SQL实例整理


下面整理了一些实例(实例主要来自网上),以提升自己写SQL的某些关键字的理解。

1、用一条SQL 语句 查询出每门课都大于80 分的学生姓名。(表结构如下图)

答案可以有如下两种:

select distinct student_name from table_test_one where student_name not in   
(select distinct student_name from table_test_one where score<=80);

或者

select student_name from table_test_one group by student_name having min(score)>80;

第二种方法是group by 、min函数 结合 having的使用,w3school教程里面也提到过(在 SQL 中增加 HAVING 子句原因是,WHERE 关键字无法与合计函数一起使用)

似乎看懂了,但是还是没有自己运行一遍深刻!!!自己能动手敲一遍就更好了! 


下面我们自己造数据,后面的例子也会用到。

建表然后倒入初始数据:

DROP TABLE IF EXISTS `table_test_one`;   
CREATE TABLE `table_test_one` (    `id` int(11) NOT NULL AUTO_INCREMENT,    `student_no` varchar(10) NOT NULL,    `student_name` varchar(10) NOT NULL,    `subject_no` varchar(10) NOT NULL,    `subject_name` varchar(10) NOT NULL,    `score` int(11) NOT NULL,    PRIMARY KEY (`id`)   ) ENGINE=InnoDB AUTO_INCREMENT=8 DEFAULT CHARSET=utf8;
INSERT INTO `table_test_one` VALUES ('1', '201601', '张三', '0001', '数学', '98');      
INSERT INTO `table_test_one` VALUES ('2', '201601', '张三', '0002', '语文', '66');      
INSERT INTO `table_test_one` VALUES ('3', '201602', '李四', '0001', '数学', '60');      
INSERT INTO `table_test_one` VALUES ('4', '201602', '李四', '0003', '英语', '78');      
INSERT INTO `table_test_one` VALUES ('5', '201603', '王五', '0001', '数学', '99');    
 INSERT INTO `table_test_one` VALUES ('6', '201603', '王五', '0002', '语文', '99');      
INSERT INTO `table_test_one` VALUES ('7', '201603', '王五', '0003', '英语', '98');

可以运行一下上面两个语句试试结果是不是你想要的。

2、删除除了id不同, 其他都相同的学生冗余信息,表如下:

答案:

delete table_test_one where id not in    
(select min(id) from table_test_one group by    
student_no, student_name, subject_no, subject_name, score);

是否有看懂?如果没能看懂的话,继续往下看:

先来造数据,题1中的数据只需要执行如下SQL就变成题2中的数据了:

update table_test_one set subject_no = '0001', subject_name = '数学' where id = 6;

然后我们先执行这个看看:

select min(id) from table_test_one group by    
student_no, student_name, subject_no, subject_name, score

这个的执行结果如下:

如果还不懂就再看看几次吧。

PS:GROUP BY 语句用于结合合计函数,根据一个或多个列对结果集进行分组。刚刚就是GROUP BY 对多列的使用场景。

3、行转列:

表数据如下:



希望查询到结果如下:

答案:

select year,   
(select amount from table_test_two t where t.month = 1 and t.year = table_test_two.year) as month1,   
(select amount from table_test_two t where t.month = 2 and t.year = table_test_two.year) as month2,   
(select amount from table_test_two t where t.month = 3 and t.year = table_test_two.year) as month3   
from table_test_two group by year;

利用group by 实现行转列,这种场景在数据统计的时候经常用到。

猿友可以造数据自己运行试试:

-- ----------------------------   -- Table structure for `table_test_two`   -- ----------------------------   

DROP TABLE IF EXISTS `table_test_two`;  
CREATE TABLE `table_test_two` (    `year` int(11) NOT NULL,    `month` int(11) NOT NULL,    `amount` decimal(10,1) NOT NULL,    PRIMARY KEY (`year`,`month`,`amount`)   ) ENGINE=InnoDB DEFAULT CHARSET=latin1;  
-- ----------------------------  
-- Records of table_test_two   -
- ----------------------------  
INSERT INTO `table_test_two` VALUES ('1991', '1', '1.1');  
INSERT INTO `table_test_two` VALUES ('1991', '2', '1.2');  
INSERT INTO `table_test_two` VALUES ('1991', '3', '1.3');  
INSERT INTO `table_test_two` VALUES ('1992', '1', '2.1');  
INSERT INTO `table_test_two` VALUES ('1992', '2', '2.2');  
INSERT INTO `table_test_two` VALUES ('1992', '3', '2.3');

4、复制表( 只复制结构, 源表名:table_test_two 新表名:table_test_three)

答案:

create table table_test_three as    
select * from table_test_two where 1=2;

PS:如果需要将数据也复制过去,则上面改成where 1=1

5、复制表数据(将表 table_test_two 的数据复制到表table_test_three 里面)

答案:

insert into table_test_three (year,month,amount)    select year,month,amount from table_test_two;

6、两张关联表,删除主表中已经在副表中没有的信息

答案:

delete from table_test_student where not exists    
(select * from table_test_class where table_test_student.class_id = table_test_class.calss_id);

我们先造点数据吧:

-- ----------------------------   
-- Table structure for `table_test_class`  
-- ----------------------------  
DROP TABLE IF EXISTS `table_test_class`;  
CREATE TABLE `table_test_class` (    `calss_id` int(11) NOT NULL AUTO_INCREMENT,    `calss_name` varchar(10) CHARACTER SET utf8 NOT NULL,    PRIMARY KEY (`calss_id`)   ) ENGINE=InnoDB AUTO_INCREMENT=2 DEFAULT CHARSET=latin1;  
-- ----------------------------  
-- Records of table_test_class  
-- ----------------------------  
INSERT INTO `table_test_class` VALUES ('1', '一班');
-- ----------------------------   
-- Table structure for `table_test_student`  
-- ----------------------------  
DROP TABLE IF EXISTS `table_test_student`;  
CREATE TABLE `table_test_student` (    `student_id` int(11) NOT NULL AUTO_INCREMENT,    `student_name` varchar(10) CHARACTER SET utf8 NOT NULL,    `class_id` int(11) NOT NULL,    PRIMARY KEY (`student_id`)   ) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=latin1;  
-- ----------------------------  
-- Records of table_test_student  
-- ----------------------------  
INSERT INTO `table_test_student` VALUES ('1', '罗国辉', '1');  
INSERT INTO `table_test_student` VALUES ('2', '小宝鸽', '2');

执行后数据如下:


显然副表student中小宝鸽这条数据的calss_id,主表没有对应的class_id.

执行对应SQL语句就会把小宝鸽这条数据删除掉了。

阅读原文

原文地址:https://www.cnblogs.com/276815076/p/5892003.html