postgresql write a materialized view query to include base record and no of records matching - postgresql

Question

I have two tables one is users and another is orders in postgresql.
users table
userid | username | usertype
1 | John | F
2 | Bob | P
orders table
userid | orderid | ordername
1 | 001 | Mobile
1 | 002 | TV
1 | 003 | Laptop
2 | 001 | Book
2 | 002 | Kindle
Now I want to write a query for postgresql materialized view it will give me output like below
userid | username | Base Order Name |No of Orders | User Type
1 | John | Mobile | 3 | F - Free
2 | Bob | Book | 2 | P- Premium
I have tried below query but it's giving five records as output instead of two records and didn't figure out how to show usertype F - Free / P - Premium
CREATE MATERIALIZED VIEW userorders
TABLESPACE pg_default
AS
SELECT
u.userid,
username,
(select count(orderid) from orders where userid = u.userid)
as no_of_orders,
(select ordername from orders where orderid=1 and userid = u.userid)
as baseorder
FROM users u
INNER JOIN orders o ON u.userid = o.userid
WITH DATA;
It's giving result like below
userid | username | no_of_orders | baseorder
1 | John | 3 | Mobile
1 | John | 3 | Mobile
1 | John | 3 | Mobile
2 | Bob | 2 | Book
2 | Bob | 2 | Book
Assume base order id is always 001. In the final materialized view user type will return F - Free/ P - Premium by some mapping in query.

Answer 1

Use a group by and this becomes pretty trivial. The only slightly complex part is getting the base order name, but this can be accomplished using FILTER:
select users.userid,
username,
max(ordername) FILTER (WHERE orderid='001') as "Base Order Name",
count(orderid) as "No of Orders",
CASE WHEN usertype = 'F' THEN 'F - Free'
WHEN usertype = 'P' THEN 'P- Premium'
END as "User Type"
FROM users
JOIN orders on users.userid = orders.userid
GROUP BY users.userid, users.username, users.usertype;