MySQL MySQL 8.0进阶操作：JSON

MySQL - MySQL 8.0进阶操作：JSON

2019年08月05日 16:43:11 写虫师 β 阅读数 1105

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文章目录

• 插入JSON
• 检索JSON
• JSON函数
  • 1. 优雅浏览
  • 2. 查找
  • 3. 修改
    • 3.1. JSON_SET()
    • 3.2. JSON_INSERT()
    • 3.3. JSON_REPLACE()
  • 4. 删除
  • 5. 其他函数

此学习文是基于MySQL 8.0写的
得益于大神朋友的悉心指导解决不少坑，才写出此文，向大神奉上膝盖

  要在MySQL中存储数据，就必须定义数据库和表结构(schema)，这是一个主要的限制。为了应对这一点，从MySQL 5.7开始，MySQL支恃了 JavaScript对象表示(JavaScriptObject Notation，JSON) 数据类型。之前，这类数据不是单独的数据类型，会被存储为字符串。新的JSON数据类型提供了自动验证的JSON文档以及优化的存储格式。

JSON文档以二进制格式存储，它提供以下功能：

• 对文档元素的快速读取访问。
• 当服务器再次读取JSON文档时，不需要重新解析文本获取该值。
• 通过键或数组索引直接查找子对象或嵌套值，而不需要读取文档中的所有值。

创建一个测试表

mysql> create table employees.emp_details (
    -> emp_no int primary key,
    -> details json
    -> );
Query OK, 0 rows affected (0.17 sec)

mysql> desc employees.emp_details;
+---------+---------+------+-----+---------+-------+
| Field   | Type    | Null | Key | Default | Extra |
+---------+---------+------+-----+---------+-------+
| emp_no  | int(11) | NO   | PRI | NULL    |       |
| details | json    | YES  |     | NULL    |       |
+---------+---------+------+-----+---------+-------+
2 rows in set (0.00 sec)

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插入JSON

mysql> insert into employees.emp_details (emp_no, details)
    -> values ('1',
    -> '{"location":"IN","phone":"+11800000000","email":"abc@example.com","address":{"line1":"abc","line2":"xyz street","city":"Bangalore","pin":"560103"}}'
    -> );
Query OK, 1 row affected (0.13 sec)

mysql> select emp_no, details from employees.emp_details;
+--------+-------------------------------------------------------------------------------------------------------------------------------------------------------------------+
| emp_no | details                                                                                                                                                           |
+--------+-------------------------------------------------------------------------------------------------------------------------------------------------------------------+
|      1 | {"email": "abc@example.com", "phone": "+11800000000", "address": {"pin": "560103", "city": "Bangalore", "line1": "abc", "line2": "xyz street"}, "location": "IN"} |
+--------+-------------------------------------------------------------------------------------------------------------------------------------------------------------------+
1 row in set (0.00 sec)

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检索JSON

可以使用->和->>运算符检索JSON列的字段：

mysql> select emp_no, details -> '$.address.pin' pin 
    -> from employees.emp_details;
+--------+----------+
| emp_no | pin      |
+--------+----------+
|      1 | "560103" |
+--------+----------+
1 row in set (0.00 sec)

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如果不用引号检索数据，可以使用->> 运算符（推荐此方式）

mysql> select emp_no, details ->> '$.address.pin' pin 
    -> from employees.emp_details;
+--------+--------+
| emp_no | pin    |
+--------+--------+
|      1 | 560103 |
+--------+--------+
1 row in set (0.00 sec)

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JSON函数

MySQL提供了许多处理JSON数据的函数，让我们看看最常用的几种函数。

1. 优雅浏览

想要以优雅的格式显示JSON值，请使用JSON_PRETTY()函数

mysql> select emp_no, json_pretty(details) 
    -> from employees.emp_details\G
*************************** 1. row ***************************
              emp_no: 1
json_pretty(details): {
  "email": "abc@example.com",
  "phone": "+11800000000",
  "address": {
    "pin": "560103",
    "city": "Bangalore",
    "line1": "abc",
    "line2": "xyz street"
  },
  "location": "IN"
}
1 row in set (0.00 sec)

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2. 查找

可以在WHERE子句中使用col ->> path运算符来引用JSON的某一列

mysql> select emp_no, details 
    -> from employees.emp_details 
    -> where details ->> '$.address.pin' = "560103";
+--------+-------------------------------------------------------------------------------------------------------------------------------------------------------------------+
| emp_no | details                                                                                                                                                           |
+--------+-------------------------------------------------------------------------------------------------------------------------------------------------------------------+
|      1 | {"email": "abc@example.com", "phone": "+11800000000", "address": {"pin": "560103", "city": "Bangalore", "line1": "abc", "line2": "xyz street"}, "location": "IN"} |
+--------+-------------------------------------------------------------------------------------------------------------------------------------------------------------------+
1 row in set (0.00 sec)

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也可以使用JSON_CONTAINS函数查询数据。如果找到了数据，则返回1，否则返回0

mysql> select json_contains(details ->> '$.address.pin',"560103") 
    -> from employees.emp_details;
+-----------------------------------------------------+
| json_contains(details ->> '$.address.pin',"560103") |
+-----------------------------------------------------+
|                                                   1 |
+-----------------------------------------------------+
1 row in set (0.00 sec)

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如何查询一个key？使用JSON_CONTAINS_PATH函数检查address. line1是否存在

mysql> select json_contains_path(details, 'one', "$.address.line1") 
    -> from employees.emp_details;
+-------------------------------------------------------+
| json_contains_path(details, 'one', "$.address.line1") |
+-------------------------------------------------------+
|                                                     1 |
+-------------------------------------------------------+
1 row in set (0.00 sec)

