So I have two tables, one is companies orders (order id, customer name, sum) and other is debts (order id) which includes orders that haven't been paid. I need to get customers with biggest total debt. Some customers have made more than 1 order. What is the best solution to do this? Thank you very much!
Orders table:
CREATE TABLE orders( order_id INT PRIMARY KEY, name VARCHAR(30), sum INT );
INSERT INTO orders VALUES
(1, 'Jack Smith', 123),
(2, 'Mary Jane', 61),
(3, 'John McCane', 90),
(4, 'Jack Smith', 512),
(5, 'Mary Jane', 33);
Debts table:
CREATE TABLE debts( order_id INT PRIMARY KEY );
INSERT INTO debts VALUES
(1),(4),(5);
Right now I have something like this:
SELECT name,SUM(sum) FROM orders INNER JOIN debts ON orders.order_id = debts.order_id GROUP BY name;
+------------+----------+
| name | SUM(sum) |
+------------+----------+
| Jack Smith | 635 |
| Mary Jane | 33 |
+------------+----------+
Desired result would look like this:
name sum
Jack Smith 635
Mary Jane 94
John McCane 90
select name,sum from orders o inner join debts d on o.order_id = d.order_id order by o.sum desc
Will give you all customers with debt ordered in descending order. Add limit 1 if you want the customer with highest debt
select name,sum from orders o inner join debts d on o.order_id = d.order_id order by o.sum desc limit 1
SELECT SUM(sum) FROM orders WHERE id IN (select order_id FROM debt)
This might help you
This is what I was looking for:
SELECT name,SUM(sum) FROM orders INNER JOIN debts ON orders.order_id = debts.order_id GROUP BY name;
Thank you everyone for answers.
Related
I want to Join to table. the condition is I want to only join those rows which have only one row to match. eg.
books:
id | name | price
1 | book1 | 19
2 | book2 | 19
3 | book3 | 30
price_offer:
id | offer | price
1 | offer1 | 19
2 | offer2 | 30
so now if I do select query on these table:
SELECT * FROM price_offer
JOIN books ON price_offer.price = books.price
I only want to join book with id 3 as it have only one match with price_offer table.
You could use a self join for books table to pick a book with only single match
select po.*, b1.*
from price_offer po
join books b1 on po.price = b1.price
join (
select price,max(id) id
from books
group by price
having count(*) = 1
) b2 on b1.id = b2.id
Demo
Try following query:
Sample data:
create table books(id int, name varchar(10), price int);
insert into books values
(1, 'book1', 19),
(2, 'book2', 19),
(3, 'book3', 30);
create table price_offer(id int, offer varchar(10), price int);
insert into price_offer values
(1, 'offer1', 19),
(2, 'offer2', 30);
Query:
select max(b.id)
from price_offer p
left join books b on b.price = p.price
where p.id is not null
group by b.price
having count(*) = 1;
If you want to avoid nesting queries where you have to use self-joins, you can use window-functions of MySQL 8.0.11, which are exactly for cases like this
I have two tables on MySql, the first contains an ID and the name of some products. I have to get the cheapest combination of brand/market for each product. So, I've inserted some itens into both tables:
UPDATE: Inserted new product (bed) with no 'Product_Brand_Market' to test LEFT JOIN.
UPDATE: Changed some product prices for better testing.
CREATE TABLE Product(
id INT UNSIGNED AUTO_INCREMENT PRIMARY KEY,
name VARCHAR(20) NOT NULL);
CREATE TABLE Product_Brand_Market(
product INT UNSIGNED,
market INT UNSIGNED, /*this will be a FOREIGN KEY*/
brand INT UNSIGNED, /*this will be a FOREIGN KEY*/
price DECIMAL(10,2) UNSIGNED NOT NULL,
PRIMARY KEY(product, market, brand),
CONSTRAINT FOREIGN KEY (product) REFERENCES Product(id));
INSERT INTO Product
(name) VALUES
('Chair'), /*will get id=1*/
('Table'), /*will get id=2*/
('Bed'); /*will get id=3*/
INSERT INTO Product_Brand_Market
(product, market, brand, price) VALUES
(1, 1, 1, 8.00), /*cheapest chair (brand=1, market=1)*/
(1, 1, 2, 8.50),
(1, 2, 1, 9.00),
(1, 2, 2, 9.50),
(2, 1, 1, 11.50),
(2, 1, 2, 11.00),
(2, 2, 1, 10.50),
(2, 2, 2, 10.00); /*cheapest table (brand=2, market=2)*/
/*no entries for bed, must return null*/
And tried the following code to get the desired values:
UPDATE: Changed INNER JOIN for LEFT JOIN.
