一、前言
mysql5.7版本开始支持JSON类型字段json_extract可以完全简写为 ->json_unquote(json_extract())可以完全简写为 ->>下面介绍中大部分会利用简写
二、创建示例表
CREATE TABLE `test_json` ( `id` int(11) NOT NULL AUTO_INCREMENT, `content` json DEFAULT NULL, PRIMARY KEY (`id`) ) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=utf8mb4;
# 插入两条测试用的记录
INSERT INTO `test_json` (`content`) VALUES ('{\"name\":\"tom\",\"age\":18,\"score\":[100,90,87],\"address\":{\"province\":\"湖南\",\"city\":\"长沙\"}}');
INSERT INTO `test_json` (`content`) VALUES ('[1, "apple", "red", {"age": 18, "name": "tom"}]');| id | content |
|---|---|
| 1 | {“age”: 18, “name”: “tom”, “score”: [100, 90, 87], “address”: {“city”: “长沙”, “province”: “湖南”}} |
| 2 | [1, “apple”, “red”, {“age”: 18, “name”: “tom”}] |
三、基本语法
- 获取JSON对象中某个key对应的value值
json_extract函数中，第一个参数content表示json数据，第二个参数为json路径，其中$表示该json数据本身，$.name就表示获取json中key为name的value值
可以利用 -> 表达式来代替json_extract
若获取的val本身为字符串，那么获取的val会被引号包起来，比如"tom"，这种数据被解析到程序对象中时，可能会被转义为\“tom\”。为了解决这个问题了，可以在外面再包上一层json_unquote函数，或者使用 ->> 代替->
content:
{“age”: 18, “name”: “tom”, “score”: [100, 90, 87], “address”: {“city”: “长沙”, “province”: “湖南”}}
# 得到"tom" select json_extract(content,'$.name') from test_json where id = 1; # 简写方式：字段名->表达式等价于json_extract(字段名,表达式) select content->'$.name' from test_json where id = 1; # 结果: +--------------------------------+ | json_extract(content,'$.name') | +--------------------------------+ | "tom" | +--------------------------------+ +-------------------+ | content->'$.name' | +-------------------+ | "tom" | +-------------------+ # 解除双引号，得到tom select json_unquote(json_extract(content,'$.name')) from test_json where id = 1; # 简写方式：字段名->>表达式等价于json_unquote(json_extract(字段名,表达式)) select content->>'$.name' from test_json where id = 1; # 结果: +----------------------------------------------+ | json_unquote(json_extract(content,'$.name')) | +----------------------------------------------+ | tom | +----------------------------------------------+ +--------------------+ | content->>'$.name' | +--------------------+ | tom | +--------------------+
- 获取JSON数组中某个元素
json_extract函数中，第一个参数content表示json数据，第二个参数为json路径，其中$表示该json数据本身，$[i]表示获取该json数组索引为i的元素（索引从0开始）
与获取key-val一样，若获取的元素为字符串，默认的方式也会得到双引号包起来的字符，导致程序转义，方法也是利用json_unquote函数，或者使用 ->> 代替->
content:
[1, “apple”, “red”, {“age”: 18, “name”: “tom”}]
# 得到"apple" select json_extract(content,'$[1]') from test_json where id = 2; # 简写，效果同上 select content->'$[1]' from test_json where id = 2; # 结果: +------------------------------+ | json_extract(content,'$[1]') | +------------------------------+ | "apple" | +------------------------------+ +-----------------+ | content->'$[1]' | +-----------------+ | "apple" | +-----------------+ # 解除双引号，得到apple select json_unquote(json_extract(content,'$[1]')) from test_json where id = 2; # 简写，效果同上 select content->>'$[1]' from test_json where id = 2; # 结果: +--------------------------------------------+ | json_unquote(json_extract(content,'$[1]')) | +--------------------------------------------+ | apple | +--------------------------------------------+ +------------------+ | content->>'$[1]' | +------------------+ | apple | +------------------+
- 获取JSON中的嵌套数据
结合前面介绍的两种获取方式，可以获取json数据中的嵌套数据
content: id=1
{“age”: 18, “name”: “tom”, “score”: [100, 90, 87], “address”: {“city”: “长沙”, “province”: “湖南”}}
content: id=2
[1, “apple”, “red”, {“age”: 18, “name”: “tom”}]
