I have a table that contains employee names, sales from current year, and sales from last year. Lets call the two sales columns 2022 and 2021. Im looking to sort my table by the highest difference between this year and last years sales. For example, the highest difference would be at the top.
Currently I have it as
SELECT
DISTINCT customerid,
full_name,
"2012 Sales",
"2013 Sales"
FROM
customer_loyalty
ORDER BY "2013 Sales" DESC
limit 10;
Can i just insert a where conditon like 2013-2012 ASC?
This should work:
SELECT
DISTINCT customerid,
full_name,
`2012 Sales`,
`2013 Sales`
FROM
customer_loyalty
ORDER BY
`2013 Sales` - `2012 Sales` DESC
LIMIT 10;
Related
i WanT a query to find the number of days between the longest and least tenured employee still working for the company. The output should include the number of employees with the longest-tenure, the number of employees with the least-tenure, and the number of days between both the longest-tenured and least-tenured hiring dates.
this is what i have so far,i am having trouble getting the difference to appear in output
select a.count, date_tenure
from
(
select distinct hire_date ,current_date - hire_date as date_tenure, count(id) over (partition by hire_date) as count, rank() over (order by current_date-hire_date) as rank_asc,
rank() over (order by current_date-hire_date desc) as rank_desc
from employees
where termination_date is null
order by 2 desc) a
where rank_desc = 1
or rank_asc = 1
table : employees
id
hire_date
termination_date
I have 2 tables
sales table
weekly sales, store, date
store table
store, type, size
my sales table has multiple years, multiple stores and multiple types. I'm trying to get the avg sales by sqft for each store type per year. I have a sub query that shows the sales by sqft for each store but Im having trouble then rolling it up into my main query to get the avg by type
Anything jumps out with my final query?
SELECT
date_part('year', sales.date) AS year,
stores.type,
AVG(sales_by_sqft)
FROM
(SELECT
SUM((sales.weekly_sales)/stores.size) AS sales_by_sqft
FROM SALES
INNER JOIN stores ON sales.store = stores.store
GROUP BY sales.store) AS sq
FROM sales
INNER JOIN stores ON sales.store = stores.store
WHERE date_part('year', date) = 2012
GROUP BY year, stores.type;
getting a syntax error on the second FROM statement
I figured it out. AVG doesn't work on money. Once I changed that data type to integer, it all fell in place
SELECT
year,
type,
ROUND(AVG(sales_by_sqft),2)AS avg_sales_by_sqft
FROM
(SELECT
date_part('year', sales.date) AS year,
stores.type,
sales.store,
stores.size,
SUM(sales.weekly_sales) AS total_sales,
SUM(sales.weekly_sales)/ AVG(stores.size) AS sales_by_sqft
FROM sales
INNER JOIN stores ON sales.store = stores.store
GROUP BY year, stores.type, sales.store, stores.size) AS sq
GROUP BY 1,2
ORDER BY 1,3 DESC;
Suppose, I have following tables
product_prices
product|price|date
-------+-----+----------
apple |10 |2014-03-01
-------+-----+----------
apple |20 |2014-05-02
-------+-----+----------
egg |2 |2014-03-03
-------+-----+----------
egg |4 |2015-10-12
purchases:
user|product|date
----+-------+----------
John|apple |2014-03-02
----+-------+----------
John|apple |2014-06-03
----+-------+----------
John|egg |2014-08-13
----+-------+----------
John|egg |2016-08-13
What I need is table similar to this:
name|product|purchase date |price date|price
----+-------+--------------+----------+-----
John|apple |2014-03-02 |2014-03-01|10
----+-------+--------------+----------+-----
John|apple |2014-06-03 |2014-05-02|20
----+-------+--------------+----------+-----
John|egg |2014-08-13 |2014-08-13|2
----+-------+--------------+----------+-----
John|egg |2016-08-13 |2015-10-12|4
Or "what is the price for product at this day". Where price is calculated based on date from products table.
On real DB I tried to use something similar to:
SELECT name, product, pu.date, pp.date, pp.price
FROM purchases AS pu
LEFT JOIN product_prices AS pp
ON pu.date = (
SELECT date
FROM product_prices
ORDER BY date DESC LIMIT 1);
But I keep either getting only left part of table (with (null) instead of product dates and prices) or many rows with all the combinations of prices and dates.
