一、前言
mysql5.7版本开始支持JSON类型字段，本文详细介绍json_extract函数如何获取mysql中的JSON类型数据
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
六、参考文档
- mysql5.7官方文档
- https://www.sjkjc.com/mysql-ref/json_extract
总结
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