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one表示至少应该存在一个键，检查address.line1或者address.line2是否存在

mysql> select json_contains_path(details, 'one', "$.address.line1", "$.address.line2") 
    -> from employees.emp_details;
+--------------------------------------------------------------------------+
| json_contains_path(details, 'one', "$.address.line1", "$.address.line2") |
+--------------------------------------------------------------------------+
|                                                                        1 |
+--------------------------------------------------------------------------+
1 row in set (0.00 sec)

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如果要检查address.line1或者address.line5是否同时存在，可以使用all，而不是one

mysql> select json_contains_path(details, 'all', "$.address.line1", "$.address.line5") 
    -> from employees.emp_details;
+--------------------------------------------------------------------------+
| json_contains_path(details, 'all', "$.address.line1", "$.address.line5") |
+--------------------------------------------------------------------------+
|                                                                        0 |
+--------------------------------------------------------------------------+
1 row in set (0.00 sec)

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3. 修改

  可以使用三种不同的函数来修改数据：JSON_SET()、JSON_INSERT()和JSON _REPLACE()。 在MySQL 8之前的版本中，我们还需要对整个列进行完整的更新，这并不是最佳的方法。

3.1. JSON_SET()

替换现有值并添加不存在的值

mysql> update employees.emp_details 
    -> set details = json_set(details, "$.address.pin", "560100", "$.nickname","kai") 
    -> where emp_no = 1;
Query OK, 1 row affected (0.01 sec)
Rows matched: 1  Changed: 1  Warnings: 0

mysql> select emp_no, json_pretty(details) 
    -> from employees.emp_details\G
*************************** 1. row ***************************
              emp_no: 1
json_pretty(details): {
  "email": "abc@example.com",
  "phone": "+11800000000",
  "address": {
    "pin": "560100",
    "city": "Bangalore",
    "line1": "abc",
    "line2": "xyz street"
  },
  "location": "IN",
  "nickname": "kai"
}
1 row in set (0.00 sec)

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3.2. JSON_INSERT()

插入值，但不替换现有值
在这种情况下，$.address.pin不会被更新，只会添加一个新的字段$.address.line4

mysql> update employees.emp_details 
    -> set details = json_insert(details, "$.address.pin", "560132", "$.address.line4","A Wing") 
    -> where emp_no = 1;
Query OK, 1 row affected (0.01 sec)
Rows matched: 1  Changed: 1  Warnings: 0

mysql> select emp_no, json_pretty(details) 
    -> from employees.emp_details\G
*************************** 1. row ***************************
              emp_no: 1
json_pretty(details): {
  "email": "abc@example.com",
  "phone": "+11800000000",
  "address": {
    "pin": "560100",
    "city": "Bangalore",
    "line1": "abc",
    "line2": "xyz street",
    "line4": "A Wing"
  },
  "location": "IN",
  "nickname": "kai"
}
1 row in set (0.01 sec)

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3.3. JSON_REPLACE()

仅替换现有值
在这种情况下，$.address.line5不会被添加， 只有$.address.pin会被更新

mysql> update employees.emp_details 
    -> set details = json_replace(details, "$.address.pin", "560132", "$.address.line5","Landmark") 
    -> where emp_no = 1;
Query OK, 1 row affected (0.00 sec)
Rows matched: 1  Changed: 1  Warnings: 0

mysql> select emp_no, json_pretty(details) 
    -> from employees.emp_details\G
*************************** 1. row ***************************
              emp_no: 1
json_pretty(details): {
  "email": "abc@example.com",
  "phone": "+11800000000",
  "address": {
    "pin": "560132",
    "city": "Bangalore",
    "line1": "abc",
    "line2": "xyz street",
    "line4": "A Wing"
  },
  "location": "IN",
  "nickname": "kai"
}
1 row in set (0.00 sec)

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4. 删除

JSON_REMOVE能从JSON文档中删除数据

mysql> update employees.emp_details 
    -> set details = json_remove(details, "$.address.line4") 
    -> where emp_no = 1;
Query OK, 1 row affected (0.01 sec)
Rows matched: 1  Changed: 1  Warnings: 0

mysql> select emp_no, json_pretty(details) 
    -> from employees.emp_details\G
*************************** 1. row ***************************
              emp_no: 1
json_pretty(details): {
  "email": "abc@example.com",
  "phone": "+11800000000",
  "address": {
    "pin": "560132",
    "city": "Bangalore",
    "line1": "abc",
    "line2": "xyz street"
  },
  "location": "IN",
  "nickname": "kai"
}
1 row in set (0.00 sec)

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5. 其他函数

• JSON_KEYS()：获取JSON文档中的所有键

mysql> select json_keys(details),json_keys(details ->> "$.address") 
    -> from employees.emp_details 
    -> where emp_no= 1;
+-------------------------------------------------------+------------------------------------+
| json_keys(details)                                    | json_keys(details ->> "$.address") |
+-------------------------------------------------------+------------------------------------+
| ["email", "phone", "address", "location", "nickname"] | ["pin", "city", "line1", "line2"]  |
+-------------------------------------------------------+------------------------------------+
1 row in set (0.00 sec)

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• JSON_LENGTH()：给出JSON文档中的元素数

mysql> select json_length(details), json_length(details ->> "$.address") 
    -> from employees.emp_details 
    -> where emp_no= 1;
+----------------------+--------------------------------------+
| json_length(details) | json_length(details ->> "$.address") |
+----------------------+--------------------------------------+
|                    5 |                                    4 |
+----------------------+--------------------------------------+
1 row in set (0.00 sec)

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延伸阅读： https://dev.mysql.com/doc/refman/8.0/en/json-function-reference.html