I have a table with 5 columns like this:
| ID | NAME | PO_NUMBER | DATE | STATS |
| 1 | Jhon | 160101-001 | 2016-01-01 | 7 |
| 2 | Jhon | 160101-002 | 2016-01-01 | 7 |
| 3 | Jhon | 160102-001 | 2016-01-02 | 7 |
| 4 | Jane | 160101-001 | 2016-01-01 | 7 |
| 5 | Jane | 160102-001 | 2016-01-02 | 7 |
| 6 | Jane | 160102-002 | 2016-01-02 | 7 |
| 7 | Jane | 160102-003 | 2016-01-02 | 7 |
I need to display all values, but stats fields without duplicate according from date field.
Like this
| ID | NAME | PO_NUMBER | DATE | STATS |
| 1 | Jhon | 160101-001 | 2016-01-01 | 7 |
| 2 | Jhon | 160101-002 | 2016-01-01 | null |
| 3 | Jhon | 160102-001 | 2016-01-02 | 7 |
| 4 | Jane | 160101-001 | 2016-01-01 | 7 |
| 5 | Jane | 160102-001 | 2016-01-02 | 7 |
| 6 | Jane | 160102-002 | 2016-01-02 | null |
| 7 | Jane | 160102-003 | 2016-01-02 | null |
I've had trouble getting the hoped. Thanks
From your sample data, it appears you only want to show the stats for po_number ending with 001. If so, this should be the easiest approach:
select id, name, po_number, date,
case when right(po_number, 3) = '001' then stats else null end as stats
from yourtable
If instead you want to order by the po_number, then here's one option using row_number:
select id, name, po_number, date,
case when rn = 1 then stats else null end as stats
from (
select *, row_number() over (partition by name, date order by po_number) as rn
from yourtable
) t
SQL Fiddle Demo
since you are using SQL 2012, you can use the LEAD() or LAG() window function to compare the DATE value
select *,
STATS = case when t.DATE = LAG(DATE) OVER(ORDER BY ID)
then NULL
else STATS
end
from yourtable t
Use below code
;with temp as (
select id,name ,PO_NUMBER ,DATE, STATS,
LAG (STATS, 1, 0)
OVER (PARTITION BY name ,PO_NUMBER ,DATE ORDER BY id) AS PrevSTATS
from tableName
)
select id,name ,PO_NUMBER ,DATE,
case when STATS = PrevSTATS then null
else STATS end as STATS
from temp
Related
I want to calculate the number of people who also had occurrence the previous day on a daily basis, but I'm not sure how to do this?
Sample Table:
| ID | Date |
+----+-----------+
| 1 | 1/10/2020 |
| 1 | 1/11/2020 |
| 2 | 2/20/2020 |
| 3 | 2/20/2020 |
| 3 | 2/21/2020 |
| 4 | 2/23/2020 |
| 4 | 2/24/2020 |
| 5 | 2/22/2020 |
| 5 | 2/23/2020 |
| 5 | 2/24/2020 |
+----+-----------+
Desired Output:
| Date | Count |
+-----------+-------+
| 1/11/2020 | 1 |
| 2/21/2020 | 1 |
| 2/23/2020 | 1 |
| 2/24/2020 | 2 |
+-----------+-------+
Edit: Added desired output. The output count should be unique to the ID, not the number of date occurrences. i.e. an ID 5 can appear on this list 10 times for dates 2/23/2020 and 2/24/2020, but that would count as "1".
Use lag():
select date, count(*)
from (select t.*, lag(date) over (partition by id order by date) as prev_date
from t
) t
where prev_date = dateadd(day, -1, date)
group by date;
Suppose I have a table sorted by date as so:
+-------------+--------+
| DATE | VALUE |
+-------------+--------+
| 01-09-2020 | 5 |
| 01-15-2020 | 5 |
| 01-17-2020 | 5 |
| 02-03-2020 | 8 |
| 02-13-2020 | 8 |
| 02-20-2020 | 8 |
| 02-23-2020 | 5 |
| 02-25-2020 | 5 |
| 02-28-2020 | 3 |
| 03-13-2020 | 3 |
| 03-18-2020 | 3 |
+-------------+--------+
I want to group by changes in value within that given date range, and add a value that increments each time as an added column to denote that.
