标签:where col name dual customer 讲解 man _id employees
1. 基本描述
本章主要讲解各种子查询。
2. 基本样例
SELECT account_id, product_cd, cust_id, avail_balance FROM account WHERE account_id = (SELECT MAX(account_id) FROM account); SELECT account_id, product_cd, cust_id, avail_balance FROM account WHERE open_emp_id <> (SELECT e.emp_id FROM employee e INNER JOIN branch b ON e.assigned_branch_id = b.branch_id WHERE e.title = "Head Teller" AND b.city = "Woburn"); SELECT branch_id, name, city FROM branch WHERE name IN ("Headquarters", "Quincy Branch"); SELECT branch_id, name, city FROM branch WHERE name = "Headquarters" OR name = "Quincy Branch"; SELECT emp_id, fname, lname, title FROM employee WHERE emp_id IN (SELECT superior_emp_id FROM employee); SELECT emp_id, fname, lname, title FROM employee WHERE emp_id NOT IN (SELECT superior_emp_id FROM employee WHERE superior_emp_id IS NOT NULL); SELECT emp_id, fname, lname, title FROM employee WHERE emp_id <> ALL (SELECT superior_emp_id FROM employee WHERE superior_emp_id IS NOT NULL); SELECT emp_id, fname, lname, title FROM employee WHERE emp_id NOT IN (1, 2, NULL); SELECT account_id, cust_id, product_cd, avail_balance FROM account WHERE avail_balance < ALL (SELECT a.avail_balance FROM account a INNER JOIN individual i ON a.cust_id = i.cust_id WHERE i.fname = ‘Frank‘ AND i.lname = ‘Tucker‘); SELECT account_id, cust_id, product_cd, avail_balance FROM account WHERE avail_balance > ANY (SELECT a.avail_balance FROM account a INNER JOIN individual i ON a.cust_id = i.cust_id WHERE i.fname = ‘Frank‘ AND i.lname = ‘Tucker‘); SELECT account_id, product_cd, cust_id FROM account WHERE open_branch_id = (SELECT branch_id FROM branch WHERE name = ‘Woburn Branch‘) AND open_emp_id IN (SELECT emp_id FROM employee WHERE title = ‘Teller‘ OR title = ‘Head Teller‘); SELECT c.cust_id, c.cust_type_cd, c.city FROM customer c WHERE 2 = (SELECT COUNT(*) FROM account a WHERE a.cust_id = c.cust_id); SELECT c.cust_id, c.cust_type_cd, c.city FROM customer c WHERE (SELECT SUM(a.avail_balance) FROM account a WHERE a.cust_id = c.cust_id) BETWEEN 5000 AND 10000; SELECT a.account_id, a.product_cd, a.cust_id FROM account a WHERE NOT EXISTS (SELECT 1 FROM business b WHERE b.cust_id = a.cust_id); SELECT d.dept_id, d.name, e_cnt.how_many num_employees FROM department d INNER JOIN (SELECT dept_id, COUNT(*) how_many FROM employee GROUP BY dept_id) e_cnt ON d.dept_id = e_cnt.dept_id; SELECT open_emp_id, COUNT(*) how_many FROM account GROUP BY open_emp_id HAVING COUNT(*) = (SELECT MAX(emp_cnt.how_many) FROM (SELECT COUNT(*) how_many FROM account GROUP BY open_emp_id) emp_cnt); SELECT emp.emp_id, CONCAT(emp.fname, " ", emp.lname) emp_name, (SELECT CONCAT(boss.fname, ‘ ‘, boss.lname) FROM employee boss WHERE boss.emp_id = emp.superior_emp_id) boss_name FROM employee emp WHERE emp.superior_emp_id IS NOT NULL ORDER BY (SELECT boss.lname FROM employee boss WHERE boss.emp_id = emp.superior_emp_id), emp.lname;
标签:where col name dual customer 讲解 man _id employees
原文地址:https://www.cnblogs.com/LuckPsyduck/p/11933802.html