SELECT xxx AS xxxx FROM WHERE GROUP BY HAVING
SELECT column_name(s) FROM table_name AS alias_name
-- JOIN: 若是表中有至少一个匹配,则返回行; 和INNRE JION 相同 SELECT column_name(s) FROM table_name1 INNER JOIN table_name2 ON table_name1.column_name=table_name2.column_name -- LEFT JOIN: 即便右表中没有匹配,也从左表返回全部的行;会返回左表(table1)中全部的行 SELECT column_name(s) FROM table_name1 LEFT JOIN table_name2 ON table_name1.column_name=table_name2.column_name -- RIGHT JOIN: 即便左表中没有匹配,也从右表返回全部的行 SELECT column_name(s) FROM table_name1 RIGHT JOIN table_name2 ON table_name1.column_name=table_name2.column_name -- FULL JOIN: 只要其中一个表中存在匹配,就返回行; 会从左右表中返回全部的行,无论左右匹不匹配 SELECT column_name(s) FROM table_name1 FULL JOIN table_name2 ON table_name1.column_name=table_name2.column_name -- CROSS JOIN: 会把左右表个列相乘,返回N*M 的组合
用菜鸟教程里的几张图形象的表示出来html
+-------------+---------+ | 列名 | 类型 | +-------------+---------+ | PersonId | int | | FirstName | varchar | | LastName | varchar | +-------------+---------+
PersonId 是上表主键mysql
表2: Addresssql
+-------------+---------+ | 列名 | 类型 | +-------------+---------+ | AddressId | int | | PersonId | int | | City | varchar | | State | varchar | +-------------+---------+
AddressId 是上表主键数据库
编写一个 SQL 查询,知足条件:不管 person 是否有地址信息,都须要基于上述两表提供 person 的如下信息:FirstName, LastName, City, State学习
-- 建立表 CREATE TABLE person( PersonID int PRIMARY KEY, FirstName varchar(20), SecondName varchar(20) ); INSERT INTO person VALUES(1,'Gali','Ma'); INSERT INTO person VALUES(2,'Yao','Xiao'); INSERT INTO person VALUES(3,'Tao','Baiwan'); CREATE TABLE address( AddressId int PRIMARY KEY, PersonId int, City varchar(20), State varchar(20) ); INSERT INTO address VALUES(1,2,'Huangpu','Shanghai'); INSERT INTO address VALUES(2,1,'Huangpu','Shanghai'); INSERT INTO address VALUES(3,3,'Changsha','Hunan'); INSERT INTO address VALUES(4,4,'Haidian','Beijing'); SELECT Person.FirstName, Person.SecondName, Address.City,Address.State FROM Person LEFT JOIN Address ON Person.PersonID = Address.PersonId
结果:code
+----+---------+ | Id | Email | +----+---------+ | 1 | a@b.com | | 2 | c@d.com | | 3 | a@b.com | +----+---------+
Id 是这个表的主键。
例如,在运行你的查询语句以后,上面的 Email表应返回如下几行:htm
+----+------------------+ | Id | Email | +----+------------------+ | 1 | a@b.com | | 2 | c@d.com | +----+------------------+
解答:blog
DELETE FROM email WHERE Id NOT IN ( SELECT minId FROM ( SELECT MIN(Id) as minId FROM email GROUP BY Email ) table_tmp ); SELECT * FROM email;
My SQL 不能直接删除 You can't specify target table 'table name' for update in FROM clause; 因此须要加一个临时表,还须要给临时表和列起别名。教程
结果:ci