mysql存json数据 mysql存json格式

1.JSON类型数据存储

 新建表

create table json_user (
 uid int auto_increment,
 data json,
 primary key(uid)
 );

插入数据

insert into json_user values (
null, '{
   "name":"lison",
"age":18,
"address":"enjoy"
   }' );

 

insert into json_user values (
 null,
 '{
  "name":"james",
  "age":28,
  "mail":"[e-mail]"
 }');

 

1.1 JSON函数

1.1.1json_extract 抽取

select json_extract('[10, 20, [30, 40]]', '$[1]');

 　　-- 10

 

 

 

 select

 json_extract(data, '$.name'),

 json_extract(data, '$.address')

 from json_user;

 　　--取出json类型字段中，name跟address

 

 

1.1.2 JSON_OBJECT 将对象转为json

select json_object("name", "enjoy", "email", "enjoy.com", "age",35);

 

insert into json_user values (

 null,

   json_object("name", "王五", "email", "[e-mail]", "age",18) );

 

 

 

 

1.1.3 json_insert 插入数据

语法：JSON_INSERT(json_doc, path, val[, path, val] ...)

 

 set @json = '{ "a": 1, "b": [2, 3]}';

　　-- @json局部   ；    -- @@json全局

 select json_insert(@json, '$.a', 10, '$.c', '[true, false]');

　　-- 当前语句：更新或者修改（数据已存在更新，没有的插入）

 

 update json_user set data = json_insert(data, "$.address_2", "xiangxue") where uid = 1;

 

 

 

1.1.4 json_merge 合并数据并返回

select json_merge('{"name": "enjoy"}', '{"id": 47}');
 
 select
 json_merge(
    json_extract(data, '$.address'),
    json_extract(data, '$.address_2')
)
 from json_user where uid = 1;

 　　-- 将当前用户的两个地址合并

 

 

 

1.1.5 其他函数：

https://dev.mysql.com/doc/refman/5.7/en/json-function-reference.html

 

2.JSON索引

JSON 类型数据本身 无法直接 创建索引，需要将需要索引的 JSON数据 重新 生成虚拟列(Virtual Columns) 之后，对 该列 进行 索引

 

create table test_inex_1(
  data json,
  gen_col varchar(10) generated always as (json_extract(data, '$.name')),
index idx (gen_col)
 );

insert into test_inex_1(data) values ('{"name":"king", "age":18, "address":"cs"}');

insert into test_inex_1(data) values ('{"name":"peter", "age":28, "address":"zz"}');

 

select * from test_inex_1;

 

 

 

疑问：这条sql查询的结果是？

select json_extract(data,"$.name") as username from test_inex_1 where gen_col="king";

 

select json_extract(data,"$.name") as username from test_inex_1 where gen_col='"king"';

 

 

 

explain select json_extract(data,"$.name") as username from test_index_1 where gen_col="king"

 

 

 

查阅官方文档，建立虚拟列，这个列查询的时候不需要加上“”符号

create table test_index_2 (
 data json,
 gen_col varchar(10) generated always as (
 json_unquote(
 json_extract(data, "$.name")
 )),
 key idx(gen_col)
 );
 
insert into test_index_2(data) values ('{"name":"king", "age":18, "address":"cs"}');
insert into test_index_2(data) values ('{"name":"peter", "age":28, "address":"zz"}');
 
select json_extract(data,"$.name") as username from test_index_2 where gen_col="king";