I have a data in this order
Id Value
-- ----
1 a
1 b
1 c
2 a
2 c
3 b
4 c
4 b
4 a
I want to sort data in this order
Id Value
-- ----
1 a
2 a
3 b
4 c
1 b
2 c
4 b
1 c
4 a
You seem to want to intersperse the numbers. For this purpose, you can use row_number():
order by row_number() over (partition by id order by value),
id
Related
So I have two tables (A and B) that have a relation of n-n.
So there is a third table (C) that is used to connect both tables.
Table A and B both have an Id and a name.
Table C has IDA, IDB and an Order, the number that is used to sort and that is user given.
My issue is that I need to migrate table C since I just added that order column and so I need to give every line an ordering number, according to the B name.
So if table A has:
Id Name
1 A
2 B
3 C
And Table B has:
Id Name
1 J
2 L
3 M
And table C has:
IdA IdB Order
1 2 0
1 1 0
1 3 0
2 1 0
2 3 0
I need a query that updates table C to be more like:
IdA IdB Order
1 2 2
1 1 1
1 3 3
2 1 1
2 3 2
I have a query that can basically do what i want but it leaves me with "gaps"
reading my results above i get:
IdA IdB Order
1 2 2
1 1 1
1 3 3
2 1 1
2 3 3
I think this should work for what you need:
With ToUpdate As
(
Select C.*,
Row_Number() Over (Partition By C.IdA Order By B.Name) As NewOrder
From C
Join B On B.Id = C.IdB
)
Update C
Set "Order" = U.NewOrder
From ToUpdate U
Where U.IdA = C.IdA
And U.IdB = C.IdB
(In full disclosure, I'm not terribly familiar with postgres, but I think this should be valid).
I have a table
ID GROUPID NAME
== ======= ========
1 100 A
2 100 B
3 200 C
4 200 D
5 300 E
6 100 F
I would like to create a table containing the permutation pairs within a group without any pairs that are the same on both first and second that looks like this:
PAIRID FIRST SECOND
====== ===== ======
1 1 2
2 1 6
3 2 1
4 2 6
5 3 4
6 4 3
7 6 1
8 6 2
I would like to do it in PL/SQL or straight SQL inserts if possible. I did this through Java already using a recursive function to go through the permutations.
You could self join the table:
SELECT ROW_NUMBER() OVER (ORDER BY a.id, b.id) AS pairid,
a.id AS FIRST, b.id AS second
FROM mytable a
JOIN mytable b ON a.groupid = b.groupid AND a.id <> b.id
ORDER BY 1 ASC;
How to write a sql query
A 1
B 1
C 1
A 1
A 1
B 1
B 1
C 1
B 1
to make it look like
A 3
B 4
C 2
A 3
A 3
B 4
B 4
C 2
B 4
SELECT field1, SUM(field2) OVER (PARTITION BY field1) as total
FROM table1
More info about window/analytic functions in Oracle : http://docs.oracle.com/cd/E11882_01/server.112/e41084/functions004.htm
thanks in advance for the help and sorry for how the "table" looks. Here's my question...
Let's say I have a subquery with this table (imagine the bold as column headers) as its output -
id 1 1 2 3 3 3 3 4 5 6 6 6
action o c o c c o c o o c c c
I would like my new query to output -
id 1 1 2 3 3 3 3 4 5 6 6 6
action o c o c c o c o o c c c
ct 1 2 1 1 2 3 4 1 1 1 2 3
#c 0 1 0 1 2 2 3 0 0 1 2 3
#o 1 1 1 0 0 1 1 1 1 0 0 0
where ct stands for count. Basically, I want to count (for each id) the occurrences of consecutive id and action as they happen. Let me know if this makes sense, and if not, how I can clarify my question.
Note: I realize the lag/lead functions may be helpful in this situation, along with the row_number() function. Looking for as many creative solutions as possible!
You are looking for the row_number() analytic function:
select id, action, row_number() over (partition by id order by id) as ct
from table t;
For #c and #o, you want cumulative sum:
select id, action, row_number() over (partition by id order by id) as ct,
sum(case when action = 'c' then 1 else 0 end) over
(partition by id order by <some column here>) as "#c",
sum(case when action = 'c' then 1 else 0 end) over
(partition by id order by <some column here>) as "#o"
from table t;
The one caveat is that you need a way to specify the order of the rows -- an id or date time stamp or something. SQL result sets and tables are inherently unordered, so there is no idea that one row comes before or after another.
SQL> select id, action,
2 row_number() over(partition by id order by rowid) ct,
3 sum(decode(action,'c',1,0)) over(partition by id order by rowid) c#,
4 sum(decode(action,'o',1,0)) over(partition by id order by rowid) o#
5 from t1
6 /
ID A CT C# O#
---------- - ---------- ---------- ----------
1 o 1 0 1
1 c 2 1 1
2 o 1 0 1
3 c 1 1 0
3 c 2 2 0
3 o 3 2 1
3 c 4 3 1
4 o 1 0 1
5 o 1 0 1
6 c 1 1 0
6 c 2 2 0
6 c 3 3 0
P.S. Sorry Gordon, didn't see your post.
I am working on a query for SQL Server 2005 that needs to return data with two 'index' fields. The first index 't_index' should increment every time the 'shade' column changes, whilst the second index increments within the partition of the values in the 'shade' column:
t_index s_index shade
1 1 A
1 2 A
1 3 A
1 4 A
1 5 A
2 1 B
2 2 B
2 3 B
2 4 B
2 5 B
To get the s_index column I am using the following:
Select ROW_NUMBER() OVER(PARTITION BY [shade] ORDER BY [shade]) as s_index
My question is how to get the first index to only increment when the value in the 'shade' column changes?
That can be accomplished with the DENSE_RANK() function:
DENSE_RANK() OVER(Order By [shade]) as t_index
You can try to use DENSE_RANK() for that:
SELECT
shade,
s_index = ROW_NUMBER() OVER(PARTITION BY [shade] ORDER BY [shade]),
t_index = DENSE_RANK() OVER (ORDER BY [shade])
FROM dbo.YourTableNameHEre
Gives output:
shade s_index t_index
A 1 1
A 2 1
A 3 1
A 4 1
A 5 1
B 1 2
B 2 2
B 3 2
B 4 2
B 5 2