I would like to add a column that will count according to the max number of series.
I have generated the SERIES_NO column using below:
MOD(ROW_NUMBER() OVER (ORDER BY item_code, loc_code, cargo_sts) - 1, 3) + 1
In this case the Max series no is 3. Below is the sample result set. Now, I want to generate the SHEET_NO column. Any suggestion? Thanks.
CARGO_STS LOC_CODE ITEM_CODE AVAIL_QTY SERIES_NO SHEET_NO
NORMAL D1867BD1 0000044500 6 1 1
NORMAL D1947GD1 0000055401 2 2 1
NORMAL D3351AA1 0000058000 2 3 1
NORMAL D1945DC2 0000058201 1 1 2
STO-DAMAGE 205-12BB 0000058300 1 2 2
NORMAL D3446FB1 0000058300 1 3 2
NORMAL Q00-37CA 0000060401 128 1 3
NORMAL D1158FA1 0000079901 36 2 3
something like following will do:
ceil (ROW_NUMBER() OVER (ORDER BY item_code, loc_code, cargo_sts) /3 )
Related
I want to group records by row numbers.
Like from row 1-3 in group 1 , 4-6 in group 2 , 7-9 in group 3 and so on.
Suppose below is the table structure:
Row NumberDataValue
1 A 10
2 A 5
3 A 1
4 A 33
5 A 2
6 A 127
1 B 1
2 B 0
3 B 7
4 B 7
5 B 5
6 B 8
7 B 1
8 B 0
I want a output like this:
GroupValue
1 10
1 5
1 1
2 33
2 2
2 127
1 1
1 0
1 7
2 7
2 5
2 8
3 1
3 0
I am using Oracle 11G.
I can achieve this using PL/SQL. But I have to use SQL only. As I have to use this query in a reporting tool.
If this is a duplicate question please provide the link of the answered question.
Subtract 1 from the column "RowNumber" and divide by 3.
Then use TRUNC() to get the integer part:
SELECT TRUNC(("RowNumber" - 1) / 3) + 1 "Group",
"Value"
FROM tablename
See the demo.
I would assume the name of the first column is ordering.
You can do:
select
1 + trunc(row_number() over(partition by data order by ordering) - 1) / 3,
value
from t
What you show looks like the output from something like this:
select ceil(rn/3) as grp, value
from your_table
order by rn;
Note that "row number" and "group" are reserved words/phrases which should not be used as column names. I used rn and grp instead.
I think the ceiling function is the simplest way to arrive at what you want. If you want to base it on the RowNumber column:
select ceil( RowNumber / 3.0) as grouping
If you want to calculate it yourself using row_number():
select ceil( row_number() over (order by RowNumber) / 3.0 ) as grouping
I have data that looks like this:
ID num_of_days
1 0
2 0
2 8
2 9
2 10
2 15
3 10
3 20
I want to add another column that increments in value only if the num_of_days column is divisible by 5 or the ID number increases so my end result would look like this:
ID num_of_days row_num
1 0 1
2 0 2
2 8 2
2 9 2
2 10 3
2 15 4
3 10 5
3 20 6
Any suggestions?
Edit #1:
num_of_days represents the number of days since the customer last saw a doctor between 1 visit and the next.
A customer can see a doctor 1 time or they can see a doctor multiple times.
If it's the first time visiting, the num_of_days = 0.
SQL tables represent unordered sets. Based on your question, I'll assume that the combination of id/num_of_days provides the ordering.
You can use a cumulative sum . . . with lag():
select t.*,
sum(case when prev_id = id and num_of_days % 5 <> 0
then 0 else 1
end) over (order by id, num_of_days)
from (select t.*,
lag(id) over (order by id, num_of_days) as prev_id
from t
) t;
Here is a db<>fiddle.
If you have a different ordering column, then just use that in the order by clauses.
I have a table with two columns personid and taskid and want to use the ROW_NUMBER function to add a row that counts up to 3 but will duplicate the number as it counts if there are multiple rows for a personid.
The code below is only ordering by personid and repeating after the number 3, but I need it to order by personid and only go to the next number after all the taskid's for the personid are assigned to one number, or essentially any duplicate personid's I want to make sure they all only get one number assigned to it.
