Summing up of columns from different tables - sql

here is my data,
Table 1:
STORAGE HANDLING TOTAL BILLING
--------------------------------------
1300 10900
0 10950
0 6000
0 5950
Table 2:
LINER REVENUE
---------------
1300
250
3000
200
I need to calculate Total Billing:
Total Billing = Storage+Handling+Liner Revenue.
Can someone guide me a query for this.

Hope this helps,
SELECT h.Storage+h.Handling+j.[Liner Revenue]
From(SELECT
STORAGE
,HANDLING
,[TOTAL BILLING]
,ROW_NUMBER () over(order by rand()) as p
FROM [Table 1] )h
INNER JOIN
(SELECT
[LINER REVENUE]
,ROW_NUMBER () over(order by rand()) as p
FROM [Table 2] )j
on h.p=j.p

Related

SQL | How to sum over partition group of 3 items?

I'm trying to get the percent of a day's revenue for top 3 product categories but struggling with the percentage. I have already the revenue per product ranked 1 to 3 but cant wrap my head on how to get the percentage.
Any pointers will be appreciated.
SELECT * FROM (
SELECT date,
category_name,
revenue,
row_number() OVER(PARTITION BY DATE(date) ORDER BY revenue DESC) AS category_rank,
(revenue / (select sum(revenue) from a group by 1)) * 100 percentage AS percentage_of_daily_total -- this is the wrong one
FROM (
SELECT DATE(orders.order_timestamp) AS date,
products.product_cat AS category_name,
ROUND(SUM(payments.payment)) AS revenue
FROM table1.orders orders
JOIN table1.t_payments payments ON orders.order_id = payments.order_id
JOIN table1.t_items items ON orders.order_id = items.order_id
JOIN table1.t_products products ON items.product_id = products.product_id
GROUP BY 1 ,2) a) b
WHERE category_rank <= 3;
Sample data is as follow: date, category_name, revenue, category_rank
2016-10-05 jeans 20 1
2016-10-05 shirts 15 2
2016-10-05 shoes 10 3
2016-10-06 skirts 50 1
2016-10-06 sports_wear 30 2
2016-10-06 accesories 10 3
Desired outcome:date, category_name, revenue, category_rank, percentage_of_daily_total
2016-10-05 jeans 30 1 50
2016-10-05 shirts 20 2 33
2016-10-05 shoes 10 3 17
2016-10-06 skirts 20 1 50
2016-10-06 sports_wear 16 2 40
2016-10-06 accessories 4 3 10
Use CTEs
WITH a AS (
SELECT DATE(orders.order_timestamp) AS date,
products.product_cat AS category_name,
ROUND(SUM(payments.payment)) AS revenue
FROM table1.orders orders
JOIN table1.t_payments payments ON orders.order_id = payments.order_id
JOIN table1.t_items items ON orders.order_id = items.order_id
JOIN table1.t_products products ON items.product_id = products.product_id
GROUP BY 1 ,2
)
SELECT * FROM (
SELECT a.date,
a.category_name,
a.revenue,
row_number() OVER(PARTITION BY DATE(a.date) ORDER BY a.revenue DESC) AS category_rank,
(a.revenue / b.revenue_sum) * 100 percentage AS percentage_of_daily_total
FROM a
JOIN (SELECT date, sum(revenue) AS revenue_sum FROM a GROUP BY 1) AS b
ON a.date = b.date)
WHERE category_rank <= 3;
Your original query is very close. Calculate the percent in the OUTERMOST sql. So drop your percentage calc and then in the outer Select:
Select *, 100*revenue/(sum(revenue) Over (Partition by Date)) as percentage_of_daily_total
Remember that by the time you get to calculating the windowing functions the Where clause has already been executed and you're down to 3 records per day so any total will only be based on the top 3.

