group_concat()函数
前言:在有group by的查询语句中,select指定的字段要么就包含在group by语句的后面,作为分组的依据,要么就包含在聚合函数中。(有关group by的知识请戳:浅析SQL中Group By的使用)。
例5:
该例查询了name相同的的人中最小的id。如果我们要查询name相同的人的所有的id呢?
当然我们可以这样查询:
例6:
但是这样同一个名字出现多次,看上去非常不直观。有没有更直观的方法,既让每个名字都只出现一次,又能够显示所有的名字相同的人的id呢?——使用group_concat()
1、功能:将group by产生的同一个分组中的值连接起来,返回一个字符串结果。
2、语法:group_concat( [distinct] 要连接的字段 [order by 排序字段 asc/desc ] [separator '分隔符'] )
说明:通过使用distinct可以排除重复值;如果希望对结果中的值进行排序,可以使用order by子句;separator是一个字符串值,缺省为一个逗号。
3、举例:
例7:使用group_concat()和group by显示相同名字的人的id号:
例8:将上面的id号从大到小排序,且用'_'作为分隔符:
例9:上面的查询中显示了以name分组的每组中所有的id。接下来我们要查询以name分组的所有组的id和score:
以id分组,把price字段的值在一行打印出来,分号分隔
+------+----------------------------------+
| id| group_concat(price separator ';') |
+------+----------------------------------+
|1 | 10;20;20 |
|2 | 20|
|3 | 200;500 |
+------+----------------------------------+
3 rows in set (0.00 sec)
=============================================
CREATE TABLE `grade1` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`stuName` varchar(22) DEFAULT NULL,
`course` varchar(22) DEFAULT NULL,
`score` int(11) DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=10 DEFAULT CHARSET=utf8;
-- ----------------------------
-- Records of grade1
-- ----------------------------
INSERT INTO `grade1` VALUES ('1', '张三', '语文', '91');
INSERT INTO `grade1` VALUES ('2', '张三', '数学', '90');
INSERT INTO `grade1` VALUES ('3', '张三', '英语', '87');
INSERT INTO `grade1` VALUES ('4', '李四', '语文', '79');
INSERT INTO `grade1` VALUES ('5', '李四', '数学', '95');
INSERT INTO `grade1` VALUES ('6', '李四', '英语', '80');
INSERT INTO `grade1` VALUES ('7', '王五', '语文', '77');
INSERT INTO `grade1` VALUES ('8', '王五', '数学', '81');
INSERT INTO `grade1` VALUES ('9', '王五', '英语', '89');
表内容如上图
先看看group_concat语法:
group_concat([DISTINCT] 要连接的字段 [Order BY ASC/DESC 排序字段] [Separator ‘分隔符’])
以stuName分组,把score字段的值打印在一行,逗号分隔(默认)
select GROUP_CONCAT(score),stuName from grade1 GROUP BY stuName;
1
2
其结果是:
那比如现在要查询出 语数外三门课的最低分,还有哪个学生考的?该怎么写??
select GROUP_CONCAT(stuName ORDER BY score ASC),
min(score) as score,
course
from
grade1
group by
course;
其结果是:
在结果中的第一列 ,有很多姓名并且以逗号隔开,其实这里的姓名就是按照score 升序排的(GROUP_CONCAT(stuName ORDER BY score ASC)),比如第一行的”王五,张三,李四”,就是按照数学的分数由低到高排序的,所以王五是数学分数最低的,那么我们只需要把这个字符串截取第一个人的名字就可以了,我们使用SUBSTRING_INDEX
/*SUBSTRING_INDEX以逗号分隔,取第一个值*/
select SUBSTRING_INDEX(GROUP_CONCAT(stuName ORDER BY score ASC),',',1),
min(score) as score,
course
from
grade1
group by
course;
其结果是:
如果有并列最低分只能取到一个学生,可以这样修改,并列最低都可以查出:
SELECT
stuName,
score,
course
FROM
grade1
WHERE
(score, course) IN (
SELECT
min(score),
course
FROM
grade1
GROUP BY
course
);
第一种如果有并列最低分只能取到一个学生,第二种没问题,不过效率可能会稍差点,数据量少就无所谓了
还有一种写法(不知道效率怎么样):
SELECT
g.`id`,g.`course`,g.`score`,g.`stuName`
FROM
(SELECT
course,
SUBSTRING_INDEX(
GROUP_CONCAT(score
ORDER BY score ASC),
',',
1
) AS score
FROM
grade1
GROUP BY course) AS t
LEFT JOIN grade1 AS g
