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  • mysql数据库中实现内连接、左连接、右连接

    内连接:把两个表中数据对应的数据查出来 
    外连接:以某个表为基础把对应数据查出来

    首先创建数据库中的表,数据库代码如下:

    /*
    Navicat MySQL Data Transfer
    Source Server         : localhost_3306
    Source Server Version : 50150
    Source Host           : localhost:3306
    Source Database       : store
    Target Server Type    : MYSQL
    Target Server Version : 50150
    File Encoding         : 65001
    Date: 2010-12-15 16:27:53
    */
    SET FOREIGN_KEY_CHECKS=0;
    -- ----------------------------
    -- Table structure for `grade`
    -- ----------------------------
    DROP TABLE IF EXISTS `grade`;
    CREATE TABLE `grade` (
      `no` int(11) NOT NULL AUTO_INCREMENT,
      `grade` int(11) NOT NULL,
      PRIMARY KEY (`no`)
    ) ENGINE=InnoDB AUTO_INCREMENT=4 DEFAULT CHARSET=utf8;
    -- ----------------------------
    -- Records of grade
    -- ----------------------------
    INSERT INTO grade VALUES ('1', '90');
    INSERT INTO grade VALUES ('2', '80');
    INSERT INTO grade VALUES ('3', '70');
    -- ----------------------------
    -- Table structure for `student`
    -- ----------------------------
    DROP TABLE IF EXISTS `student`;
    CREATE TABLE `student` (
      `no` int(11) NOT NULL AUTO_INCREMENT,
      `name` varchar(20) NOT NULL,
      PRIMARY KEY (`no`)
    ) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8;
    -- ----------------------------
    -- Records of student
    -- ----------------------------
    INSERT INTO student VALUES ('1', 'a');
    INSERT INTO student VALUES ('2', 'b');
    INSERT INTO student VALUES ('3', 'c');
    INSERT INTO student VALUES ('4', 'd');

    student表中的字段分别是no和name,grade表中的字段是no和grade。两张表中的no都代表的是学生的学号。

     查询student表的结果:

    mysql> select * from grade;
    +----+-------+
    | no | grade |
    +----+-------+
    |  1 |    90 |
    |  2 |    80 |
    |  3 |    70 |
    +----+-------+
    3 rows in set

    查询grade表的结果:

    mysql> select * from student;
    +----+------+
    | no | name |
    +----+------+
    |  1 | a    |
    |  2 | b    |
    |  3 | c    |
    |  4 | d    |
    +----+------+
    4 rows in set

    内连接 inner join(查找条件中对应的数据,no4没有数据不列出来) 

    mysql> select * from student s inner join grade g on s.no=g.no;
     
    +----+------+----+-------+
    | no | name | no | grade |
    +----+------+----+-------+
    |  1 | a    |  1 |    90 |
    |  2 | b    |  2 |    80 |
    |  3 | c    |  3 |    70 |
    +----+------+----+-------+
    3 rows in set

    左连接(左表中所有数据,右表中对应数据) 

    mysql> select * from student as s left join grade as 
    g on s.no=g.no; 
    +----+------+------+-------+
    | no | name | no   | grade |
    +----+------+------+-------+
    |  1 | a    |    1 |    90 |
    |  2 | b    |    2 |    80 |
    |  3 | c    |    3 |    70 |
    |  4 | d    | NULL | NULL  |
    +----+------+------+-------+
    4 rows in set

    右连接(右表中所有数据,左表中对应数据) 

    mysql> select * from student as s right
     join grade as g on s.no=g.no; 
    +----+------+----+-------+
    | no | name | no | grade |
    +----+------+----+-------+
    |  1 | a    |  1 |    90 |
    |  2 | b    |  2 |    80 |
    |  3 | c    |  3 |    70 |
    +----+------+----+-------+
    3 rows in set
    作者:xwdreamer
    欢迎任何形式的转载,但请务必注明出处。
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  • 原文地址:https://www.cnblogs.com/xwdreamer/p/2297058.html
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