MySQL字段内容拆分及合并
#头条创作挑战赛#
1. 创建测试表及数据 -- 创建一张tb_stu表 CREATE TABLE tb_user( id INT PRIMARY KEY AUTO_INCREMENT, NAME VARCHAR(10) COMMENT "人名", c_no VARCHAR(64) COMMENT "持剑ID,以逗号分隔" ); INSERT INTO tb_user(NAME,c_no) VALUES("蘧伯玉","1,3,5,7"); INSERT INTO tb_user(NAME,c_no) VALUES("高渐离","1,2,4,8,5"); INSERT INTO tb_user(NAME,c_no) VALUES("樗里疾","2,9"); INSERT INTO tb_user(NAME,c_no) VALUES("澹台灭明","1,2"); INSERT INTO tb_user(NAME,c_no) VALUES("钟子期","1,2,6,8,7,3,5"); INSERT INTO tb_user(NAME,c_no) VALUES("柳下惠","2,4,3,5"); INSERT INTO tb_user(NAME,c_no) VALUES("百里奚","1,9"); INSERT INTO tb_user(NAME,c_no) VALUES("阚止","1,6,7"); INSERT INTO tb_user(NAME,c_no) VALUES("霍去病","1,8,5"); INSERT INTO tb_user(NAME,c_no) VALUES("慕容白曜","1,2,3,4,5,7"); INSERT INTO tb_user(NAME,c_no) VALUES("鱼幼薇","7,8,9"); INSERT INTO tb_user(NAME,c_no) VALUES("宋玉","6,5"); -- 创建一张剑名 create table tb_sword( id int primary key AUTO_INCREMENT, c_name varchar(4)) comment "剑名"; insert into tb_sword(c_name)values("轩辕"); insert into tb_sword(c_name)values("湛卢"); insert into tb_sword(c_name)values("赤霄"); insert into tb_sword(c_name)values("太阿"); insert into tb_sword(c_name)values("七星龙渊"); insert into tb_sword(c_name)values("干将"); insert into tb_sword(c_name)values("莫邪"); insert into tb_sword(c_name)values("鱼肠"); insert into tb_sword(c_name)values("纯钧");
两张表内容如下:
tb_user
tb_sword
2. 数据拆分及合并
需求: 使用一条SQL获得tb_user表中每个人持有的剑名(剑名用"|"分隔),即得到如下结果
拆解需求:
1) 先将tb_user表中的c_no按逗号拆分
2)将拆分后c_no中的各个id与tb_sword中的id关联,获取剑名
3) 最后将每一个user对应的剑名合并成一个字段
分段SQL如下:
步骤1:
每一个user的c_no按逗号拆分为对应的c_id,此方法需借助于mysql.help_topic表 SELECT a.id,a.name,a.c_no, SUBSTRING_INDEX( SUBSTRING_INDEX( a.c_no, ",", b.help_topic_id + 1 ), ",",- 1 ) c_id FROM tb_user a JOIN mysql.help_topic b ON b.help_topic_id < ( LENGTH( a.c_no ) - LENGTH( REPLACE ( a.c_no, ",", "" ) ) + 1 ) ORDER BY a.id
结果如下:
步骤2:关联获取每个id对应的剑名 SELECT a2.id,a2.name,a2.c_no,a2.c_id,b2.c_name FROM ( SELECT a.id,a.name,a.c_no, SUBSTRING_INDEX( SUBSTRING_INDEX( a.c_no, ",", b.help_topic_id + 1 ), ",",- 1 ) c_id FROM tb_user a JOIN mysql.help_topic b ON b.help_topic_id < ( LENGTH( a.c_no ) - LENGTH( REPLACE ( a.c_no, ",", "" ) ) + 1 ) ORDER BY a.id ) a2, -- a2表即步骤1中拆分的结果 tb_sword b2 WHERE a2.c_id =b2.id -- 关联,相当于inner join(或者join)
结果如下
步骤3:
将每个人的剑名合并为1个字段显示,并用"|" 符合合并 SELECT a2.id,a2.name,a2.c_no, GROUP_CONCAT(b2.c_name SEPARATOR "|" ) sword_name -- SEPARATOR 指定分隔富,不加默认为逗号分隔 FROM (SELECT a.id,a.name,a.c_no, SUBSTRING_INDEX( SUBSTRING_INDEX( a.c_no, ",", b.help_topic_id + 1 ), ",",- 1 ) c_id FROM tb_user a JOIN mysql.help_topic b ON b.help_topic_id < ( LENGTH( a.c_no ) - LENGTH( REPLACE ( a.c_no, ",", "" ) ) + 1 ) ORDER BY a.id) a2,tb_sword b2 WHERE a2.c_id =b2.id GROUP BY a2.id
结果如下:
实现需求