I am trying to find max count of product. The result must only display the brands which have max number of products in it. Can anyone suggest a better way to display result.
Here are the table details:
create table Brand (
Id integer PRIMARY KEY,
Name VARCHAR(40) NOT NULL
)
create table Product (
ID integer PRIMARY KEY,
Name VARCHAR(40) NOT NULL,
Price integer NOT NULL,
BrandId integer NOT NULL,
CONSTRAINT BrandId FOREIGN KEY (BrandId)
REFERENCES Brand(Id) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE NO ACTION
)
I have used this SQL query below but I am seeing an error below. The result must display the list of the brands that have max number of products as compared to other brands.
select count(p.id) as count, b.name AS brand_name from product p join brand b on b.Id = p.BrandId
where count(p.id) = (select max(count) from product)
group by b.name
ERROR: aggregate functions are not allowed in WHERE
LINE 2: where count(p.id) = (select max(count) from product)
^
SQL state: 42803
Character: 105
In Postgres 13+, you can use FETCH WITH TIES:
select count(*) as count, b.name AS brand_name
from product p join
brand b
on b.Id = p.BrandId
group by b.name
order by count(*) desc
fetch first 1 row with ties;
In older versions you can use window functions.
Related
I have two tables
create table AUTHORS
(
id integer primary key,
name varchar(30)
)
create table BOOKS
(
id integer primary key,
name varchar(30),
year integer,
sold_copies integer,
author_id integer references AUTHORS (id)
)
How would I create a table with authors in descending order per sold copies of books per year in pgAdmin4 (PostgreSQL) ?
im a beginner and I am not sure how to go about this
I don't see why you would need to create a new table. A simple SELECT will do (maybe stored inside a view)
select a.id, a.name, b.year, sum(b.sold_copies) as total_copies
from author a
join books b on b.author_id = a.id
group by a.id, a.name, b.year
order by sum(b.sold_copies) desc;
I have 2 tables
CREATE TABLE BOOKS
(
numbk INT PRIMARY KEY IDENTITY,
nombk NVARCHAR(60),
_numrub INT FOREIGN KEY REFERENCES CLASSIFICATION(numrub)
)
CREATE TABLE TARIFER
(
_numbk INT FOREIGN KEY REFERENCES BOOKS(numbk),
_nomed NVARCHAR(60) FOREIGN KEY REFERENCES EDITEURS(nomed),
_date DATE,
price DECIMAL(20,2),
PRIMARY KEY (_numouv, _nomed)
)
The question is: how do I list all titles of books (nombk) that have the max price?
PS: TRAFIER has the price columns, and a foreign key from BOOKS which is _numbk
I tried this:
select
o.nombk, max(prix)
from
TARIFER tr, books o
where
o.numbk = tr._numbk
group by
o.nombk
This lists all, but when I execute this:
select max(prix)
from TARIFER tr, books o
where o.numbk = tr._numbk
It returns only the max price. I don't know why. Could someone please explain?
In SQL Server, you can use TOP (1) WITH TIES:
select top (1) with ties b.nombk, t.prix
from books b join
TARIFER t
on b.numbk = t._numbk
order by t.prix desc;
Why not use just a subquery the get the max(prix) and then use that one to list all records with that prix:
select o.nombk ,prix
from TARIFER tr , books o
where o.numbk = tr._numbk
and tr.prix in (select max(prix) from TARIFER tr)
Both queries to aggregation, but not at the same level:
The first query has group by o.nombk, so it generate one record per book, and gives you the maximum price of this book accross all tarifers.
The second query has no group by clause, hence it gives you the maximum price of all books over all tarifers.
If you want the book with the higher price, there is no need to aggregate: you can join and sort the results by price:
select top (1) with ties b.*, t.*
from books b
inner join join tarifer t on b.numbk = t._numbk
order by t.prix desc;
top (1) with ties gives you the first record; if there are several records with the same, top price, the query returns them all.
I have some database tables containing some documents that people need to sign. The tables are defined (somewhat simplified) as follows.
create table agreement (
id integer NOT NULL,
name character varying(50) NOT NULL,
org_id integer NOT NULL,
CONSTRAINT agreement_pkey PRIMARY KEY (id)
CONSTRAINT org FOREIGN KEY (org_id) REFERENCES org (id) MATCH SIMPLE
)
create table version (
id integer NOT NULL,
content text NOT NULL,
publish_date timestamp NOT NULL,
agreement_id integer NOT NULL,
CONSTRAINT version_pkey PRIMARY KEY (id)
CONSTRAINT agr FOREIGN KEY (agreement_id) REFERENCES agreement (id) MATCH SIMPLE
)
I skipped the org table, to reduce clutter. I have been trying to write a query that would give me all the right agreement information for a given org. So far, I can do
SELECT a.id, a.name FROM agreement AS a
JOIN version as v ON (a.id = v.agreement_id)
JOIN org as o ON (o.id = a.org_id)
WHERE o.name = $1
GROUP BY a.id
This seems to give me a single record for each agreement that belongs to the org I want and has at least one version. But I need to also include content and date published of the latest version available. How do I do that?
