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"}]');
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.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[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,'.score') from test_json where id = 1;
结果:
+-----------------------------------------+ | json_extract(content,'.score') | +-----------------------------------------+ | [18, [100, 90, 87]] | +-----------------------------------------+
select json_extract(content,'.address.province','$.address.city') from test_json where id = 1;
结果:
+----------------------------------------------------------------------+ | json_extract(content,'.address.province','$.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' | +----------------------------------+ | ["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' | +---------------------------------------------------------+ | ["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' | +-----------------------------+ | ["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') | +---------------------------------+ | NULL | +---------------------------------+
-
路径中有NULL
存在任意路径为NULL则返回NULL
select 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
稳涨内容来源:https://blog.csdn.net/AJakey/article/details/128113607