How to aggregate different CTEs in outer query SQL - sql

i am trying to join two ctes to get the difference in performance of different countries and group on id here is my example
every campaign can be done in different countries, so how can i group by at the end to have 1 row per campaign id ?
CTE 1: (planned)
select
country
, campaign_id
, sum(sales) as planned_sales
from table x
group by 1,2
CTE 2: (Actual)
select
country
, campaign_id
, sum(sales) as actual_sales
from table y
group by 1,2
outer select
select
country,
planned_sales,
actual_sales
planned - actual as diff
from cte1
join cte2
on campaign_id = campaign_id


This should do it:
select
cte1.campaign_id,
sum(cte1.planned_sales),
sum(cte2.actual_sales)
sum(cte1.planned_sales) - sum(cte2.actual_sales) as diff
from cte1
join cte2
on cte1.campaign_id = cte2.campaign_id and cte1.country = cte2.country
group by 1


I would suggest using full join, so all data is included in both tables, not just data in one or the other. Your query is basically correct but it needs a group by.
select campaign_id,
sum(cte1.planned_sales) as planned_sales
sum(cte2.actual_sales) as actual_sales,
(coalesce(sum(cte1.planned_sales), 0) -
coalesce(sum(cte2.actual_sales), 0)
) as diff
from cte1 full join
cte2
using (campaign_id, country)
group by campaign_id;
That said, there is no reason why the CTEs should aggregate by both campaign and country. They could just aggregate by campaign id -- simplifying the query and improving performance.

Related

How to write SQL query without join?

Recently during an interview I was asked a question: if I have a table like as below:
The requirement is: how many orders and how many shipments per day (based on date column) - output needs to be like this:
I have written the following code, but interviewer ask me to write a SQL query without JOIN and UNION, achieve the same output.
SELECT
COALESCE(a.order_date, b.ship_date), orders, shipments
FROM
(SELECT
order_date, COUNT(1) AS orders
FROM
table
GROUP BY 1) a
FULL JOIN
(SELECT
ship_date, COUNT(1) AS shipments
FROM table) b ON a.order_date = b.ship_date
Is this possible? Could you guys please advice?

You can use UNION and GROUP BY with conditional aggregation as follows:
SELECT DATE_,
COUNT(CASE WHEN FLAG = 'ORDER' THEN 1 END) AS ORDERS,
COUNT(CASE WHEN FLAG = 'SHIP' THEN 1 END) AS SHIPMENTS
FROM (SELECT ORDER_DATE AS DATE_, 'ORDER' AS FLAG FROM YOUR_TABLE
UNION ALL
SELECT SHIP_DATE AS DATE_, 'SHIP' AS FLAG FROM YOUR_TABLE) T

In BigQuery, I would express this as:
select date, countif(n = 0) as orders, countif(n = 1) as numships
from t cross join
unnest(array[order_date, ship_date]) date with offset n
group by 1
order by date;
The advantage of this approach (over union all) is two-fold. First, it only scans the table once. More importantly, the unnest() is all on the same node where the data resides -- so data does not need to be moved for the unpivot.

SQL Total Distinct Count on Group By Query

Trying to get an overall distinct count of the employees for a range of records which has a group by on it.
I've tried using the "over()" clause but couldn't get that to work. Best to explain using an example so please see my script below and wanted result below.
EDIT:
I should mention I'm hoping for a solution that does not use a sub-query based on my "sales_detail" table below because in my real example, the "sales_detail" table is a very complex sub-query.
Here's the result I want. Column "wanted_result" should be 9:
Sample script:
CREATE TEMPORARY TABLE [sales_detail] (
[employee] varchar(100),[customer] varchar(100),[startdate] varchar(100),[enddate] varchar(100),[saleday] int,[timeframe] varchar(100),[saleqty] numeric(18,4)
);
INSERT INTO [sales_detail]
([employee],[customer],[startdate],[enddate],[saleday],[timeframe],[saleqty])
VALUES
('Wendy','Chris','8/1/2019','8/12/2019','5','Afternoon','1'),
('Wendy','Chris','8/1/2019','8/12/2019','5','Morning','5'),
('Wendy','Chris','8/1/2019','8/12/2019','6','Morning','6'),
('Dexter','Chris','8/1/2019','8/12/2019','2','Mid','2.5'),
('Jennifer','Chris','8/1/2019','8/12/2019','4','Morning','2.75'),
('Lila','Chris','8/1/2019','8/12/2019','2','Morning','3.75'),
('Rita','Chris','8/1/2019','8/12/2019','2','Mid','1'),
('Tony','Chris','8/1/2019','8/12/2019','4','Mid','2'),
('Tony','Chris','8/1/2019','8/12/2019','1','Morning','6'),
('Mike','Chris','8/1/2019','8/12/2019','4','Mid','1.5'),
('Logan','Chris','8/1/2019','8/12/2019','3','Morning','6.25'),
('Blake','Chris','8/1/2019','8/12/2019','4','Afternoon','0.5')
;
SELECT
[timeframe],
SUM([saleqty]) AS [total_qty],
COUNT(DISTINCT [s].[employee]) AS [employee_count1],
SUM(COUNT(DISTINCT [s].[employee])) OVER() AS [employee_count2],
9 AS [wanted_result]
FROM (
SELECT
[employee],[customer],[startdate],[enddate],[saleday],[timeframe],[saleqty]
FROM
[sales_detail]
) AS [s]
GROUP BY
[timeframe]
;