SELECT p.id product, MIN(pbm.price) price, pbm.brand, pbm.market
FROM Product p
LEFT JOIN Product_Brand_Market pbm
ON p.id = pbm.product
GROUP BY p.id;
The returned price is OK, but I'm getting the wrong keys:
| product | price | brand | market |
|---------|-------|-------|--------|
| 1 | 8 | 1 | 1 |
| 2 | 10 | 1 | 1 |
| 3 | null | null | null |
So the only way I could think to solve it is with subqueries, but I had to use two subqueries to get both brand and market:
SELECT
p.id product,
(
SELECT pbm.brand
FROM Product_Brand_Market pbm
WHERE p.id = pbm.product
ORDER BY pbm.price
LIMIT 1
) as brand,
(
SELECT pbm.market
FROM Product_Brand_Market pbm
WHERE p.id = pbm.product
ORDER BY pbm.price
LIMIT 1
) as market
FROM Product p;
It returns the desired table:
| product | brand | market |
|---------|-------|--------|
| 1 | 1 | 1 |
| 2 | 2 | 2 |
| 3 | null | null |
But I want to know if I really should use these two similar subqueries or there is a better way to do that on MySql, any ideas?
Use a correlated subquery with LIMIT 1 in the WHERE clause:
SELECT product, brand, market
FROM Product_Brand_Market pbm
WHERE (pbm.brand, pbm.market) = (
SELECT pbm1.brand, pbm1.market
FROM Product_Brand_Market pbm1
WHERE pbm1.product = pbm.product
ORDER BY pbm1.price ASC
LIMIT 1
)
This will return only one row per product, even if there are two or many of them with the same lowest price.
Demo: http://rextester.com/UIC44628
Update:
To get all products even if they have no entries in the Product_Brand_Market table, you will need a LEFT JOIN. Note that the condition should be moved to the ON clause.
SELECT p.id as product, pbm.brand, pbm.market
FROM Product p
LEFT JOIN Product_Brand_Market pbm
ON pbm.product = p.id
AND (pbm.brand, pbm.market) = (
SELECT pbm1.brand, pbm1.market
FROM Product_Brand_Market pbm1
WHERE pbm1.product = pbm.product
ORDER BY pbm1.price ASC
LIMIT 1
);
Demo: http://rextester.com/MGXN36725
The follwing query might make a better use of your PK for the JOIN:
SELECT p.id as product, pbm.brand, pbm.market
FROM Product p
LEFT JOIN Product_Brand_Market pbm
ON (pbm.product, pbm.market, pbm.brand) = (
SELECT pbm1.product, pbm1.market, pbm1.brand
FROM Product_Brand_Market pbm1
WHERE pbm1.product = p.id
ORDER BY pbm1.price ASC
LIMIT 1
);
An index on Product_Brand_Market(product, price) should also help to improve the performance of the subquery.
I have two tables: orders and oldorders. Both are structured the same way. I want to union these two tables and then join them to another table: users. Previously I only had orders and users, I am trying to shoehorn oldorders into my current code.
SELECT u.username, COUNT(user) AS cnt
FROM orders o
LEFT JOIN users u
ON u.userident = o.user
WHERE shipped = 1
AND total != 0
GROUP BY user
This finds the number of nonzero total orders all users have made in table orders, but I want to this in the union of orders and oldorders. How can I accomplish this?
create table orders (
user int,
shipped int,
total decimal(4,2)
);
insert into orders values
(5, 1, 28.21),
(5, 1, 24.12),
(5, 1, 19.99),
(5, 1, 59.22);
create table users (
username varchar(100),
userident int
);
insert into users values
("Bob", 5);
Output for this is:
+----------+-----+
| username | cnt |
+----------+-----+
| Bob | 4 |
+----------+-----+
After creating the oldorders table:
create table oldorders (
user int,
shipped int,
total decimal(4,2)
);
insert into oldorders values
(5, 1, 62.94),
(5, 1, 53.21);
The expected output when run on the union of the two tables is:
+----------+-----+
| username | cnt |
+----------+-----+
| Bob | 6 |
+----------+-----+
Just not sure where or how to shoehorn a union into there. Instead of running the query on orders, it needs to be on orders union oldorders. It can be assumed there is no intersect between the two tables.
You just need to union this way:
SELECT u.username, COUNT(user) AS cnt
FROM
(
SELECT * FROM orders
UNION
SELECT * FROM oldorders
) o
LEFT JOIN users u ON u.userident = o.user
WHERE shipped = 1
AND total != 0
GROUP BY user;
First get the combined orders using UNION between orders and oldorders table.
The rest of the work is exactly same what you did.
SEE DEMO
Note:
Left join doesn't make sense in this case. Orders for which the users don't exist then you will get NULL 0 as output. This doesn't hold any value.
