I have a sample DB below. I'm looking to see how many TV and Internet bundles we sold. In the sample data, only Bob and Trevor sold that bundle so we sold 2.
How do I write the query for the number of bundles sold by each Sales rep and the total price of the bundles sold?
Thanks
I imagine that, for a bundle to happen, the same sales person needs to have sold both products to the same customer.
I would approach this with two levels of aggregation. First group by sales person and customer in a subquery to identify the bundles, then, in an outer query, count how many such bundles happened for each sales person:
SELECT sales_person, COUNT(*) bundles_sold, SUM(total_price) total_price
FROM (
SELECT sales_person, customer_name, SUM(total_price) total_price
FROM mytable
WHERE product_name in ('TV', 'Phone')
GROUP BY sales_person
HAVING COUNT(DISTINCT product_name) = 2
) x
You can simply group the salesman's by counting the distinct products they sold -
SELECT Sales_Person, FLOOR(COUNT(DISTINCT product_name)/2) NO_OF_BUNDLES, sum(total_price)
FROM YOUR_TAB
WHERE product_name IN ('TV', 'Internet')
GROUP BY Sales_Person
HAVING COUNT(DISTINCT product_name) >= 2
Using cte as below:
with cte1(sales_person, customer_name, product_count) as
(
select sales_person, customer_name, count(product_name)
from sales
where product_name in ('TV', 'Internet')
group by sales_person, customer_name
having count(product_name) = 2
)
select sales_person, count(product_count)
from cte1
group by sales_person
I would suggest two levels of aggregation:
select sales_person, count(*), sum(total_price)
from (select sales_person, customer_name,
sum(total_price) as total_price,
max(case when product_name = 'tv' then 1 else 0 end) as has_tv,
max(case when product_name = 'phone' then 1 else 0 end) as has_phone,
max(case when product_name = 'internet' then 1 else 0 end) as has_internet
from t
group by sales_person, customer_name
) sc
where has_phone = 0 and
has_tv = 1 and
has_internet = 1
group by sales_person;
I recommend this structure because it is pretty easy to change the conditions in the where clause to return this for any bundle -- or even to aggregate by the three flags and return the totals for all bundles in one query.
Related
I have a sales table in SQLite:
purchase_date
units_sold
customer_id
15
1
1
17
1
1
30
3
1
I want to get the total unit_solds for each customer on the first date and last date of their purchases. My query is:
select customer_id,
sum(units_sold) total_units_sold
from sales
group by customer_id
having purchase_date = min(purchase_date)
or purchase_date = max(purchase_date)
I was expecting results like:
customer_id
total_units_sold
1
4
but I got:
customer_id
total_units_sold
1
5
I would like to know why this solution doesn't work.
The order of the phrase is incorrect
Note: The having statement is executed after compilation.
You need to get the results as partial queries
For example, I arranged to know the first line of the date according to each customer
as well as the last line of the date (by getting the first line after descending order)
and then execute the group statement
The example is complete
select customer_id,sum(units_sold) from (
select customer_id, units_sold,purchase_date,
ROW_NUMBER() over(partition by customer_id order by purchase_date) As RowDatefirst,
ROW_NUMBER() over(partition by customer_id order by purchase_date desc)As RowDatelast
from sales
) t where t.RowDatefirst = 1 or t.RowDatelast=1
group by customer_id
Try this:
SELECT a.customer_id, SUM(a.units_sold) as total_units_sold
FROM sales a
INNER JOIN (
SELECT customer_id, MIN(purchase_date) as _first ,MAX(purchase_date) as _last
FROM sales
GROUP BY customer_id
) b ON a.customer_id = b.customer_id AND
(a.purchase_date = b._first OR a.purchase_date = b._last)
GROUP BY a.customer_id
http://sqlfiddle.com/#!7/0a4a4/7
I have this table called table a
I need to get the CustomerID, sum(Income) of 2015, sum(Income) of 2016, did he ever bought productId A (boolean), is the total sum(income)> 1000 (boolean), number of total InvoiceID
all that in one query and the results should be with 1 row per customer.
please help I don't even know how to start!
This is basically conditional aggregation:
select customerid,
sum(case when extract(year from date) = 2015 then sales end) as sales_2015,
sum(case when extract(year from date) = 2016 then sales end) as sales_2016,
max( product = 'A' ) as ever_bought_a,
sum(income) > 1000 as sum_exceeds_1000,
count(*) as num_invoices
from t
group by customerid;
You haven't specified a database, so this is really psuedocode. You'll need to adapt it for your particular database.
Help me please build PostgreSQL query.
There are 2 tables: products(id, title) and prices(id, product_id, price_type, moment, value)
moment - timestamp, can be in past or future
Assume that price_type has only two option: retail or purchase
But one product may has many retail prices with different moments.
I need select all products with actual retail and purchase prices, where moment less than now.