If I understand correctly, you are simply looking for a COUNT(DISTINCT) for all employees in the table? I believe this query will return the results you are looking for:
SELECT
[timeframe],
SUM([saleqty]) AS [total_qty],
COUNT(DISTINCT [s].[employee]) AS [employee_count1],
(SELECT COUNT(DISTINCT [employee]) FROM [sales_detail]) AS [employee_count2],
9 AS [wanted_result]
FROM #sales_detail [s]
GROUP BY
[timeframe]

You can try this below option-
SELECT
[timeframe],
SUM([saleqty]) AS [total_qty],
COUNT(DISTINCT [s].[employee]) AS [employee_count1],
SUM(COUNT(DISTINCT [s].[employee])) OVER() AS [employee_count2],
[wanted_result]
-- select count form sub query
FROM (
SELECT
[employee],[customer],[startdate],[enddate],[saleday],[timeframe],[saleqty],
(select COUNT(DISTINCT [employee]) from [sales_detail]) AS [wanted_result]
--caculate the count with first sub query
FROM [sales_detail]
) AS [s]
GROUP BY
[timeframe],[wanted_result]

Use a trick where you only count each person on the first day they are seen:
select timeframe, sum(saleqty) as total_qty),
count(distinct employee) as employee_count1,
sum( (seqnum = 1)::int ) as employee_count2
9 as wanted_result
from (select sd.*,
row_number() over (partition by employee order by startdate) as seqnum
from sales_detail sd
) sd
group by timeframe;
Note: From the perspective of performance, your complex subquery is only evaluated once.

SQL Oracle Find Max of count

I have this table called item:
| PERSON_id | ITEM_id |
|------------------|----------------|
|------CP2---------|-----A03--------|
|------CP2---------|-----A02--------|
|------HB3---------|-----A02--------|
|------BW4---------|-----A01--------|
I need an SQL statement that would output the person with the most Items. Not really sure where to start either.

I advice you to use inner query for this purpose. the inner query is going to include group by and order by statement. and outer query will select the first statement which has the most items.
SELECT * FROM
(
SELECT PERSON_ID, COUNT(*) FROM TABLE1
GROUP BY PERSON_ID
ORDER BY 2 DESC
)
WHERE ROWNUM = 1
here is the fiddler link : http://sqlfiddle.com/#!4/4c4228/5

Locating the maximum of an aggregated column requires more than a single calculation, so here you can use a "common table expression" (cte) to hold the result and then re-use that result in a where clause:
with cte as (
select
person_id
, count(item_id) count_items
from mytable
group by
person_id
)
select
*
from cte
where count_items = (select max(count_items) from cte)
Note, if more than one person shares the same maximum count; more than one row will be returned bu this query.

How to join three select queries which has one common column

I have three select queries as below which gives a respective output
select DATE_FORMAT(table1.value_date,'%b')as Month,
DATE_FORMAT(table1.value_date,'%Y') as Year,
table1.open as Open
from index_main as table1
join ( select min(`value_date`) `value_date`
from index_main
group by month(`value_date`), year( `value_date`)
) as table2 on table1.`value_date` = table2.`value_date`
Output columns - Month,year,open
select DATE_FORMAT(table1.value_date,'%b')as Month,
DATE_FORMAT(table1.value_date,'%Y') as Year,
table1.close as Open
from index_main as table1
join ( select max(`value_date`) `value_date`
from index_main group by month(`value_date`), year( `value_date`)
) as table2 on(table1.`value_date` = table2.`value_date`)
Output columns - Month,year,close
select DATE_FORMAT(table1.value_date,'%b')as Month,
DATE_FORMAT(table1.value_date,'%Y') as Year,
max(table1.high) as High
FROM `index_main` as table1
GROUP BY table1.month,table1.year
ORDER BY year(table1.value_date) desc, month(table1.value_date) desc
Output columns - Month,year,high,low
I want to join these three select queries based on the common columns i.e month & year.
My final result should have the following columns - month,year,open,close,high,low.

Try this.
First create 3 views, one with each query (vw1, vw2 and vw3). Then use a query like this:
SELECT vw1.Month, vw1.Year, Open, Close High FROM vw1 LEFT join vw2 on vw1.Year=vw2.Year and vw1.Month=vw2.Month LEFT JOIN vw3 on vw1.Year=vw3.Year and vw1.Month=vw3.Month
Hope this helps you.

Creating an accumulating rollup

I have records in a table that have codes specific to a certain level and an amount attached to that level. They do not add up and that is not the issue.
I wish to create a query that sums up all the values by the level code plus those in the levels below it. I would also like the amount per level in the same query, but it is not necessary. I have create a sample table and output below. Does anyone have a good way of doing this? Also, is there an actual definition for this kind of roll up?
CREATE TABLE LEVEL_AMOUNTS(
LEVEL_CODE char(1)
AMOUNT integer
)
INSERT INTO LEVEL_AMOUNTS VALUES
('A',1),('A',1),('A',1),('A',1),
('B',1),('B',1),('B',1),('B',1),
('C',1),('C',1),('C',1),('C',1)
Output:
A | 12
B | 8
C | 4

with cte as (
select distinct level_code
from level_amounts
)
select l.level_code, sum(l.amount)
from cte
inner join level_amounts l on l.level_code <= cte.level_code
group by l.level_code
or
select l.level_code, sum(l.amount)
from level_amounts l
inner join (select distinct level_code
from level_amounts) l1
on l.level_code <= l1.level_code
group by l.level_code;
sqlfiddle

SQL Server 2008 does not have cumulative sums. You can do this with subqueries or joins:
with cte as (
select level_code, sum(amount) as amount
from amounts
group by level_code
)
select level_code,
(select sum(amount) from cte cte2 where cte.level_code <= cte2.level_code) as cumamount
from cte;