SQL query to calculate totals

I'm quite new to SQL querying so please go easy on me if what I've done so far is really odd :)
I have two tables - A for Income and B for Expenditure:
Business_ID Income_Desc Income_Amount
1 Income A 1000
1 Income B 3000
1 Income C 2000
Business_ID Expen_Amount
1 2500
I'd like to produce a table that shows each of the income amounts, the one expenditure amount, the total income, the total expenditure and a Grand Total of total income-total expenditure.
Something like this if possible
Business_ID Income Description Income Amount Expenditure Amount Total
1 Income A 1000 2500 -
1 Income B 3000 - -
1 Income C 2000 - -
1 All Amounts 6000 2500 3500
This is what I've tried so far
SELECT a. Business_ID, COALESCE (a.Income_Desc, 'All Amounts') AS 'Income Description', SUM(a.Income_Amount) AS 'Income Amount', SUM(b.Expen_Amount) AS Expenditure Amount', (sum(a.Income_Amount)-SUM(b.Expen_Amount)) AS 'Total'
FROM Income AS a LEFT JOIN Expenditure AS b ON a.Business_ID = b. Business_ID
GROUP BY a. Business_ID, a.Income_Desc WITH ROLLUP
The result I'm getting is this
Business_ID Income Description Income Amount Expenditure Amount Total
1 Income A 1000 2500 -1500
1 Income B 3000 2500 500
1 Income C 2000 2500 -500
1 All Amounts 6000 7500 -1500
All Amounts 6000 7500 -1500
Is it possible to get an output like the one I provided above? Could you show me how to achieve it (or something very close) please?
Thanks
You can use row_number() for the join:
with ie as (
select i.business_id, i.income_desc, i.income_amount,
e.expen_amount
from (select i.*,
row_number() over (partition by business_id order by income_desc) as seqnum
from income i
) i left join
(select e.*,
row_number() over (partition by business_id order by expen_amount) as seqnum
from expenditure e
) e
on i.business_id = e.business_id and i.seqnum = e.seqnum
)
select ie.*
from ie
union all
select business_id, 'Total', sum(income_amount), sum(expen_amount)
from ie
group by business_id;
You could make a sub query out of your original query and only select values where the business ID is not null. Furthermore, use CASE WHEN to identify those values < 0 and replace it with "-":
SELECT x.Business_ID
, x.`Income Description`
, x.`Income Amount`
, x.`Expenditure Amount`
, x.Total
FROM
(SELECT a. Business_ID
, COALESCE (a.Income_Desc, 'All Amounts') AS 'Income Description'
, SUM(a.Income_Amount) AS 'Income Amount'
, SUM(b.Expen_Amount) AS 'Expenditure Amount'
, CASE WHEN (sum(a.Income_Amount)-SUM(b.Expen_Amount)) < 0
THEN '-'
ELSE (sum(a.Income_Amount)-SUM(b.Expen_Amount))
END AS 'Total'
FROM Income AS a
LEFT JOIN Expenditure AS b ON a.Business_ID = b.Business_ID
WHERE a.Business_ID is not null and b.Business_ID is not null
GROUP BY a.Business_ID, a.Income_Desc WITH ROLLUP) as x
where x.Business_ID is not null
DB Fiddle

calculating percentage of sales profit in SQL

Product Group Product ID Sales Profit
A 6797 1,000 200
A 6745 500 90
B 1278 200 60
B 1245 1,500 350
C 7890 650 80
D 4587 350 50
Q1). Filter out product IDs that contribute to top 80% of the total profit of their respective group.
Not sure what rdbms you are using you can get the output in SQL server in this way. You can get profit for a group and use aggregate function to compare and filter the rows.
select 'A' as Product_group, 6797 as ProductID, 1000 as Sales , 200 as Profit into #temp1 union all
select 'A' as Product_group, 6745 as ProductID, 500 as Sales , 90 as Profit union all
select 'B' as Product_group, 1278 as ProductID, 200 as Sales , 60 as Profit union all
select 'B' as Product_group, 1245 as ProductID, 1500 as Sales , 350 as Profit union all
select 'C' as Product_group, 7890 as ProductID, 650 as Sales , 80 as Profit union all
select 'D' as Product_group, 4587 as ProductID, 350 as Sales , 50 as Profit
select t.Product_group, t.ProductID, sum(t.sales) totalsles, sum(t.profit) totalProfit, sum(Profit_grp) Groupprofit from #temp1 t
join (select Product_group, sum(sales) totalsles_group, sum(profit) Profit_grp from #temp1 t1 group by Product_group) t1 on t1.Product_group = t.Product_group
group by t.Product_group, t.ProductID
having sum(t.profit) *1.0/ sum(t1.Profit_grp) *1.0 >= 0.8
Output: I added group profit just to compare. You can remove the aggregate and add in group by if you would like
Product_group ProductID totalsles totalProfit Groupprofit
B 1245 1500 350 410
C 7890 650 80 80
D 4587 350 50 50
I think this may works out:
with CTE as(
select [Product Group], sum([Sales]) as Tolsum from Table
group by [Product Group]
select prod.*,
sum(prod.[Profit]/cte.[Tolsum]) over (Partition by prod.[Product Group] Order by prod.[Product ID]) as contribution
from CTE cte
inner join
Table prod
on
cte.[Product Group] = prod.[Product Group]
having
sum(prod.[Profit]/cte.[Tolsum]) over (Partition by prod.[Product Group] Order by prod.[Product ID]) < 0.8

T-SQL calculate the percent increase or decrease between the earliest and latest for each project