Also, I have a separate table called signatures that links to a user and a version. If possible, I would like to extend this query to only include agreements where a given user didn't yet sign the latest version.
Edit: reflected the need for the org join, since I select orgs by name rather than by id
You can use a correlated subquery:
SELECT a.id, a.name, v.*
FROM agreement a JOIN
version v
ON a.id = v.agreement_id
WHERE a.org_id = $1 AND
v.publish_date = (SELECT MAX(v2.publish_date) FROM version v2 WHERE v2.agreement_id = v.agreement_id);
Notes:
The org table is not needed because agreement has an org_id.
No aggregation is needed for this query. You are filtering for the most recent record.
The correlated subquery is one method that retrieves the most recent version.
Postgresql has Window Functions.
Window functions allow you to operate a sort over a specific column or set of columns. the rank function returns the row's place in the results for the sort. If you filter to just where the rank is 1 then you will always get just one row and it will be the highest sorted for the partition.
select u.id, u.name, u.content, u.publish_date from (
SELECT a.id, a.name, v.content, v.publish_date, rank() over (partition by a.id order by v.id desc) as pos
FROM agreement AS a
JOIN version as v ON (a.id = v.agreement_id)
JOIN org as o ON (o.id = a.org_id)
WHERE o.id = $1
) as u
where pos = 1
SELECT a.id, a.name, max(v.publish_date) publish_date FROM agreement AS a
JOIN version as v ON (a.id = v.agreement_id)
JOIN org as o ON (o.id = a.org_id)
WHERE o.id = $1
GROUP BY a.id, a.name
I have this database schema:
CREATE TABLE users (
id SERIAL PRIMARY KEY,
name char(50) NOT NULL UNIQUE
);
CREATE TABLE products (
id SERIAL PRIMARY KEY,
name char(50) NOT NULL,
);
CREATE TABLE orders (
id SERIAL PRIMARY KEY,
uid INTEGER REFERENCES users (id) NOT NULL,
pid INTEGER REFERENCES products (id) NOT NULL,
quantity INTEGER NOT NULL,
price FLOAT NOT NULL CHECK (price >= 0)
);
I am trying to write a query that will give me all combinations of users and products, as well as the total amount spent by the user on that product. Specifically, if I have 5 products and 5 users, there should be 25 rows in the table. Right now I have a query that almost gets the job done, however, if the user has never purchased that product then there is no row printed at all.
Here's what I've written so far:
SELECT u.name as username, p.name as productname, SUM(o.quantity * o.price) as totalPrice
FROM users u, orders o, products p
WHERE u.id = o.uid
AND p.id = o.pid
GROUP BY u.name, p.name
ORDER BY u.name, p.name
I figure that this requires some sort of join, but my SQL knowledge is limited and I am not sure what would be the best way to go about doing this. I think if somebody can help me figure this out then I will have a much better understanding.
You can do this using cross join and left join:
select u.name as username, p.name as productname,
sum(o.quantity * o.price) as totalPrice
from users u cross join
products p left join
orders o
on o.uid = u.id and o.pid = p.id
group by u.name, p.name;
The cross join generates all the rows. The left join brings in the matching rows. A simple rule when using SQL is: Never use commas in the FROM clause. Always use explicit JOIN syntax.
I have got Two Tables
Product (Id, Name, CCode)
Category (CCode, CatName) - No Primary Key
Insert Into ProductNew (DW_Prod_Id, ProdId, ProdName, CC, CName)
Select Dw_Prod_Id.Nextval, Id, Name, CCode,
(Select CatName
From Category cc, Product p
Where cc.CCode IN p.CatCode
Group By CatName )
From Product;
SQL Error: ORA-01427: single-row subquery returns more than one row
01427. 00000 - "single-row subquery returns more than one row"
I am getting the above error Because my SubQuery returns more than one row.
I would like to Match the CatCode of each row from Product table to the Category Table so that I can obtain the CatName and then Insert rows into my New Table :)
if product can have only one category :
INSERT INTO ProdcutNew (DW_Prod_Id, ProdId, ProdName, CC, CName)
(SELECT Dw_Prod_Id.Nextval, p.Id, p.Name, cc.CCode, cc.CName
FROM Product p
INNER JOIN Category cc on p.CatCode = cc.CCode)
And you can correct your table name
ProdcutNew
to ProductNew ;)
EDIT :
But if, as #Gordon Linoff pointed, you have duplicates CCode, this won't work.
If you don't want a primary key on Category table, add at least a unique constraint (you'll have to clean your datas first)
ALTER TABLE Category ADD CONSTRAINT Unique_code UNIQUE(CCode);
EDIT 2 :
But the proper way would be :
Add an Id in Category as PK, and use it as Category_ID FK in Product (if CCode can change)
With the unique constraint on CCode.
You appear to have dulicates in your category table; otherwise a simple join would suffice:
select p.*, c.ccode
from Category c join
Product p
on c.ccode = p.catcode
To choose one category arbitrarily, do something like:
select p.*, c.ccode
from (select c.*
from (select c.*, row_number() over (partition by c.ccode order by c.ccode) as seqnum
from Category c
) c
where seqnum = 1
) c join
Product p
on c.ccode = p.catcode