ON (
t.course = g.`course`
AND t.score = g.`score`
)
---------------------
MySQL中group_concat函数
完整的语法如下:
group_concat([DISTINCT] 要连接的字段 [Order BY ASC/DESC 排序字段] [Separator '分隔符'])
基本查询
- select * from aa;
+------+------+
| id| name |
+------+------+
|1 | 10|
|1 | 20|
|1 | 20|
|2 | 20|
|3 | 200 |
|3 | 500 |
+------+------+
6 rows in set (0.00 sec)
以id分组,把name字段的值打印在一行,逗号分隔(默认)
- select id,group_concat(name) from aa group by id;
+------+--------------------+
| id| group_concat(name) |
+------+--------------------+
|1 | 10,20,20|
|2 | 20 |
|3 | 200,500|
+------+--------------------+
3 rows in set (0.00 sec)
以id分组,把name字段的值打印在一行,分号分隔
- select id,group_concat(name separator ';') from aa group by id;
+------+----------------------------------+
| id| group_concat(name separator ';') |
+------+----------------------------------+
|1 | 10;20;20 |
|2 | 20|
|3 | 200;500 |
+------+----------------------------------+
3 rows in set (0.00 sec)
以id分组,把去冗余的name字段的值打印在一行,
逗号分隔
- select id,group_concat(distinct name) from aa group by id;
+------+-----------------------------+
| id| group_concat(distinct name) |
+------+-----------------------------+
|1 | 10,20|
|2 | 20 |
|3 | 200,500 |
+------+-----------------------------+
3 rows in set (0.00 sec)
以id分组,把name字段的值打印在一行,逗号分隔,以name排倒序
- select id,group_concat(name order by name desc) from aa group by id;
+------+---------------------------------------+
| id| group_concat(name order by name desc) |
+------+---------------------------------------+
|1 | 20,20,10 |
|2 | 20|
|3 | 500,200|
+------+---------------------------------------+
3 rows in set (0.00 sec)
测试sql,项目中用到的。
- SELECT
- EMPLOYEES.EMPID
- ,EMPLOYEES.EMPNAME
- ,DEPARTMENTS.DEPARTMENTNAME
- ,EMPLOYEES.DEPTID
- ,EMPLOYEES.EMPPWD
- ,EMPLOYEES.INSIDEEMAIL
- ,EMPLOYEES.OUTSIDEEMAIL
- ,EMPLOYEES.DELEFLAG
- ,EMPLOYEES.EMPCLASS
- ,(CONCAT('[', <span style="color: #ff0000;">GROUP_CONCAT</span>
- (ROLE.Role_Name SEPARATOR '],['), ']')) AS ROLENAME
- ,(concat( '[', (
- SELECT
- <span style="color: #ff0000;">GROUP_CONCAT</span>
- (DEPARTMENTS.DEPARTMENTNAME separator '],[')
- FROM
- EMP_ROLE_DEPT
- LEFT JOIN DEPARTMENTS
- ON (
- DEPARTMENTS.DEPARTMENTID = EMP_ROLE_DEPT.DEPTID
- AND DEPARTMENTS.DELEFLAG = 0
- )
- GROUP BY
- EMP_ROLE_DEPT.EMPID
- HAVING
- EMP_ROLE_DEPT.EMPID = EMPLOYEES.EMPID
- ),']')) AS DEPARTMENTRIGHT
- FROM
- EMPLOYEES
- LEFT JOIN DEPARTMENTS
- ON (
- DEPARTMENTS.DEPARTMENTID = EMPLOYEES.DEPTID
- AND DEPARTMENTS.DELEFLAG = 0
- )
- LEFT JOIN ROLE_EMP
- ON (ROLE_EMP.EMP_ID = EMPLOYEES.EMPID)
- LEFT JOIN ROLE
- ON (ROLE_EMP.ROLE_ID = ROLE.ROLE_ID)
- <span style="color: #ff0000;"> GROUP BY
- EMPLOYEES.EMPID</span>
- HAVING
- EMPLOYEES.EMPID LIKE '%%'
- AND EMPLOYEES.EMPNAME LIKE '%%'
- AND EMPLOYEES.DELEFLAG = 0
- AND (
- EMPLOYEES.EMPCLASS = '1'
- OR EMPLOYEES.EMPCLASS = '2'
- )
- AND EMPLOYEES.DEPTID = '001' LIMIT 0
- ,16