This is hard to explain so I will give an example.
I need SQL (ms server), I assume its with row_number over partition but can't get it to work.
I have this table:
ID
PreviousID
Data
1
a
2
1
b
3
2
c
4
d
5
4
e
6
f
I want these results:
ID
NewID
Data
1
1
a
2
1
b
3
1
c
4
2
d
5
2
e
6
3
f
And another with just the new IDs of each sequence:
NewID
Data
1
a
2
d
3
f
Instead of a row number new id, it could also have the first id of the sequence, whatever is easier, as long as it identifies the sequence.
Seems you want a windowed COUNT of rows where the value of PreviousID is NULL.
SELECT ID,
COUNT(CASE WHEN PreviousID IS NULL THEN 1 END) OVER (ORDER BY ID) AS NewID,
Data
FROM dbo.YourTable;
Related
Say I have the following table:
ID
T
R
1
2
1
3
Y
1
4
1
5
1
6
Y
1
7
I would like to add a column which equals the value from column T based on the last non-null value from column R. This means the following:
ID
T
R
GOAL
1
2
1
3
Y
1
4
Y
3
1
5
4
1
6
Y
4
1
7
6
I do have many ID's so I need to make use of the OVER (PARTITION BY ...) clause. Also, if possible, I would like to use a single statement, like
SELECT *
, GOAL
FROM TABLE
So without any extra select statement.
T is in ascending order so just null it out according to R and take the maximum looking backward.
select *,
max(case when R is not null then T end)
over (
partition by id
order by T
rows between unbounded preceding and 1 preceding
) as GOAL
from TBL
http://sqlfiddle.com/#!18/c927a5/5
So I've been trying to change the order of rows according to my own logic.
Let's say I have a set of numbers from 1 to 4. If the rows have n entries of different values from 1 to 4 and I want to rearrange all the rows such that I have only rows with 4 at the top followed by a grouped combination of two 2's and a 1, followed by the 3's.
How can it be done using Sql server?
ID ROWS --> ID NEW ORDER
-- ---- -- ---------
A 1 G 4
B 1 G 4
C 2 G 4
C 2 G 4
D 2 C 2
D 2 C 2
E 2 A 1
E 2 E 2
F 3 E 2
F 3 B 1
F 3 D 2
G 4 D 2
G 4 F 3
G 4 F 3
G 4 F 3
Code till now:
SELECT * FROM table
ORDER BY
CASE
WHEN ROW = 4 THEN '1'
WHEN ROW = 1 THEN '2'
WHEN ROW = 2 THEN '3'
WHEN ROW = 3 THEN '4'
END ASC
Assuming ROWS is exactly the number of rows with a paticular ID,
order the rows first by a group according to ROWS interval (4),(1..2), (all the rest) then within the second group row_number() sequences of IDs having ROWS = 1 and 2 independently. Within the same sequence number order by ROWS reverse order.
select ID, ROWS
from (
select *
, case when ROWS = 4 then 1
when ROWS between 1 and 2 then 2
else 3 end grp1
, row_number() over(partition by ROWS order by ROWS) - row_number() over(partition by ROWS, ID order by ID) pos2
from yourtable) t
order by grp1, case ROWS when 1 then pos2 else pos2/2 end, -ROWS, ID
If you want you can try doing it using order by clause. You can use your logic to sort values accordingly. Default order by ASC. So i have tried below query with order by if this gives you some idea of doing it:
select id from test order by case when id%2 = 0 then id end desc
Will give;
ID
8
6
4
5
So you can try ordering with some logic inside.
We can easily get a sql result as following:
SQL>select Name, Value from table order by Name;
Name Value
------------
A 1
A 2
B 1
C 5
C 6
C 7
However, is there a way to link the name to a number so that an index of different names can be formed? Suppose we don't know how many different names are in the table and don't know what they are.
Name Value idx
-----------------
A 1 0
A 2 0
B 1 1
C 5 2
C 6 2
C 7 2
This can easily be done using a window function:
select Name,
Value,
dense_rank() over (order by name) - 1 as idx
from table
order by Name;
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
I have the following table
Id col1 col2 col3
1 c 2 m
2 c 3 6
2 b d u
3 e 6 9
4 1 v 8
4 2 b t
4 4 5 g
As you can see, there are duplicate value in id column, 2 and 4. I only want to select rows with unique id value and ignore the following rows with duplicate id value. I just want to keep the first of the rows with duplicate values
1 c 2 m
2 c 3 6
3 e 6 9
4 1 v 8
There is FK constraint, so I cannot delete rows with duplicate values.
I am using SQL SERVER 2008 R2
Any reply will be appreciated.
You can use row_number to number each row with the same id. Then you can select only the first row per id:
select *
from (
select row_number() over (partition by id order by col1, col2, col3) rn
from YourTable
) as SubQueryAlias
where rn = 1
The subquery is required because SQL Server doesn't allow row_number directly in the where clause.