I have a table like below, I am trying to run a query in T-SQL to get the earliest and latest costs for each project_id according to the date column and calculate the percent cost increase or decrease and return the data-set show in the second table (I have simplified the table in this question).
project_id date cost
-------------------------------
123 7/1/17 5000
123 8/1/17 6000
123 9/1/17 7000
123 10/1/17 8000
123 11/1/17 9000
456 7/1/17 10000
456 8/1/17 9000
456 9/1/17 8000
876 1/1/17 8000
876 6/1/17 5000
876 8/1/17 10000
876 11/1/17 8000
Result:
(Edit: Fixed the result)
project_id "cost incr/decr pct"
------------------------------------------------
123 80% which is (9000-5000)/5000
456 -20%
876 0%
Whatever query I run I get duplicates.
This is what I tried:
select distinct
p1.Proj_ID, p1.date, p2.[cost], p3.cost,
(nullif(p2.cost, 0) / nullif(p1.cost, 0)) * 100 as 'OVER UNDER'
from
[PROJECT] p1
inner join
(select
[Proj_ID], [cost], min([date]) min_date
from
[PROJECT]
group by
[Proj_ID], [cost]) p2 on p1.Proj_ID = p2.Proj_ID
inner join
(select
[Proj_ID], [cost], max([date]) max_date
from
[PROJECT]
group by
[Proj_ID], [cost]) p3 on p1.Proj_ID = p3.Proj_ID
where
p1.date in (p2.min_date, p3.max_date)
Unfortunately, SQL Server does not have a first_value() aggregation function. It does have an analytic function, though. So, you can do:
select distinct project_id,
first_value(cost) over (partition by project_id order by date asc) as first_cost,
first_value(cost) over (partition by project_id order by date desc) as last_cost,
(first_value(cost) over (partition by project_id order by date desc) /
first_value(cost) over (partition by project_id order by date asc)
) - 1 as ratio
from project;
If cost is an integer, you may need to convert to a representation with decimal places.
You can use row_number and OUTER APPLY over top 1 ... prior to SQL 2012
select
min_.projectid,
latest_.cost - min_.cost [Calculation]
from
(select
row_number() over (partition by projectid order by date) rn
,projectid
,cost
from projectable) min_ -- get the first dates per project
outer apply (
select
top 1
cost
from projectable
where
projectid = min_.projectid -- get the latest cost for each project
order by date desc
) latest_
where min_.rn = 1
This might perform a little better
;with costs as (
select *,
ROW_NUMBER() over (PARTITION BY project_id ORDER BY date) mincost,
ROW_NUMBER() over (PARTITION BY project_id ORDER BY date desc) maxcost
from table1
)
select project_id,
min(case when mincost = 1 then cost end) as cost1,
max(case when maxcost = 1 then cost end) as cost2,
(max(case when maxcost = 1 then cost end) - min(case when mincost = 1 then cost end)) * 100 / min(case when mincost = 1 then cost end) as [OVER UNDER]
from costs a
group by project_id

How can I sum cost items, grouped by invoice?

I have two SQL Server tables below:
Invoice
InvoiceId Amount [Date]
1 10 2015-05-28 21:47:50.000
2 20 2015-05-28 21:47:50.000
3 25 2015-05-28 23:25:50.000
InvoiceItem
Id InvoiceId Cost
1 1 8
2 1 3
3 1 7
4 2 15
5 2 17
6 3 20
7 3 22
Now I want to JOIN these two tables ON InvoiceId and retrieve the following:
COUNT of DISTINCT InvoiceId from Invoice table AS [Count]
SUM of Amount from Invoice table AS Amount
SUM of Cost from InvoiceItem table AS Cost
HOUR part of [Date]
and GROUP them BY HOUR part of [Date].
Desired Output wil be:
[Count] Amount Cost HourOfDay
2 30 50 22
1 25 42 23
How can I do this?
one approach is to use a derived table:
SELECT CAST([Date] AS DATE) AS [Date],
DATEPART(HOUR,i.[Date]) AS HourOfDay,
COUNT(i.InvoiceId) AS NumberOfInvoices,
SUM(i.Amount) AS Amount,
SUM(it.Cost) AS Cost
FROM invoice i
INNER JOIN
(SELECT InvoiceId, SUM(Cost) AS Cost
FROM invoiceitem
GROUP BY InvoiceId) it ON i.InvoiceId = it.InvoiceId
GROUP BY [Date],DATEPART(HOUR,i.[Date])
or a CTE (Common Table Expression)
WITH InvoiceCosts (InvoiceId, Cost)
AS
(
SELECT InvoiceId, SUM(Cost) AS Cost
FROM invoiceitem
GROUP BY InvoiceId
)
SELECT CAST([Date] AS DATE) AS [Date],
DATEPART(HOUR,i.[Date]) AS HourOfDay,
COUNT(i.InvoiceId) AS NumberOfInvoices,
SUM(i.Amount) AS Amount,
SUM(ic.Cost) AS Cost
FROM invoice i
INNER JOIN
InvoiceCosts ic ON i.InvoiceId = ic.InvoiceId
GROUP BY [Date],DATEPART(HOUR,i.[Date])
SELECT COUNT (DISTINCT inv.InvoiceId) [Count],
SUM (Amount) Amount,
SUM (Cost) Cost,
datepart(HOUR, inv.[Date]) HourOfDay
FROM Invoice inv
INNER JOIN InvoiceItem itm
ON inv.InvoiceId = itm.InvoiceId
GROUP BY datepart(HOUR, inv.[Date]);