With Max A. And the smallest B, in case there is more than one record with MAX A. This is simply:
Select *
from (MY SELECT STATEMENT order by A desc, B)
where ROWNUM = 1;
This sorts by A descending first, so you get all maximal A records first. Then it sorts by B, so inside each A group you get the least B first. This gives you the desired A record first, no matter if the found A is unique or not.
or avoid the vagaries of rownun and go for row_number() instead:
SELECT
*
FROM (
SELECT
*
, ROW_NUMBER (ORDER BY A DESC) adesc
, ROW_NUMBER (ORDER BY B ASC) basc
FROM SomeQuery
)
WHERE adesc = 1
OR basc = 1
footnote: select * is a convenience only, please replace with the actual columns required along with table names etc.
Try this if that works
Select *
from (MY SELECT STATEMENT order by A desc)
where ROWNUM <= 1
union
Select *
from (MY SELECT STATEMENT order by A asc)
where ROWNUM <= 1
SELECT * FROM
(Select foo.*, 0 as union_order
from (MY SELECT STATEMENT order by A desc) foo
where ROWNUM <= 1
UNION
Select foo.*, 1
from (MY SELECT STATEMENT order by B asc) foo
where ROWNUM <= 1)
ORDER BY
union_order
I have two tables with user rankings.
Table rankingA and rankingB.
Each table has the columns:
user_id
points
group_id
Higher the points so higher the rank of the user/group...
Now i try to get the group ranking for the question which rank has my group.
So far i have this SQL:
select sum(ra.points) as rapoints, sum(rb.points) as rbpoints from public.rankinga ra
LEFT JOIN public.rankingb rb ON ra.group_id=rb.group_id and ra.user_id=rb.user_id where
ra.group_id=200;
It returns the points from rankinga and rankinb for the group 200.
How can i get the rankings of the group? I tryd it with:
row_number() OVER (ORDER BY sum(rb.points) DESC) AS rankb
but got a wrong result.
My expected result for group_id 200 is:
rapoints,rbpoints,rarank, rbrank
420, 10, 3, same points as group_id 300 so rbrank 2 or 3
How can i get this?
Setup
CREATE TABLE rankinga
(
user_id bigint,
group_id bigint,
points integer
)
CREATE TABLE rankingb
(
user_id bigint,
group_id bigint,
points integer
)
insert into public.rankinga (user_id,group_id,points) values (1,100,120),(2,100,300), (3,100,20),(4,200,300),(5,200,120),(6,300,600);
insert into public.rankingb (user_id,group_id,points) values (1,100,5),(2,100,3),(3,100,10),(4,200,2),(5,200,8),(6,300,10);
I think you want to do this with union all, aggregation, and the window function. Joining the tables is likely to miss rows (if users are in one table but not the other) or over count (if you join on group). So this may do what you want:
select group_id, sum(rapoints) as rapoints, sum(rbpoints) as rbpoints,
sum(rapoints) + sum(rbpoints) as points,
dense_rank() over (order by sum(rapoints) + sum(rbpoints) desc) as ranking
from ((select ra.group_id, sum(ra.points) as rapoints, 0 as rbpoints
from public.rankinga ra
group by ra.group_id
) union all
(select rb.group_id, 0, sum(rb.points) as rbpoints
from public.rankingb rb
group by rb.group_id
)
) ab
group by group_id;
If you want to select just one group, then put this in a subquery (or CTE) and then select the group.
Here is a SQL Fiddle.
EDIT:
If you want just the result for one group, you still need to calculate the values for all groups. So:
select ab.*
from (<above query here>) ab
where group_id = 200;
You have three fields ID, Date and Total. Your table contains multiple rows for the same day which is valid data however for reporting purpose you need to show only one row per day. The row with the highest ID per day should be returned the rest should be hidden from users (not returned).
To better picture the question below is sample data and sample output:
ID, Date, Total
1, 2011-12-22, 50
2, 2011-12-22, 150
The correct result is:
2, 2012-12-22, 150
The correct output is single row for 2011-12-22 date and this row was chosen because it has the highest ID (2>1)
Assuming that you have a database that supports window functions, and that the date column is indeed just date (and not datetime), then something like:
SELECT
* --TODO - Pick columns
FROM
(
SELECT ID,[Date],Total,ROW_NUMBER() OVER (PARTITION BY [Date] ORDER BY ID desc) rn
FROM [Table]
) t
WHERE
rn = 1
Should produce one row per day - and the selected row for any given day is that with the highest ID value.
SELECT *
FROM table
WHERE ID IN ( SELECT MAX(ID)
FROM table
GROUP BY Date )
This will work.
SELECT *
FROM tableName a
INNER JOIN
(
SELECT `DATE`, MAX(ID) maxID
FROM tableName
GROUP BY `DATE`
) b ON a.id = b.MaxID AND
a.`date` = b.`date`
SQLFiddle Demo
Probably
SELECT * FROM your_table ORDER BY ID DESC LIMIT 1
Select MAX(ID),Data,Total from foo
for MySQL
Another simple way is
SELECT TOP 1 * FROM YourTable ORDER BY ID DESC
And, I think this is the most simple way!
SELECT * FROM TABLE_SUM S WHERE S.ID =
(
SELECT MAX(ID) FROM TABLE_SUM
WHERE CDATE = GG.CDATE
GROUP BY CDATE
)