标签:first tab The sele without products bee distinct 解决
-- Find products that are more
-- expensive than Lettuce (id = 3)
USE sql_store;
SELECT *
FROM products
WHERE unit_price > (
SELECT unit_price
FROM products
WHERE product_id = 3
);
-- Exercise
-- in sql_hr database:
-- Find employees whose earn more than average
USE sql_hr;
SELECT *
FROM employees
WHERE salary > (
SELECT AVG(salary)
FROM employees
);
-- 子查询IN
-- Find the products that have never been ordered
USE sql_store;
SELECT *
FROM products
WHERE product_id NOT IN (
SELECT DISTINCT product_id
FROM order_items
);
-- Exercise
-- Find clients without invoices
USE sql_invoicing;
SELECT *
FROM clients
WHERE client_id NOT IN (
SELECT DISTINCT client_id
FROM invoices
);
-- 子查询和连接
SELECT *
FROM clients
LEFT JOIN invoices USING(client_id)
WHERE invoice_id IS NULL;
-- Find customers who have ordered Lettuce (id = 3)
-- Select customer_id, first_name, last_name
USE sql_store;
SELECT
customer_id,
first_name,
last_name
FROM customers
WHERE customer_id IN (
SELECT customer_id
FROM orders
WHERE order_id IN (
SELECT order_id
FROM order_items
WHERE product_id = 3
));
-- 连接
SELECT
DISTINCT customer_id,
first_name,
last_name
FROM customers
JOIN orders o USING (customer_id)
JOIN order_items oi USING(order_id)
WHERE oi.product_id = 3;
-- ALL 关键字
-- Select invoices larger than all invoices of
-- client 3
SELECT MAX(invoice_total)
FROM invoices
WHERE client_id = 3;
SELECT *
FROM invoices
WHERE invoice_total > (
SELECT MAX(invoice_total)
FROM invoices
WHERE client_id = 3
);
-- ALL 高于所有
SELECT *
FROM invoices
WHERE invoice_total > ALL (
SELECT (invoice_total)
FROM invoices
WHERE client_id = 3
);
-- ANY SOME 高于任何一个即可
SELECT *
FROM invoices
WHERE invoice_total > ANY(
SELECT (invoice_total)
FROM invoices
WHERE client_id = 3
);
-- SELECT clients with at least two invoices
USE sql_invoicing;
SELECT *
FROM clients
WHERE client_id IN (
SELECT
client_id
FROM invoices
GROUP BY client_id
HAVING COUNT(*) >= 2
);
SELECT *
FROM clients
WHERE client_id = ANY(
SELECT
client_id
FROM invoices
GROUP BY client_id
HAVING COUNT(*) >= 2
);
-- 相关子查询
-- Select employees whose salary is
-- above the average in their office
USE sql_hr;
SELECT *
FROM employees e
WHERE salary > (
SELECT
AVG(salary)
FROM employees
WHERE office_id = e.office_id
);
-- Exercise
-- Get invoices that are larger than the
-- client‘s average invoice amount
USE sql_invoicing;
SELECT *
FROM invoices i
WHERE invoice_total >(
SELECT
AVG(invoice_total)
FROM invoices
WHERE client_id = i.client_id
);
-- Exist
-- Select clients that have an invoice
SELECT *
FROM clients
WHERE client_id IN(
SELECT DISTINCT client_id
FROM invoices
);
SELECT *
FROM clients c
WHERE EXISTS(
SELECT client_id
FROM invoices
WHERE client_id = c.client_id
);
-- Find the products that have never been ordered
USE sql_store;
SELECT *
FROM products
WHERE product_id NOT IN(
SELECT product_id
FROM order_items);
USE sql_store;
SELECT *
FROM products p
WHERE NOT EXISTS (
SELECT product_id
FROM order_items
WHERE product_id = p.product_id
);
-- SELECT 子查询
USE sql_invoicing;
SELECT
invoice_id,
invoice_total,
(SELECT AVG(invoice_total)
FROM invoices) AS invoice_average,
invoice_total - (SELECT invoice_average) AS difference
FROM invoices;
SELECT
client_id,
name,
(SELECT SUM(invoice_total)
FROM invoices
WHERE client_id = c.client_id) AS total_sales,
(SELECT AVG(invoice_total) FROM invoices) AS average,
(SELECT total_sales - average) AS difference
FROM clients c;
-- FROM 子语句查询
SELECT *
FROM (
SELECT
client_id,
name,
(SELECT SUM(invoice_total)
FROM invoices
WHERE client_id = c.client_id) AS total_sales,
(SELECT AVG(invoice_total) FROM invoices) AS average,
(SELECT total_sales - average) AS difference
FROM clients c
) AS sales_summary;
标签:first tab The sele without products bee distinct 解决
原文地址:https://www.cnblogs.com/jly1/p/12977466.html