I have tried a number of different things, such as using the lag function:
SELECT value, value - lag(value) over (order by date) as count
GROUP BY value
In short, I want to take the table above and have it look like:
+-------------+--------+-------+
| DATE | VALUE | COUNT |
+-------------+--------+-------+
| 01-09-2020 | 5 | 1 |
| 01-15-2020 | 5 | 1 |
| 01-17-2020 | 5 | 1 |
| 02-03-2020 | 8 | 2 |
| 02-13-2020 | 8 | 2 |
| 02-20-2020 | 8 | 2 |
| 02-23-2020 | 5 | 3 |
| 02-25-2020 | 5 | 3 |
| 02-28-2020 | 3 | 4 |
| 03-13-2020 | 3 | 4 |
| 03-18-2020 | 3 | 4 |
+-------------+--------+-------+
I want to eventually have it all in one small table with the earliest date for each.
+-------------+--------+-------+
| DATE | VALUE | COUNT |
+-------------+--------+-------+
| 01-09-2020 | 5 | 1 |
| 02-03-2020 | 8 | 2 |
| 02-23-2020 | 5 | 3 |
| 02-28-2020 | 3 | 4 |
+-------------+--------+-------+
Any help would be very appreciated
you can use a combination of Row_number and Dense_rank functions to get the required results like below:
;with cte
as
(
select t.DATE,t.VALUE
,Dense_rank() over(partition by t.VALUE order by t.DATE) as d_rank
,Row_number() over(partition by t.VALUE order by t.DATE) as r_num
from table t
)
Select t.Date,t.Value,d_rank as count
from cte
where r_num = 1
You can use a lag and cumulative sum and a subquery:
SELECT value,
SUM(CASE WHEN prev_value = value THEN 0 ELSE 1 END) OVER (ORDER BY date)
FROM (SELECT t.*, LAG(value) OVER (ORDER BY date) as prev_value
FROM t
) t
Here is a db<>fiddle.
You can recursively use lag() and then row_number() analytic functions :
WITH t2 AS
(
SELECT LAG(value,1,value-1) OVER (ORDER BY date) as lg,
t.*
FROM t
)
SELECT t2.date,t2.value, ROW_NUMBER() OVER (ORDER BY t2.date) as count
FROM t2
WHERE value - lg != 0
Demo
and filter through inequalities among the returned values from those functions.
I want to calculate
(1) total sales amount
(2) number of distinct stores per product
in one query, if possible. Suppose we have data:
+-----------+---------+-------+--------+
| store | product | month | amount |
+-----------+---------+-------+--------+
| Anthill | A | 1 | 1 |
| Anthill | A | 2 | 1 |
| Anthill | A | 3 | 1 |
| Beetle | A | 1 | 1 |
| Beetle | A | 3 | 1 |
| Cockroach | A | 1 | 1 |
| Cockroach | A | 2 | 1 |
| Cockroach | A | 3 | 1 |
| Anthill | B | 1 | 1 |
| Beetle | B | 2 | 1 |
| Cockroach | B | 3 | 1 |
+-----------+---------+-------+--------+
I have tried this with no luck:
select
[product]
,[month]
,[amount]
,cnt_distinct_stores = count(distinct(stores))
from dbo.temp
group by
[product]
,[month]
order by 1,2
Would there be possible any combination of GROUP BY clause with window functions like SUM(amount) OVER(partition by [product],[month] ORDER BY [month] ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW)
Try
SELECT product,
SUM(amount),
COUNT(DISTINCT store)
FROM dbo.temp
GROUP BY product
Can you please help me with one complicated select statement?
I have a table like this:
+----+-----------+-----------+-----------------+
| ID | User_name | Situation | Date_time |
+----+-----------+-----------+-----------------+
| 1 | Alex | 1 | 14.3.18 11:30 |
| 4 | Alex | 2 | 14.3.18 11:35 |
| 6 | Alex | 3 | 14.3.18 12:30 |
| 7 | Johnny | 1 | 15.3.18 10:01 |
| 9 | Johnny | 2 | 15.3.18 10:05 |
| 12 | Johnny | 3 | 15.3.18 10:20 |
| 14 | Alex | 1 | 20.3.18 20:00 |
| 15 | Alex | 2 | 20.3.18 20:25 |
| 17 | Alex | 3 | 20.3.18 21:25 |
+----+-----------+-----------+-----------------+
And I need a select statement, which will give me the following result:
User_name, Date_time_1 (Date_time of situation 1), Date_time_3 (Date_time of situation 3).