I would suggest changing product_prices table to use a daterange column instead (or at least a start_date and an end_date).
You can use an exclusion constraint to make sure you never have overlapping ranges for one product and an insert trigger that "closes" the "current" prices and creates a new unbounded range for the newly inserted price.
A daterange can efficiently be indexed and with that in place the query gets as easy as:
SELECT name, product, pu.date, pp.valid_during, pp.price
FROM purchases AS pu
LEFT JOIN product_prices AS pp ON pu.date <# pp.valid_during
(assuming the range column is named valid_during)
The exclusion constraint would only work however if the product was an integer (not a varchar) - but I guess your real product_purchases table uses a foreign key to some product table anyway (which is an integer).
The new table definitions could look something like this:
create table purchase_prices
(
product_id integer not null references products,
price numeric(16,4) not null,
valid_during daterange not null
);
And the constraint that prevents overlapping ranges:
alter table purchase_prices
add constraint check_price_range
exclude using gist (product_id with =, valid_during with &&);
The constraint needs the btree_gist extension.
As always improving query speed comes with a price and in this case it's the higher maintenance costs for the GiST index. You would need to run some tests to see if the easier (and most probably much faster) query outweighs the slower insert performance on purchase_prices.
Look at your scalar sub-query very closely. It is not correlated back to the outer query. In other words, it will return the same result every time: the latest date in the product_prices table. Period. Think about the query out of context:
SELECT date
FROM product_prices
ORDER BY date DESC LIMIT 1
There are two problems with it:
It will return 2015-10-12 for every row in the join and ultimately, nothing was purchased on that date, hence, null.
Your approximation of closest is that the dates are equal. Unless you have a product_prices row for every product for every single date, you'll always have misses. "Closest" implies distance and ranking.
WITH close_prices_by_purchase AS (
SELECT
p.user,
p.product,
p.date pp.date,
pp.price,
row_number() over (partition by pp.product, order by pp.date desc) as distance -- calculate distance between purchase date and price date
FROM purchases AS p
INNER JOIN product_prices AS pp on pp.product = p.product
WHERE pp.date < p.date
)
SELECT user as name, product, pu.date as purchase_date, pp.date as price_date, price
FROM close_prices_by_purchase AS cpbp
WHERE distance = 1; -- shortest distance
You can try something like this, although I am sure there's a better way:
with diffs as (
select
a.*,
b."date" as bdate,
b.price,
b."date" - a."date" as diffdays,
row_number() over (
partition by "user", a."product", a."date"
order by "user", a."product", a."date", b."date" - a."date" desc
) as sr
from purchases a
inner join product_prices b on a.product = b.product
where b."date" - a."date" < 1
)
select
"user" as "name",
product,
"date" as "purchase date",
bdate as "price date",
price
from diffs
where sr = 1
Example: https://www.db-fiddle.com/f/dwQ9EXmp1SdpNpxyV1wc6M/0
Explanation
I attempted to join both tables and find the difference between dates of purchase and price, and ranked them by closest date prior to the purchase. Rank of 1 will go to the closest date. Then, data with rank of 1 was extracted.
This is a great place to use date ranges! We know the start date of the price range and we can use a window function to get the next date. At that point, it's really easy to figure out the price on any day.
with price_ranges as
(select product,
price,
date as price_date,
daterange(date, lead(date, 1)
OVER (partition by product order by date), '[)'
) as valid_price_range from product_prices
)
select "user" as name,
purchases.product,
purchases.date,
price_date,
price
from purchases
join price_ranges on purchases.product = price_ranges.product
and purchases.date <# price_ranges.valid_price_range
order by purchases.date;
I have a table called retail which stores items and their price along with date of purchase. I want to find out total monthly count of unique items sold.
This is the sql query I tried
select date_trunc('month', date) as month, sum(count(distinct(items))) as net_result from retail group by month order by date;
But I get the following error
ERROR: aggregate function calls cannot be nested
Now I searched for similar stackoverflow posts one of which is postgres aggregate function calls may not be nested and but I am unable to replicate it to create the correct sql query.
What am I doing wrong?
