I have a script
SELECT customer_id, COUNT(*)
FROM devices
GROUP BY customer_id
ORDER BY 2 desc;
and the result is
customer_id count
1234 8
4567 7
8910 7
1112 7
1314 5
1516 5
but what I expect is
devices customer
8 1
7 3
5 2
Is there any way without using a new table?
Use your current query as a source for another:
with temp as
-- this is your current query
(select customer_id,
count(*) cnt
from devices
group by customer_id
)
select cnt as devices,
count(*) as customer
from temp
group by cnt;
The usual use of count(*) is to find out many rows have been aggregated by group function. Want you ant to do is aggregate your result in a second select, and ask for count(*) a second time to get result you've described.
Related
I have a table, consisting of 3 columns (Person, Year and Count), so for each person, there are several rows with different years and counts and the final row with total count. I want to keep the table ordered by Name, but also order it by the total count.
So the rows should be ordered by sum, but also grouped by the Person and ordered by year. When I am trying to order by sum, of course, both person and years are messed up. Is there a way to sort like this?
You've stored those "total" rows as well? Gosh! Why did you do that?
Anyway: if you
compute rank for rows whose year column is equal to 'total' and
add case expression into the order by clause,
you might get what you want:
SQL> with sorter as
2 (select name, cnt,
3 rank() over (order by cnt) rnk
4 from test
5 where year = 'total'
6 )
7 select t.*
8 from test t join sorter s on s.name = t.name
9 order by s.rnk, case when year = 'total' then '9'
10 else year
11 end;
NAME YEAR CNT
---- ----- ----------
John 2018 3
John 2019 2
John total 5
Bob 2017 2
Bob 2019 4
Bob total 6
6 rows selected.
SQL>
I am trying to develop a query to pull out the top 2 months of sales by customer id. Here is a sample table:
Customer_ID Sales Amount Period
144567 40 2
234567 50 5
234567 40 7
144567 80 10
144567 48 2
234567 23 7
desired output would be
Customer_ID Sales Sum Period
144567 80 10
144567 48 2
234567 50 5
234567 40 7
I've tried
select sum(net_sales_usd_spot), valid_period, customer_id
from sales_trans_price_output
where valid_period in (select valid_period, sum(net_sales_usd_spot)
from sales_trans_price_output
where rank<=2)
group by valid_period, customer_id
error is
too many values ORA-00913.
I see why, but not sure how to rework it.
Try:
SELECT *
FROM (
SELECT t.*,
row_number() over (partition by customer_id order by sales_amount desc ) rn
FROM sales_trans_price t
)
WHERE rn <= 2
ORDER BY 1,2 desc
Demo: http://sqlfiddle.com/#!4/882888/3
what if you change your where clause to:
where valid_period in
(
select p.valid_period from sales_trans_price_output p
join (select valid_period, sum(net_sales_usd_spot)
from sales_trans_price_output
where rank<=2) s on s.valid_period = p.valid_period
)
It might be ugly and need refactoring, but I think this is the logic you're after.
The error is because of this.
where valid_period in (select valid_period, sum(net_sales_usd_spot)
from sales_trans_price_output
where rank<=2)
The subquery can only contain one field.
You are on the right track using rank, but you might not be using it correctly. Google oracle rank to find the correct syntax.
Back to what you are looking to achieve, a derived table is the approach I would use. That's simply a subquery with an alias. Or, if you use the keyword with, it might be called a CTE - Computed Table Expression.
Try it
SELECT * FROM (
SELECT T.*,
RANK () OVER (PARTITION BY CUSTOMER_ID
ORDER BY VALID_PERIOD DESC) FN_RANK
FROM SALES_TRANS_PRICE_OUTPUT T
) A
WHERE A.FN_RANK <= 2
ORDER BY CUSTOMER_ID ASC, VALID_PERIOD DESC, FN_RANK DESC
Hi I have the following table:
crm_id | customer_id
jon 12345
jon 12346
ben 12347
sam 12348
I would like to show the following:
Crm_ID count | Number of customer_ids
1 2
2 1
Basically I want to count the number crm_ids that have 1,2,3,4,5+ customer_ids.
Thanks
One approach is to aggregate twice. First, aggregate over crm_id and generate counts. Then, aggregate over those counts themselves and generate a count of counts.
SELECT
cnt AS crm_id_cnt,
COUNT(*) AS num_customer_ids
FROM
(
SELECT crm_id, COUNT(DISTINCT customer_id) AS cnt
FROM yourTable
GROUP BY crm_id
) t
GROUP BY cnt;
Have a look at a demo below, given in MySQL as you did not specify a particular database (though my answer should run on most databases I think).
Demo
In a firebird database with a table "Sales", I need to select the first sale of all customers. See below a sample that show the table and desired result of query.
---------------------------------------
SALES
---------------------------------------
ID CUSTOMERID DTHRSALE
1 25 01/04/16 09:32
2 30 02/04/16 11:22
3 25 05/04/16 08:10
4 31 07/03/16 10:22
5 22 01/02/16 12:30
6 22 10/01/16 08:45
Result: only first sale, based on sale date.
ID CUSTOMERID DTHRSALE
1 25 01/04/16 09:32
2 30 02/04/16 11:22
4 31 07/03/16 10:22
6 22 10/01/16 08:45
I've already tested following code "Select first row in each GROUP BY group?", but it did not work.