# 得到:87 select content->'$.score[2]' from test_json where id = 1; # 结果: +-----------------------+ | content->'$.score[2]' | +-----------------------+ | 87 | +-----------------------+ # 得到:18 select content->'$[3].age' from test_json where id = 2; # 结果: +---------------------+ | content->'$[3].age' | +---------------------+ | 18 | +---------------------+
四、渐入佳境
- 获取JSON多个路径的数据
将会把多个路径的数据组合成数组返回
content: id=1
{“age”: 18, “name”: “tom”, “score”: [100, 90, 87], “address”: {“city”: “长沙”, “province”: “湖南”}}
select json_extract(content,'$.age','$.score') from test_json where id = 1; # 结果: +-----------------------------------------+ | json_extract(content,'$.age','$.score') | +-----------------------------------------+ | [18, [100, 90, 87]] | +-----------------------------------------+ select json_extract(content,'$.name','$.address.province','$.address.city') from test_json where id = 1; # 结果: +----------------------------------------------------------------------+ | json_extract(content,'$.name','$.address.province','$.address.city') | +----------------------------------------------------------------------+ | ["tom", "湖南", "长沙"] | +----------------------------------------------------------------------+
- 路径表达式*的使用
将会把多个路径的数据组合成数组返回
# 先插入一条用于测试的数据
INSERT INTO `test_json` (`id`,`content`) VALUES(3,'{"name":"tom","address":{"name":"中央公园","city":"长沙"},"class":{"id":3,"name":"一年三班"},"friend":[{"age":20,"name":"marry"},{"age":21,"name":"Bob"}]}')content: id=3
{“name”: “tom”, “class”: {“id”: 3, “name”: “一年三班”}, “friend”: [{“age”: 20, “name”: “marry”}, {“age”: 21, “name”: “Bob”}], “address”: {“city”: “长沙”, “name”: “中央公园”}}
# 获取所有二级嵌套中key=name的值 # 由于friend的二级嵌套是一个数组,所以.name获取不到其中的所有name值 select content->'$.*.name' from test_json where id = 3; +----------------------------------+ | content->'$.*.name' | +----------------------------------+ | ["一年三班", "中央公园"] | +----------------------------------+``` # 获取所有key为name值的数据，包括任何嵌套内的name select content->'$**.name' from test_json where id = 3; +---------------------------------------------------------+ | content->'$**.name' | +---------------------------------------------------------+ | ["tom", "一年三班", "marry", "Bob", "中央公园"] | +---------------------------------------------------------+ # 获取数组中所有的name值 select content->'$.friend[*].name' from test_json where id = 3; +-----------------------------+ | content->'$.friend[*].name' | +-----------------------------+ | ["marry", "Bob"] | +-----------------------------+
- 返回NULL值
content: id=1
{“age”: 18, “name”: “tom”, “score”: [100, 90, 87], “address”: {“city”: “长沙”, “province”: “湖南”}}
寻找的JSON路径都不存在
# age路径不存在，返回NULL # 若有多个路径，只要有一个路径存在则不会返回NULL select json_extract(content,'$.price') from test_json where id = 1; +---------------------------------+ | json_extract(content,'$.price') | +---------------------------------+ | NULL | +---------------------------------+
路径中有NULL
# 存在任意路径为NULL则返回NULL select json_extract(content,'$.age',NULL) from test_json where id = 1; +------------------------------------+ | json_extract(content,'$.age',NULL) | +------------------------------------+ | NULL | +------------------------------------+
- 返回错误
若第一个参数不是JSON类型的数据，则返回错误
select json_extract('{1,2]',$[0])若路径表达式不规范，则返回错误
select content->'$age' from test_json where id = 1; # 结果: ERROR 3143 (42000): Invalid JSON path expression. The error is around character position 1.
五、使用场景
JSON_EXTRACT函数通常用于要获取JSON中某个特定的数据或者要根据它作为判断条件时使用