*In this case the result will have just 3 rows (2 for Alex and 1 for Johnny). Each row will contain 3 columns as described above.
And sorry for the formatting - I posted that from a mobile. I will add the result table when I will get to PC.*
That's how the output should looks like:
+----+-----------+-------------+-----------------+
| ID | User_name |Date_time_1 | Date_time_3 |
+----+-----------+-------------+-----------------+
| 1 | Alex |14.3.18 11:30| 14.3.18 12:30 |
| 2 | Johnny |15.3.18 10:01| 15.3.18 10:20 |
| 3 | Alex |20.3.18 20:00| 20.3.18 21:25 |
+----+-----------+-------------+-----------------+
You could use conditional aggregation:
SELECT User_name,
MAX(CASE WHEN Situation = 1 THEN Date_time END) AS date_time_1,
MAX(CASE WHEN Situation = 3 THEN Date_time END) AS date_time_3
FROM tab
GROUP BY User_name;
EDIT
In this case the result will have just 3 rows (2 for Alex and 1 for Johnny)
WITH cte AS (
SELECT t.*, SUM(CASE WHEN Situation=1 THEN 1 ELSE 0 END)
OVER(PARTITION BY User_name ORDER BY id) AS s
FROM tab t
)
SELECT User_name,
MAX(CASE WHEN Situation = 1 THEN Date_time END) AS date_time_1,
MAX(CASE WHEN Situation = 3 THEN Date_time END) AS date_time_3
FROM cte
GROUP BY s, User_name;
DBFiddle Demo
Iam experiencing an issue in oracle analytic functions
I want the rank in oracle to be displayed sequentialy but require a cyclic fashion.But this ranking should happen within a group.
Say I have 10 groups
In 10 groups each group must be ranked in till 9. If greater than 9 the rank value must start again from 1 and then end till howmuch so ever
emp id date1 date 2 Rank
123 13/6/2012 13/8/2021 1
123 14/2/2012 12/8/2014 2
.
.
123 9/10/2013 12/12/2015 9
123 16/10/2013 15/10/2013 1
123 16/3/2014 15/9/2015 2
In the above example the for the group of rows of the empid 123 i have split the rank in two subgroup fashion.Sequentially from 1 to 9 is one group and for the rest of the rows the rank again starts from 1.How to achieve this in oracle rank functions.
as per suggestion from Egor Skriptunoff above:
select
empid, date1, date2
, row_number() over(order by date1, date2) as "rank"
, mod(row_number() over(order by date1, date2)-1, 9)+1 as "cycle_9"
from yourtable
example result
| empid | date1 | date2 | rn | ranked |
|-------|----------------------|----------------------|----|--------|
| 72232 | 2016-10-26T00:00:00Z | 2017-03-07T00:00:00Z | 1 | 1 |
| 04365 | 2016-11-03T00:00:00Z | 2017-07-29T00:00:00Z | 2 | 2 |
| 79203 | 2016-12-15T00:00:00Z | 2017-05-16T00:00:00Z | 3 | 3 |
| 68638 | 2016-12-18T00:00:00Z | 2017-02-08T00:00:00Z | 4 | 4 |
| 75784 | 2016-12-24T00:00:00Z | 2017-11-18T00:00:00Z | 5 | 5 |
| 72836 | 2016-12-24T00:00:00Z | 2018-09-10T00:00:00Z | 6 | 6 |
| 03679 | 2017-01-24T00:00:00Z | 2017-10-14T00:00:00Z | 7 | 7 |
| 43527 | 2017-02-12T00:00:00Z | 2017-01-15T00:00:00Z | 8 | 8 |
| 03138 | 2017-02-26T00:00:00Z | 2017-01-30T00:00:00Z | 9 | 9 |
| 89758 | 2017-03-29T00:00:00Z | 2018-04-12T00:00:00Z | 10 | 1 |
| 86377 | 2017-04-14T00:00:00Z | 2018-10-07T00:00:00Z | 11 | 2 |
| 49169 | 2017-04-28T00:00:00Z | 2017-04-21T00:00:00Z | 12 | 3 |
| 45523 | 2017-05-03T00:00:00Z | 2017-05-07T00:00:00Z | 13 | 4 |
SQL Fiddle