Mysql将查询结果集转换为JSON数据的实例代码

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”}]

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原文链接：https://blog.csdn.net/weixin_39157014/article/details/113989085