In Firebird 2.5 you can do this with the following query; this is a minor modification of the second part of the accepted answer of the question you linked to tailored to your schema and requirements:
select x.id,
x.customerid,
x.dthrsale
from sales x
join (select customerid,
min(dthrsale) as first_sale
from sales
group by customerid) p on p.customerid = x.customerid
and p.first_sale = x.dthrsale
order by x.id
The order by is not necessary, I just added it to make it give the order as shown in your question.
With Firebird 3 you can use the window function ROW_NUMBER which is also described in the linked answer. The linked answer incorrectly said the first solution would work on Firebird 2.1 and higher. I have now edited it.
Search for the sales with no earlier sales:
SELECT S1.*
FROM SALES S1
LEFT JOIN SALES S2 ON S2.CUSTOMERID = S1.CUSTOMERID AND S2.DTHRSALE < S1.DTHRSALE
WHERE S2.ID IS NULL
Define an index over (customerid, dthrsale) to make it fast.
in Firebird 3 , get first row foreach customer by min sales_date :
SELECT id, customer_id, total, sales_date
FROM (
SELECT id, customer_id, total, sales_date
, row_number() OVER(PARTITION BY customer_id ORDER BY sales_date ASC ) AS rn
FROM SALES
) sub
WHERE rn = 1;
İf you want to get other related columns, This is where your self-answer fails.
select customer_id , min(sales_date)
, id, total --what about other colums
from SALES
group by customer_id
So simple as:
select CUSTOMERID min(DTHRSALE) from SALES group by CUSTOMERID
I need to count a value (M_Id) at each change of a date (RS_Date) and create a column grouped by the RS_Date that has an active total from that date.
So the table is:
Ep_Id Oa_Id M_Id M_StartDate RS_Date
--------------------------------------------
1 2001 5 1/1/2014 1/1/2014
1 2001 9 1/1/2014 1/1/2014
1 2001 3 1/1/2014 1/1/2014
1 2001 11 1/1/2014 1/1/2014
1 2001 2 1/1/2014 1/1/2014
1 2067 7 1/1/2014 1/5/2014
1 2067 1 1/1/2014 1/5/2014
1 3099 12 1/1/2014 3/2/2014
1 3099 14 2/14/2014 3/2/2014
1 3099 4 2/14/2014 3/2/2014
So my goal is like
RS_Date Active
-----------------
1/1/2014 5
1/5/2014 7
3/2/2014 10
If the M_startDate = RS_Date I need to count the M_id and then for
each RS_Date that is not equal to the start date I need to count the M_Id and then add that to the M_StartDate count and then count the next RS_Date and add that to the last active count.
I can get the basic counts with something like
(Case when M_StartDate <= RS_Date
then [m_Id] end) as Test.
But I am stuck as how to get to the result I want.
Any help would be greatly appreciated.
Brian
-added in response to comments
I am using Server Ver 10
If using SQL SERVER 2012+ you can use ROWS with your the analytic/window functions:
;with cte AS (SELECT RS_Date
,COUNT(DISTINCT M_ID) AS CT
FROM Table1
GROUP BY RS_Date
)
SELECT *,SUM(CT) OVER(ORDER BY RS_Date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS Run_CT
FROM cte
Demo: SQL Fiddle
If stuck using something prior to 2012 you can use:
;with cte AS (SELECT RS_Date
,COUNT(DISTINCT M_ID) AS CT
FROM Table1
GROUP BY RS_Date
)
SELECT a.RS_Date
,SUM(b.CT)
FROM cte a
LEFT JOIN cte b
ON a.RS_DAte >= b.RS_Date
GROUP BY a.RS_Date
Demo: SQL Fiddle
You need a cumulative sum, easy in SQL Server 2012 using Windowed Aggregate Functions. Based on your description this will return the expected result
SELECT p_id, RS_Date,
SUM(COUNT(*))
OVER (PARTITION BY p_id
ORDER BY RS_Date
ROWS UNBOUNDED PRECEDING)
FROM tab
GROUP BY p_id, RS_Date
It looks like you want something like this:
SELECT
RS_Date,
SUM(c) OVER (PARTITION BY M_StartDate ORDER BY RS_Date ROWS UNBOUNDED PRECEEDING)
FROM
(
SELECT M_StartDate, RS_Date, COUNT(DISTINCT M_Id) AS c
FROM my_table
GROUP BY M_StartDate, RS_Date
) counts
The inline view computes the counts of distinct M_Id values within each (M_StartDate, RS_Date) group (distinctness enforced only within the group), and the outer query uses the analytic version of SUM() to add up the counts within each M_StartDate.
Note that this particular query will not exactly reproduce your example results. It will instead produce:
RS_Date Active
-----------------
1/1/2014 5
1/5/2014 7
3/2/2014 8
3/2/2014 2
This is on account of some rows in your example data with RS_Date 3/2/2014 having a later M_StartDate than others. If this is not what you want then you need to clarify the question, which currently seems a bit inconsistent.
Unfortunately, analytic functions are not available until SQL Server 2012. In SQL Server 2010, the job is messier. It could be done like this:
WITH gc AS (
SELECT M_StartDate, RS_Date, COUNT(DISTINCT M_Id) AS c
FROM my_table
GROUP BY M_StartDate, RS_Date
)
SELECT
RS_Date,
(
SELECT SUM(c)
FROM gc2
WHERE gc2.M_StartDate = gc.M_StartDate AND gc2.RS_Date <= gc.RS_Date
) AS Active
FROM gc
If you are using SQL 2012 or newer you can use LAG to produce a running total.
https://msdn.microsoft.com/en-us/library/hh231256(v=sql.110).aspx