Consider the following table?
ID COL VALUE
1 A 10
2 B 10
3 C 10
4 D 10
5 E 10
Output:
ID COL VALUE
1 A 10
2 B 20
3 C 30
4 D 40
5 E 50
Based on your (deleted) comment in output it is taking up the sum of the upper values, it sounds like you're wanting a cumulative SUM().
You can do this with a windowed function:
Select Id, Col, Sum(Value) Over (Order By Id) As Value
From YourTable
Output
Id Col Value
1 A 10
2 B 20
3 C 30
4 D 40
5 E 50
Please make use of the the below code to obtain the cumulative sum. The code is working as expected with SQL Server 2012.
DECLARE #Table TABLE (ID int, COL CHAR(2), VALUE int)
INSERT #Table
(ID,COL,[VALUE])
VALUES
(1,'A',10),
(2,'B',10),
(3,'C',10),
(4,'D',10),
(5,'E',10)
SELECT t.ID,t.COL,SUM(VALUE) OVER (ORDER BY t.ID) AS VALUE
FROM #Table t
Not really sure what you are asking for. If my assumption is correct, you want to SUM the contents of a column and group it.
Select sum(value), col
from table
group by col
Related
I am trying to build a query to calculate median of 3 column values. My table looks like below,
Item
Column 1
Column 2
Column 3
A
10
12
4
B
5
14
20
C
15
5
4
I want to be able to output,
Item
Column 1
Column 2
Column 3
Median
A
10
12
4
10
B
5
14
20
14
C
15
5
4
5
I have tried percentile_cont() but that seems to be only for values in a single column. How do i achieve this?
Consider below approach
select *,
( select distinct percentile_disc(col, 0.5) over()
from unnest([Column1, Column2, Column3]) as col
) AS Median
from your_table
if applied to sample data in your question - output is
Have you tried this:
select Col1, Col2, Col3,
PERCENTILE_CONT([Col1, Col2, Col3], 0.5) OVER() AS Median
from tableName
I would like to format the data of below table as follows. Here on the value column, I want to maintain only on value for each of the duplicated rows.
input table
code value
A 10
A 10
A 10
B 20
B 20
B 20
C 30
C 30
D 40
Expected result
code value
A 10
A
A
B 20
B
B
C 30
C
D 40
A combination of CASE and window function can solve your problem
select code,
case when t.rn = 1 then value else null end value
from (
select row_number() over (partition by code, value order by value) rn,
code, value
from your_table
) t
It's hard to formulate, so i'll just show an example and you are welcome to edit my question and title.
Suppose, i have a table
flag id value datetime
0 b 1 343 13
1 a 1 23 12
2 b 1 21 11
3 b 1 32 10
4 c 2 43 11
5 d 2 43 10
6 d 2 32 9
7 c 2 1 8
For each id i want to squeze the table by flag columns such that all duplicate flag values that follow each other collapse to one row with sum aggregation. Desired result:
flag id value
0 b 1 343
1 a 1 23
2 b 1 53
3 c 2 75
4 d 2 32
5 c 2 1
P.S: I found functions like CONDITIONAL_CHANGE_EVENT, which seem to be able to do that, but the examples of them in docs dont work for me
Use the differnece of row number approach to assign groups based on consecutive row flags being the same. Thereafter use a running sum.
select distinct id,flag,sum(value) over(partition by id,grp) as finalvalue
from (
select t.*,row_number() over(partition by id order by datetime)-row_number() over(partition by id,flag order by datetime) as grp
from tbl t
) t
Here's an approach which uses CONDITIONAL_CHANGE_EVENT:
select
flag,
id,
sum(value) value
from (
select
conditional_change_event(flag) over (order by datetime desc) part,
flag,
id,
value
from so
) t
group by part, flag, id
order by part;
The result is different from your desired result stated in the question because of order by datetime. Adding a separate column for the row number and sorting on that gives the correct result.
The SQL Server 2005 table I'm working from is really strange. Here's a simplified example:
TABLE 1:
key | a | b | c | d | e | f
z 0 1 2 3 4 5 6
y 1 8 9 10 11 12 13
x 14 15 16 17 18 19 20
w 21 22 23 24 25 26 27
TABLE 2:
id | Value
1 a
2 b
3 c
4 e
5 f
What I need to accomplish is to, in a single statement, select the column name from a different table, then get the value here. So it would be something like "Select (select colName from table2 where id=VAR1) From table1 where key = VAR2"
So table2 will return either a/b/c/d/e/f, and then the main statement will get the value of the corresponding value based on the key.
Table2 will always return a-f, and I will know the VAR1 and VAR2 ahead of time.
You can use UNPIVOT:
SELECT [key], val, col
FROM
(SELECT [key], a, b, c, d, e, f
FROM table1
WHERE [key] = #var2) AS src
UNPIVOT
(val FOR col IN
(a, b, c, d, e, f)
)AS unpvt
WHERE col = (SELECT value FROM table2 WHERE id = #var1)
UNPIVOT operation transposes table data from columns to rows creating an extra field where column name is placed. This way you can query table data using the name of the column that is obtained from table2.
Demo here
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.