It's I can done
SELECT
products.id,
products.title_translations AS title,
retail_prices.moment AS ret_moment,
pur_prices.value AS purchase,
retail_prices.value AS retail
FROM products
LEFT OUTER JOIN prices AS pur_prices ON products.id=pur_prices.product_id AND pur_prices.price_type='purchase' AND pur_prices.moment<current_timestamp
LEFT OUTER JOIN prices AS retail_prices ON products.id=retail_prices.product_id AND retail_prices.price_type='retail' AND retail_prices.moment<current_timestamp
ORDER BY products.id;
It works, but returns
product with all prices, but I need only last prices(by moment).
Just use ROW_NUMBER to find what is the last price before current time
with last_prices as (
SELECT
products.id,
products.title_translations AS title,
prices.moment,
prices.value,
prices.price_type,
ROW_NUMBER() OVER (PARTITION BY product_id, price_type
ORDER BY moment DESC) as rn
FROM products
LEFT JOIN prices
ON products.id = prices.product_id
WHERE moment < now()
)
SELECT id, title,
MAX(CASE WHEN price_type = 'retail'
THEN moment
END) as retail_moment,
MAX(CASE WHEN price_type = 'retail'
THEN value
END) as retail_price,
MAX(CASE WHEN price_type = 'purchase'
THEN moment
END) as purchase_moment,
MAX(CASE WHEN price_type = 'purchase'
THEN value
END) as purchase_price
FROM last_prices
WHERE rn = 1
GROUP BY id, title
ORDER BY id
To keep things organized, and straight in my mind, I'd use CTEs to generate two subsets of price data, one for purchase one for retail and assign a row number in ascending sequence with the lowest number having the most recent moment less than the currenttimestmap. And then when we join to these ctes, we only return the lowest number assigned.
With Pur_prices as (SELECT P.*, row_Number() over (partition by product_ID order by moment desc) RN
FROM prices P
WHERE price_Type = 'purchase'
and p.moment < current_timestamp)
, Retail_prices as (SELECT P.*, row_Number() over (partition by product_ID order by moment desc) RN
FROM prices P
WHERE price_Type = 'retail'
and p.moment < current_timestamp)
SELECT
p.id,
p.title_translations AS title,
rp.moment AS ret_moment,
rp.value AS retail,
pp.moment AS Pur_moment,
pp.value AS purchase
FROM products p
LEFT JOIN pur_prices pp
ON p.id=pp.product_id
AND pp.RN = 1 --Only show the most recent price less than current time
LEFT JOIN retail_prices rp
ON p.id=rp.product_id
AND RP.RN = 1 --Only show the most recent price less than current time
ORDER BY p.id;
The end result should be all products regardless if they have a retail or purchase price; but if they do show the retail/purchase pricing for the most recent moment before now. My only concern is this implies all pricing has a moment they start (no null values allowed!)
You may be wanting it be ordered with respect to moment in descending order.
Change
ORDER BY products.id;
to
ORDER BY product.id ASC, moment DESC;
I have a table with purchased orders data.
Each row contails the amount of certain item purchased, cost per item and the order number group. Each different item purchased is a new row with same order number.
I basically want to return the total cost for that order. I have tried the following but am getting nowhere:
SELECT order_number, SUM( sub_total ) AS `total`
FROM
SELECT order_number, SUM( SUM( amount ) * SUM( cost_per_item ) ) AS `sub_total`
FROM `ecom_orders`
WHERE member_id = '4'
GROUP BY order_number
ORDER BY purchase_date DESC
Pretty much any SQL-92 compliant RDBMS will take this:
SELECT
order_number
,SUM(amount * cost_per_item) AS total
,purchase_date
FROM
ecom_orders
WHERE member_id = '4'
GROUP BY order_number,purchase_date
ORDER BY purchase_date DESC
Let's say I have a table:
SELECT SUM(quantity) AS items_sold_since_date,
product_ID
FROM Sales
WHERE order_date >= '01/01/09'
GROUP BY product_ID
This returns a list of products with the quantity sold since a particular date. Is there a way to select not only this sum, but ALSO the sum WITHOUT the where condition? I'd like to see sales since a particular date for each product alongside all (not date limited) sales.
SELECT SUM(CASE WHEN order_date >= '01/01/09' THEN quantity ELSE 0 END) AS items_sold_since_date,
SUM(quantity) AS items_sold_total,
product_ID
FROM Sales
GROUP BY product_ID
something like this?:
SELECT SUM(quantity) AS items_sold_since_date,
total_items_sold = (SELECT SUM(quantity) from Sales GROUP BY product_ID),
product_ID
FROM Sales
WHERE order_date >= '01/01/09'
GROUP BY product_ID
If you like to see total sales alongside, then you would use sum(sale_amt), and in the group by add the sale_amt. I hope it helps.
You could use GROUP BY to split up the Sales based on date.
In Oracle you could say:
select count(*)
,case when order_date >= '01/01/09' then 'after' else 'before' end
from log
group by case when order_date >= '01/01/09' then 'after' else 'before' end;
you can write
SELECT SUM(quantity) AS items_sold_since_date,(SELECT SUM(quantity) AS items_sold_since_date FROM Sales
GROUP BY product_ID) as items_sold,
product_ID
FROM Sales
WHERE order_date >= '01/01/09'
GROUP BY product_ID