Question

您好，有人可以告诉我此查询出了什么问题。

DECLARE @json LONGTEXT;

SET @json = '[ { "name":"John Smith",  "address":"780 Mission St, San Francisco, CA 94103"}, { "name":"Sally Brown",  "address":"75 37th Ave S, St Cloud, MN 94103"}, { "name":"John Johnson",  "address":"1262 Roosevelt Trail, Raymond, ME 04071"}     ]';

##SELECT @json;

SELECT * FROM JSON_TABLE (@json, '$[*]' COLUMNS (
                name VARCHAR(40)  PATH '$.name',
                address VARCHAR(100) PATH '$.address'));

我得到的错误是

您的SQL语法有错误；检查手册对应于您的MySQL服务器版本以使用正确的语法靠近'（@json，'$ [*]'栏（名称VARCHAR（40）PATH'$ .name'， '在第1行

仅供参考，我使用TOAD作为工具将sql实例连接到云。

Answer 1

首先，在这种情况下，您无需在mysql中声明变量。只需使用“ SET”关键字即可。最后，您需要为“选择”查询放置一个别名。像这样：

SET @json = '[ { "name":"John Smith",  "address":"780 Mission St, San Francisco, CA 94103"}, { "name":"Sally Brown",  "address":"75 37th Ave S, St Cloud, MN 94103"}, { "name":"John Johnson",  "address":"1262 Roosevelt Trail, Raymond, ME 04071"}     ]';

##SELECT @json;

SELECT * FROM JSON_TABLE (@json, '$[*]' COLUMNS (
                `name` VARCHAR(40)  PATH '$.name',
                `address` VARCHAR(100) PATH '$.address')) AS T;

Answer 2

理想情况下，您应该升级到MYSQL 8，此答案更倾向于玩得开心

您需要编写棘手的SQL才能在MySQL 8以下的版本中模拟/仿真MySQL 8.0 JSON_TABLE()

查询

SET @json = '[ { "name":"John Smith",  "address":"780 Mission St, San Francisco, CA 94103"}, { "name":"Sally Brown",  "address":"75 37th Ave S, St Cloud, MN 94103"}, { "name":"John Johnson",  "address":"1262 Roosevelt Trail, Raymond, ME 04071"}     ]';

##SELECT @json;

SELECT * FROM JSON_TABLE (@json, '$[*]' COLUMNS (
                `name` VARCHAR(40)  PATH '$.name',
                `address` VARCHAR(100) PATH '$.address')) AS T;

结果

| name         | address                                 |
| ------------ | --------------------------------------- |
| John Smith   | 780 Mission St, San Francisco, CA 94103 |
| Sally Brown  | 75 37th Ave S, St Cloud, MN 94103       |
| John Johnson | 1262 Roosevelt Trail, Raymond, ME 04071 |

请参阅demo

MySQL 5.7的模拟/仿真查询涉及使用数字生成器和多个本机MySQL的JSON函数。

查询

SELECT 
   REPLACE(JSON_EXTRACT(json_record.json, CONCAT('$[',number,'].name')), '"', '') AS name
 , REPLACE(JSON_EXTRACT(json_record.json, CONCAT('$[',number,'].address')), '"', '') AS address
FROM (
   SELECT 
     @row := @row + 1 AS number
   FROM (
      SELECT 0 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION   SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
      ) row1
      CROSS JOIN (
      SELECT 0 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION  SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
    ) row2
    CROSS JOIN (
      SELECT @row := -1 
    ) init_user_params 
  ) AS number_generator
CROSS JOIN (
SELECT 
   JSON_LENGTH(json_information.json) - 1 AS json_length
 , json_information.json
FROM (
  SELECT 
    record.json
  FROM (
    SELECT 
      '
      [{
            "name": "John Smith",  
            "address": "780 Mission St, San Francisco, CA 94103"
      }, {
           "name": "Sally Brown",
           "address": "75 37th Ave S, St Cloud, MN 94103"
      }, {
           "name": "John Johnson",
           "address": "1262 Roosevelt Trail, Raymond, ME 04071"
      }]
     ' AS json
    FROM 
     DUAL   
  ) AS record  
) AS json_information

) AS json_record
WHERE 
 number BETWEEN 0 AND json_length

结果

| name         | address                                 |
| ------------ | --------------------------------------- |
| John Smith   | 780 Mission St, San Francisco, CA 94103 |
| Sally Brown  | 75 37th Ave S, St Cloud, MN 94103       |
| John Johnson | 1262 Roosevelt Trail, Raymond, ME 04071 |

请参阅demo

如何从mysql中的JSON选择值

2 个答案: