Mysql将查询结果集转换为JSON数据
Mysql将查询结果集转换为JSON数据前言学生表学生成绩表查询单个学生各科成绩(转换为对象JSON串并用逗号拼接)将单个学生各科成绩转换为数组JSON串将数组串作为value并设置key两张表联合查询(最终SQL,每个学生各科成绩)最终结果前言我们经常会有这样一种需求,一对关联关系表,一对多的关系,使用一条sql语句查询两张表的所有记录,例:一张学生表,一张学生各科成绩表,我们想要用一条SQL查
·
Mysql将查询结果集转换为JSON数据
前言
我们经常会有这样一种需求,一对关联关系表,一对多的关系,使用一条sql语句查询两张表的所有记录,例:一张学生表,一张学生各科成绩表,我们想要用一条SQL查询出每个学生各科成绩;
学生表
CREATE TABLE IF NOT EXISTS `student`(
`id` INT UNSIGNED AUTO_INCREMENT,
`name` VARCHAR(100) NOT NULL
PRIMARY KEY ( `id` )
)ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO student( id, name ) VALUES ( 1, '张三' );
INSERT INTO student( id, name ) VALUES ( 2, '李四' );
学生成绩表
CREATE TABLE IF NOT EXISTS `score`(
`id` INT UNSIGNED AUTO_INCREMENT,
`name` VARCHAR(100) NOT NULL
`student_id` INT(100) NOT NULL,
`score` VARCHAR(100) NOT NULL
PRIMARY KEY ( `id` )
)ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO score( id, name, student_id, score) VALUES ( 1, '数学', 1, '95.5' );
INSERT INTO score( id, name, student_id, score) VALUES ( 2, '语文', 1, '99.5' );
INSERT INTO score( id, name, student_id, score) VALUES ( 3, '数学', 2, '95.5' );
INSERT INTO score( id, name, student_id, score) VALUES ( 4, '语文', 2, '88' );
查询单个学生各科成绩(转换为对象JSON串并用逗号拼接)
SELECT GROUP_CONCAT(JSON_OBJECT(
'id',id,'name',name,'student_id',student_id, 'score', score)) as scores FROM scroe where student_id = 1;
## 查询结果
## {"id": 1, "name": "数学", "student_id": 1, "score": "95.5"},{"id": 2, "name": "语文", "student_id": 1, "score": "99.5"}
将单个学生各科成绩转换为数组JSON串
SELECT CONCAT('[', GROUP_CONCAT(JSON_OBJECT(
'id',id,'name',name,'student_id',student_id, 'score', score)), ']') as scores FROM scroe where student_id = 1
## 查询结果
## [{"id": 1, "name": "数学", "student_id": 1, "score": "95.5"},{"id": 2, "name": "语文", "student_id": 1, "score": "99.5"}]
将数组串作为value并设置key
SELECT CONCAT('{"scoreData":[', GROUP_CONCAT(JSON_OBJECT(
'id',id,'name',name,'student_id',student_id, 'score', score)), ']}') as scores FROM scroe where student_id = 1
## 查询结果
## {"scoreData": [{"id": 1, "name": "数学", "student_id": 1, "score": "95.5"},{"id": 2, "name": "语文", "student_id": 1, "score": "99.5"}]}
两张表联合查询(最终SQL,每个学生各科成绩)
SELECT id, name,
(SELECT CONCAT('[', GROUP_CONCAT(JSON_OBJECT(
'id',id,'name',name,'student_id',student_id, 'score', score)), ']') as scores FROM scroe WHERE student_id = stu.id) AS scores
from student stu
## [{"id": 1, "name": "数学", "student_id": 1, "score": "95.5"},{"id": 2, "name": "语文", "student_id": 1, "score": "99.5"}]
最终结果
ID | NAME | SCORES |
---|---|---|
1 | 张三 | [{“id”: 1, “name”: “数学”, “student_id”: 1, “score”: “95.5”},{“id”: 2, “name”: “语文”, “student_id”: 1, “score”: “99.5”}] |
2 | 李四 | [{“id”: 3, “name”: “数学”, “student_id”: 1, “score”: “95.5”},{“id”:4, “name”: “语文”, “student_id”: 1, “score”: “88”}] |
更多推荐
已为社区贡献3条内容
所有评论(0)