SELECT `name`,boyName FROM beauty,boys;
SELECT `name`,boyName FROM beauty,boys WHERE beauty.`boyfriend_id` = boys.`id`;

SELECT last_name,department_name FROM employees,departments WHERE employees.department_id = departments.department_id;SELECT last_name,e.job_id,job_title FROM employees AS e,jobs j WHERE e.job_id = j.job_id;SELECT last_name,commission_pct,department_name FROM employees e,departments d WHERE e.`department_id` =d.`department_id` AND e.`commission_pct` IS NOT NULL;SELECT city,department_name FROM departments d,locations l WHERE d.`location_id` = l.`location_id` AND city LIKE 'T%';SELECT city,COUNT(*) 个数 FROM locations l,departments d WHERE d.`location_id` =l.`location_id` GROUP BY city;SELECT d.manager_id,department_name,MIN(salary) FROM employees e,departments d WHERE d.`department_id` = e.`department_id` AND commission_pct IS NOT NULL GROUP BY department_name,d.manager_id;SELECT job_title,COUNT(*) FROM employees e,jobs j WHERE e.`job_id` = j.`job_id` GROUP BY job_title ORDER BY COUNT(*) DESC;SELECT last_name,department_name,city FROM employees e,departments d,locations l WHERE e.`department_id` = d.`department_id` AND d.`location_id` = l.`location_id` GROUP BY department_name ORDER BY LENGTH(city) DESC;
SELECT salary,grade_level FROM employees e,job_grades jg WHERE salary BETWEEN lowest_sal AND highest_sal AND grade_level = 'D' ORDER BY grade_level;
SELECT e.employee_id,e.last_name,m.employee_id,m.last_name FROM employees e,employees m WHERE e.manager_id = m.employee_id;
SELECT 查询列表 FROM 表1 别名 [连接类型] JOIN 表2 别名 ON 连接条件 [WHERE 筛选条件] [GROUP BY 分组] [HAVING 筛选条件] [ORDER BY 排序列表];
示例如下:
SELECT last_name,department_name FROM employees e INNER JOIN departments d ON e.department_id = d.department_id;SELECT last_name,job_title FROM employees e INNER JOIN jobs j ON e.job_id = j.job_id WHERE e.last_name LIKE '%e%';SELECT city,COUNT(*) 个数 FROM departments d INNER JOIN locations l ON d.`location_id` = l.`location_id` GROUP BY city HAVING 个数 > 3;SELECT COUNT(*),department_name FROM employees e INNER JOIN departments d ON e.`department_id` = d.`department_id` GROUP BY department_name HAVING COUNT(*) > 3 ORDER BY COUNT(*) DESC;SELECT last_name,department_name,job_title FROM employees e INNER JOIN departments d ON e.`department_id` = d.`department_id` INNER JOIN jobs j ON e.`job_id` = j.`job_id` ORDER BY department_name DESC;SELECT COUNT(*),grade_level FROM employees e JOIN job_grades g ON e.`salary` BETWEEN g.`lowest_sal` AND g.`highest_sal` GROUP BY grade_level HAVING COUNT(*) > 20 ORDER BY grade_level DESC;SELECT e.`last_name`,m.`last_name` FROM employees e INNER JOIN employees m ON e.`manager_id` = m.`employee_id` WHERE e.`last_name` LIKE '%K%';SELECT b.`name` FROM beauty b LEFT OUTER JOIN boys bo ON b.`boyfriend_id` = bo.`id` WHERE bo.`id` IS NOT NULL;SELECT b.`name`,bo.* FROM boys bo RIGHT JOIN beauty b ON b.`boyfriend_id` = bo.`id` WHERE bo.`id` IS NULL;SELECT d.*,e.`employee_id` FROM departments d LEFT JOIN employees e ON d.`department_id` = e.`department_id` WHERE e.`employee_id` IS NULL;SELECT b.*,bo.* FROM beauty b CROSS JOIN boys bo;
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