SQL MAX(column) With Additional Criteria - sql

I have a single table, where I want to return a list of the MAX(id) GROUPed by another identifier. However I have a third column that, when it meets a certain criteria, "trumps" rows that don't meet that criteria.
Probably easier to explain with an example. Sample table has:
UniqueId (int)
GroupId (int)
IsPriority (bit)
Raw data:
UniqueId GroupId IsPriority
-----------------------------------
1 1 F
2 1 F
3 1 F
4 1 F
5 1 F
6 2 T
7 2 T
8 2 F
9 2 F
10 2 F
So, because no row in groupId 1 has IsPriority set, we return the highest UniqueId (5). Since groupId 2 has rows with IsPriority set, we return the highest UniqueId with that value (7).
So output would be:
5
7
I can think of ways to brute force this, but I am looking to see if I can do this in a single query.

SQL Fiddle Demo
WITH T
AS (SELECT *,
ROW_NUMBER() OVER (PARTITION BY GroupId
ORDER BY IsPriority DESC, UniqueId DESC ) AS RN
FROM YourTable)
SELECT UniqueId,
GroupId,
IsPriority
FROM T
WHERE RN = 1

Related

sql - select single ID for each group with the lowest value

Consider the following table:
ID GroupId Rank
1 1 1
2 1 2
3 1 1
4 2 10
5 2 1
6 3 1
7 4 5
I need an sql (for MS-SQL) select query selecting a single Id for each group with the lowest rank. Each group needs to only return a single ID, even if there are two with the same rank (as 1 and 2 do in the above table). I've tried to select the min value, but the requirement that only one be returned, and the value to be returned is the ID column, is throwing me.
Does anyone know how to do this?
Use row_number():
select t.*
from (select t.*,
row_number() over (partition by groupid order by rank) as seqnum
from t
) t
where seqnum = 1;

Calculate "position in run" in SQL

I have a table of consecutive ids (integers, 1 ... n), and values (integers), like this:
Input Table:
id value
-- -----
1 1
2 1
3 2
4 3
5 1
6 1
7 1
Going down the table i.e. in order of increasing id, I want to count how many times in a row the same value has been seen consecutively, i.e. the position in a run:
Output Table:
id value position in run
-- ----- ---------------
1 1 1
2 1 2
3 2 1
4 3 1
5 1 1
6 1 2
7 1 3
Any ideas? I've searched for a combination of windowing functions including lead and lag, but can't come up with it. Note that the same value can appear in the value column as part of different runs, so partitioning by value may not help solve this. I'm on Hive 1.2.
One way is to use a difference of row numbers approach to classify consecutive same values into one group. Then a row number function to get the desired positions in each group.
Query to assign groups (Running this will help you understand how the groups are assigned.)
select t.*
,row_number() over(order by id) - row_number() over(partition by value order by id) as rnum_diff
from tbl t
Final Query using row_number to get positions in each group assigned with the above query.
select id,value,row_number() over(partition by value,rnum_diff order by id) as pos_in_grp
from (select t.*
,row_number() over(order by id) - row_number() over(partition by value order by id) as rnum_diff
from tbl t
) t

SQL Server GROUP BY COUNT Consecutive Rows Only

I have a table called DATA on Microsoft SQL Server 2008 R2 with three non-nullable integer fields: ID, Sequence, and Value. Sequence values with the same ID will be consecutive, but can start with any value. I need a query that will return a count of consecutive rows with the same ID and Value.
For example, let's say I have the following data:
ID Sequence Value
-- -------- -----
1 1 1
5 1 100
5 2 200
5 3 200
5 4 100
10 10 10
I want the following result:
ID Start Value Count
-- ----- ----- -----
1 1 1 1
5 1 100 1
5 2 200 2
5 4 100 1
10 10 10 1
I tried
SELECT ID, MIN([Sequence]) AS Start, Value, COUNT(*) AS [Count]
FROM DATA
GROUP BY ID, Value
ORDER BY ID, Start
but that gives
ID Start Value Count
-- ----- ----- -----
1 1 1 1
5 1 100 2
5 2 200 2
10 10 10 1
which groups all rows with the same values, not just consecutive rows.
Any ideas? From what I've seen, I believe I have to left join the table with itself on consecutive rows using ROW_NUMBER(), but I am not sure exactly how to get counts from that.
Thanks in advance.
You can use Sequence - ROW_NUMBER() OVER (ORDER BY ID, Val, Sequence) AS g to create a group:
SELECT
ID,
MIN(Sequence) AS Sequence,
Val,
COUNT(*) AS cnt
FROM
(
SELECT
ID,
Sequence,
Sequence - ROW_NUMBER() OVER (ORDER BY ID, Val, Sequence) AS g,
Val
FROM
yourtable
) AS s
GROUP BY
ID, Val, g
Please see a fiddle here.

SQL - Order by amount of occurrences

It's my first question here so I hope I can explain it well enough,
I want to order my data by amount of occurrences in the table.
My table is like this:
id Daynr
1 2
1 4
2 4
2 5
2 6
3 1
4 2
4 5
And I want it to sort it like this:
id Daynr
3 1
1 2
1 4
4 2
4 5
2 4
2 5
2 6
Player #3 has one day in the table, and Player #1 has 2.
My table is named "dayid"
Both id and Daynr are foreign keys, together making it a primary key
I hope this explains my problem enough, Please ask for more information it's my first time here.
Thanks in advance
You can do this by counting the number of times that things occur for each id. Most databases support window functions, so you can do this as:
select id, daynr
from (select t.*, count(*) over (partition by id) as cnt
from table t
) t
order by cnt, id;
You can also express this as a join:
select t.id, t.daynr
from table as t inner join
(select id, count(*) as cnt
from table
group by id
) as tg
on t.id = tg.id
order by tg.cnt, id;
Note that both of these include the id in the order by. That way, if two ids have the same count, all rows for the id will appear together.

Order by a column not included in a group by clause

Here is my data set:
myID MemberID SourceID Acuity
1 5 3 2
2 5 3 1
3 5 3 2
4 5 3 1
I need to return a data set for each distinct MemberID, SourceID and Acuity combination. But I need to order by myID. I am inserting this into a different table and need the rows to get inserted in the proper order. The expected result I want is this:
MemberID SourceID Acuity
5 3 2
5 3 1
The problem is that I can't order by myID unless it is included in the group by clause so what happens is the result is opposite because it orders by each column by default. How can I get the intended result and order by myID?
You problem is ambiguous, because there is more than one row with different ids. You can order by the minimum id, however:
select MemberID, SourceID, Acuity
from your_table
group by MemberID, SourceID, Acuity
order by min(myID)