MySQL - MySQL 8.0 进阶操作：JSON

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)

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)

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)

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_detailsG

*************************** 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)

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)

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)

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)

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)

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)

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_detailsG

*************************** 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)

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_detailsG

*************************** 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)

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_detailsG

*************************** 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)

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_detailsG

*************************** 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)

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)

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)

原文链接：https://blog.csdn.net/nangy2514/article/details/98490082