SQL beginner practice problems - mysql

Question

Given two tables, orders (order_id, date, $, customer_id) and customers (ID, name)
Here's my method but I'm not sure if it's working & I'd like to know if there's faster/better way of solving these problems:
1) find out number of customers who made at least one order on date 7/9/2018
Select count (distinct customer_id)
From
(
Select customer_id from orders a
Left join customer b
On a.customer_id = b.ID
Group by customer_id,date
Having date = 7/9/2018
) a
2) find out number of customers who did not make an order on 7/9/2018
Select count (customer_id) from customer where customer_id not in
(
Select customer_id from orders a
Left join customer b
On a.customer_id = b.ID
Group by customer_id,date
Having date = 7/9/2018
)
3) find the date with most sales between 7/1 and 7/30
select date, max($)
from (
Select sum($),date from orders a
Left join customer b
On a.customer_id = b.ID
Group by date
Having date between 7/1 and 7/30
)
Thanks,

Answer 1

For problem 1, a valid solution might look like this:
SELECT COUNT(DISTINCT customer_id) x
FROM orders
WHERE date = '2018-09-07'; -- or is that '2018-07-09' ??
For problem 2, a valid solution might look like this:
SELECT COUNT(*) x
FROM customer c
LEFT
JOIN orders o
ON o.customer_id = x.customer_id
AND o.date = '2018-07-09'
WHERE o.crder_id IS NULL;
Assuming there are no ties, a valid solution to problem 3 might look like this:
SELECT date
, COUNT(*) sales
FROM orders
WHERE date BETWEEN '2018-07-01' AND '2018-07-30'
GROUP
BY date
ORDER
BY sales DESC
LIMIT 1;

Answer 2

The default format for a date in MySQL is YYYY-MM-DD, although this can be customized. You have to put quotes around it, otherwise it's treated as an arithmetic expression.
And none of your queries need to join with the customer table. The customer ID is already in the orders table, and you're not returning any info about the customers (like the name or address), you're just counting them.
1) You don't need the subquery or grouping.
SELECT COUNT(DISTINCT customer_id)
FROM orders
WHERE date = '2018-07-09'
2) Again, you don't need GROUP BY in the subquery. There's also a better pattern than NOT IN to get the count of non-matching rows.
SELECT COUNT(*)
FROM customer AS c
LEFT JOIN order AS o on c.id = o.customer_id AND o.date = '2018-07-09'
WHERE o.id IS NULL
See Return row only if value doesn't exist for various patterns to do this.
3) You can't use MAX($) in the outer query because the inner query doesn't return a column with that name. But even if you fix that, it still won't work, because the date column won't necessarily come from the same row that has the maximum. See SQL select only rows with max value on a column for more explanation of this.
You don't need a subquery at all. Use a query that returns the total sales for each day, then use ORDER BY to get the highest one.
SELECT date, SUM($) AS total_sales
FROM orders
WHERE date BETWEEN '2018-07-01' AND '2017-07-30'
GROUP BY date
ORDER BY total_sales DESC
LIMIT 1
If "most sales" is supposed to mean "most number of sales", replace SUM($) with COUNT(*).