From your description, it doesn't seem like you need to nest the aggregate functions, the count(distinct item) construction will give you a count of distinct items sold, like so:
select date_trunc('month', date) as month
, count(distinct items) as unique_items_sold
, count(items) as total_items_sold
from retail
group by "month"
order by "month" ;
If you had a column called item_count (say if there was row in the table for each item sold, but a sale might include, say, three widgets)
select date_trunc('month', date) as month
, count(distinct items) as unique_items_sold
, sum(item_count) as total_items_sold
from retail
group by "month"
order by "month" ;
Use subqueries:
Select month, sum(citems) as net_result
from
(select
date_trunc('month', date) as month,
count(distinct(items)) as citems
from
retail
group by month
order by date
)
I am suspect your group by statement will throw an Error because your month column are condition column and you cannot put in the same level in your query so put your full expression instead.
select
month,
sum(disct_item) as net_results
from
(select
date_trunc('month', date) as month,
count(distinct items) as disct_item
from
retail
group by
date_trunc('month', date)
order by
date) as tbl
group by
month;
You cannot make nested aggregate so you wrap first count to subquery and after that in outer you make sum to do the operation.
Let's say I have an orders table with customer_id, order_total, and order_date columns. I'd like to build a report that shows all customers who haven't placed an order in the last 30 days, with a column for the total amount their last order was.
This gets all of the customers who should be on the report:
select customer, max(order_date), (select order_total from orders o2 where o2.customer = orders.customer order by order_date desc limit 1)
from orders
group by 1
having max(order_date) < NOW() - '30 days'::interval
Is there a better way to do this that doesn't require a subquery but instead uses a window function or other more efficient method in order to access the total amount from the most recent order? The techniques from How to select id with max date group by category in PostgreSQL? are related, but the extra having restriction seems to stop me from using something like DISTINCT ON.
demo:db<>fiddle
Solution with row_number window function (https://www.postgresql.org/docs/current/static/tutorial-window.html)
SELECT
customer, order_date, order_total
FROM (
SELECT
*,
first_value(order_date) OVER w as last_order,
first_value(order_total) OVER w as last_total,
row_number() OVER w as row_count
FROM orders
WINDOW w AS (PARTITION BY customer ORDER BY order_date DESC)
) s
WHERE row_count = 1 AND order_date < CURRENT_DATE - 30
Solution with DISTINCT ON (https://www.postgresql.org/docs/9.5/static/sql-select.html#SQL-DISTINCT):
SELECT
customer, order_date, order_total
FROM (
SELECT DISTINCT ON (customer)
*,
first_value(order_date) OVER w as last_order,
first_value(order_total) OVER w as last_total
FROM orders
WINDOW w AS (PARTITION BY customer ORDER BY order_date DESC)
ORDER BY customer, order_date DESC
) s
WHERE order_date < CURRENT_DATE - 30
Explanation:
In both solutions I am working with the first_value window function. The window function's frame is defined by customers. The rows within the customers' groups are ordered descending by date which gives the latest row first (last_value is not working as expected every time). So it is possible to get the last order_date and the last order_total of this order.
The difference between both solutions is the filtering. I showed both versions because sometimes one of them is significantly faster
The window function style is creating a row count within the frames. Every first row can be filtered later. This is done by adding a row_number window function. The benefit of this solution comes out when you are trying to filter the first two or three data sets. You simply have to change the filter from WHERE row_count = 1 to WHERE row_count = 2
But if you want only one single row per group you just need to ensure that the expected row per group is ordered to be the first row in the group. Then the DISTINCT ON function can delete all following rows. DISTINCT ON (customer) gives the first (ordered) row per customer group.
Try to join table on itself
select o1.customer, max(order_date),
from orders o1
join orders o2 on o1.id=o2.id
group by o1.customer
having max(o1.order_date) < NOW() - '30 days'::interval
Subqueries in select is a bad idea, because DB will execute a query for each row
If you use postgres you can also try to use CTE
https://www.postgresql.org/docs/9.6/static/queries-with.html
WITH t as (
select id, order_total from orders o2 where o2.customer = orders.customer
order by order_date desc limit 1
) select o1.customer, max(order_date),
from orders o1
join t t.id=o2.id
group by o1.customer
having max(order_date) < NOW() - '30 days'::interval