Select
personid,
taskid,
1 + ( (row_number() over (order by personid) - 1) % 3) as numberCount
from taskTable
Current Table Being Queried From:
PersonId Taskid
1 1
1 2
1 6
2 3
3 8
3 10
4 9
4 4
4 5
5 7
5 11
5 12
Expected Results After Query:
PersonId Taskid numberCount
1 1 1
1 2 1
1 6 1
2 3 2
3 8 3
3 10 3
4 9 1
4 4 1
4 5 1
5 7 2
5 11 2
5 12 2
Try this below script using DENSE_RANK -
SELECT *,
(DENSE_RANK() OVER(ORDER BY PersonId)-1)%3 + 1 AS numberCount
FROM your_table
I think you want dense_rank() and modulo arithmetic:
select t.*,
(dense_rank() over (order by personId) - 1) % 3) + 1 as numberCount
from t;
Note: The syntax for modulo arithmetic may vary in your database. Typically it is one of mod(), the % operator, or using mod as an operator.
Data Looks like -
1
2
3
1
2
2
2
3
1
5
4
1
2
So whenever there is a 1, it marks the beginning of a group which includes all the elements until it hits the next 1. So here,
1 2 3 - group 1
1 2 2 2 3 - group 2
and so on..
What would be the SQL query to show the average for every such group.
I could not figure out how to group them without using for loops or PLSQL code.
Result should look like two columns, one with the actual data and col 2 with the average value-
1 - avg value of 1,2 3
2
3
1 - avg value of 1,2,2,2,3
2
2
2
3
1 - avg value of 1,5,4
5
4
1 - avg value of 1,2
2
SQL tables represent unordered sets. There is no ordering, unless a column specifies the ordering. Let me assume that you have such a column.
You can identify the groups using a cumulative sum:
select t.*,
sum(case when t.col = 1 then 1 else 0 end) over (order by ?) as grp
from t;
? is the column that specifies the ordering.
You can then calculate the average using aggregation:
select grp, avg(col)
from (select t.*,
sum(case when t.col = 1 then 1 else 0 end) over (order by ?) as grp
from t
) t
group by grp;
I am working with SQL Server 2008
If I have a Table as such:
Code Value
-----------------------
4 240
4 299
4 210
2 NULL
2 3
6 30
6 80
6 10
4 240
2 30
How can I find the median AND group by the Code column please?
To get a resultset like this:
Code Median
-----------------------
4 240
2 16.5
6 30
I really like this solution for median, but unfortunately it doesn't include Group By:
https://stackoverflow.com/a/2026609/106227
The solution using rank works nicely when you have an odd number of members in each group, i.e. the median exists within the sample, where you have an even number of members the rank method will fall down, e.g.
1
2
3
4
The median here is 2.5 (i.e. half the group is smaller, and half the group is larger) but the rank method will return 3. To get around this you essentially need to take the top value from the bottom half of the group, and the bottom value of the top half of the group, and take an average of the two values.
WITH CTE AS
( SELECT Code,
Value,
[half1] = NTILE(2) OVER(PARTITION BY Code ORDER BY Value),
[half2] = NTILE(2) OVER(PARTITION BY Code ORDER BY Value DESC)
FROM T
WHERE Value IS NOT NULL
)
SELECT Code,
(MAX(CASE WHEN Half1 = 1 THEN Value END) +
MIN(CASE WHEN Half2 = 1 THEN Value END)) / 2.0
FROM CTE
GROUP BY Code;
Example on SQL Fiddle
In SQL Server 2012 you can use PERCENTILE_CONT
SELECT DISTINCT
Code,
Median = PERCENTILE_CONT(0.5) WITHIN GROUP (ORDER BY Value) OVER(PARTITION BY Code)
FROM T;
Example on SQL Fiddle
SQL Server does not have a function to calculate medians, but you could use the ROW_NUMBER function like this:
WITH RankedTable AS (
SELECT Code, Value,
ROW_NUMBER() OVER (PARTITION BY Code ORDER BY VALUE) AS Rnk,
COUNT(*) OVER (PARTITION BY Code) AS Cnt
FROM MyTable
)
SELECT Code, Value
FROM RankedTable
WHERE Rnk = Cnt / 2 + 1
To elaborate a bit on this solution, consider the output of the RankedTable CTE:
Code Value Rnk Cnt
---------------------------
4 240 2 3 -- Median
4 299 3 3
4 210 1 3
2 NULL 1 2
2 3 2 2 -- Median
6 30 2 3 -- Median
6 80 3 3
6 10 1 3
Now from this result set, if you only return those rows where Rnk equals Cnt / 2 + 1 (integer division), you get only the rows with the median value for each group.