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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