目录

一、前言二、创建示例表三、基本语法- 获取JSON对象中某个key对应的value值- 获取JSON数组中某个元素- 获取JSON中的嵌套数据四、渐入佳境- 获取JSON多个路径的数据- 路径表达式*的使用- 返回NULL值- 返回错误五、使用场景六、参考文档总结

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”}}

【资料图】

# 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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