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
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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
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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)
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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
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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
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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)
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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 |
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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