If you want <user,total orders> for all users including users who might not have ordered yet then you need to change the order of the LEFT JOIN
Consider a database of accounts and deposits:
CREATE TABLE accounts (
id int not null primary key,
name varchar(63)
);
CREATE TABLE deposits (
id int not null primary key,
account int references accounts(id),
dollars decimal(15, 2),
status enum('pending','complete')
);
insert into accounts values
(0, 'us'),
(1, 'europe'),
(2, 'asia');
insert into deposits values
(0, 0, 10, 'pending'),
(1, 0, 20, 'complete'),
(2, 1, 100, 'complete'),
(3, 1, 200, 'pending'),
(4, 1, 300, 'complete'),
(5, 2, 1000, 'pending');
I would like to get a total of all the complete deposits per bank, this is the expected result:
+--------+-----+
| us | 20 |
| europe | 400 |
| asia | 0 |
+--------+-----+
This is the SQL that I tried, but it does not work as expected:
SELECT
a.name, SUM(d.dollars)
FROM
accounts a
INNER JOIN
deposits d ON (a.id = d.account AND d.status='complete');
This is the result that it gave:
+--------+-----+
| us | 420 |
+--------+-----+
Here is an SQLfiddle of the current code.
What have I done wrong, and how can I get the expected sum?
try this
SELECT
a.name, coalesce(SUM(d.dollars),0) as sums
FROM
accounts a
left JOIN
deposits d ON (a.id = d.account AND d.status='complete')
group by a.name
order by sums desc
you should use LEFT JOIN , and you should use GROUP BY also.
LOOK DEMO
You should use grouping by a.name (or maybe even a.id) and LEFT OUTER JOIN (if you want to get non-present values).
EDIT:
SELECT
a.name, SUM(d.dollars)
FROM
accounts a
LEFT OUTER JOIN
deposits d ON (a.id = d.account AND d.status='complete')
GROUP BY a.name;
Try this query:
SELECT
a.name, IF(SUM(d.dollars) IS NULL, 0, SUM(d.dollars) )
FROM
accounts a
LEFT JOIN
deposits d ON (a.id = d.account AND d.status='complete')
GROUP BY a.name ORDER BY a.id;
By joining accounts with deposits you only make sure that you sum dollars for deposits that are linked to an account. If you would also Group by the account name, or even account Id then you will get a Sum/bank.
I have a list of product IDs and I want to find out which orders contain all those products. Orders table is structured like this:
order_id | product_id
----------------------
1 | 222
1 | 555
2 | 333
Obviously I can do it with some looping in PHP but I was wondering if there is an elegant way to do it purely in mysql.
My ideal fantasy query would be something like:
SELECT order_id
FROM orders
WHERE (222,555) IN GROUP_CONCAT(product_id)
GROUP BY order_id
Is there any hope or should I go read Tolkien? :) Also, out of curiosity, if not possible in mysql, is there any other database that has this functionality?
You were close
SELECT order_id
FROM orders
WHERE product_id in (222,555)
GROUP BY order_id
HAVING COUNT(DISTINCT product_id) = 2
Regarding your "out of curiosity" question in relational algebra this is achieved simply with division. AFAIK no RDBMS has implemented any extension that makes this as simple in SQL.
I have a preference for doing set comparisons only in the having clause:
select order_id
from orders
group by order_id
having sum(case when product_id = 222 then 1 else 0 end) > 0 and
sum(case when product_id = 555 then 1 else 0 end) > 0
What this is saying is: get me all orders where the order has at least one product 222 and at least one product 555.
I prefer this for two reasons. The first is generalizability. You can arrange more complicated conditions, such as 222 or 555 (just by changing the "and" to and "or"). Or, 333 and 555 or 222 without 555.
Second, when you create the query, you only have to put the condition in one place, in the having clause.
Assuming your database is properly normalized, i.e. there's no duplicate Product on a given Order
Mysqlism:
select order_id
from orders
group by order_id
having sum(product_id in (222,555)) = 2
Standard SQL:
select order_id
from orders
group by order_id
having sum(case when product_id in (222,555) then 1 end) = 2
If it has duplicates:
CREATE TABLE tbl
(`order_id` int, `product_id` int)
;
INSERT INTO tbl
(`order_id`, `product_id`)
VALUES
(1, 222),
(1, 555),
(2, 333),
(1, 555)
;
Do this then:
select order_id
from tbl
group by order_id
having count(distinct case when product_id in (222,555) then product_id end) = 2
Live test: http://www.sqlfiddle.com/#!2/fa1ad/5
CREATE TABLE orders
( order_id INTEGER NOT NULL
, product_id INTEGER NOT NULL
);
INSERT INTO orders(order_id,product_id) VALUES
(1, 222 ) , (1, 555 ) , (2, 333 )
, (3, 222 ) , (3, 555 ) , (3, 333 ); -- order#3 has all the products
CREATE TABLE products AS (SELECT DISTINCT product_id FROM orders);
SELECT *
FROM orders o1
--
-- There should not exist a product
-- that is not part of our order.
--
WHERE NOT EXISTS (
SELECT *
FROM products pr
WHERE 1=1
-- extra clause: only want producs from a literal list
AND pr.product_id IN (222,555,333)
-- ... that is not part of our order...
AND NOT EXISTS ( SELECT *
FROM orders o2
WHERE o2.product_id = pr.product_id
AND o2.order_id = o1.order_id
)
);
Result:
order_id | product_id
----------+------------
3 | 222
3 | 555
3 | 333
(3 rows)