正如问题的 cmets 中所述,这似乎是 MySQL 解析包含非整数数值的 JSON 文字的方式的问题,而 DOUBLE 不是最佳数据类型。如果您想将 JSON 作为字符串文字提供以加快批量插入过程,那么一种可能的解决方法是将数值作为 JSON 字符串插入,然后在批量插入完成后更新行:
mysql> SELECT VERSION();
+-------------------------+
| VERSION() |
+-------------------------+
| 5.7.20-0ubuntu0.16.04.1 |
+-------------------------+
1 row in set (0.01 sec)
mysql> CREATE TABLE json_test (id INT PRIMARY KEY, jv JSON);
Query OK, 0 rows affected (0.20 sec)
将数值作为字符串插入 ...
mysql> INSERT INTO json_test (id, jv) VALUES (1, '{"value": "212765.700000000010000"}');
Query OK, 1 row affected (0.11 sec)
mysql> INSERT INTO json_test (id, jv) VALUES (2, '{"whatever": "foo"}');
Query OK, 1 row affected (0.06 sec)
mysql> INSERT INTO json_test (id, jv) VALUES (3, '{"value": "212765.700000000010000", "whatever": "bar"}');
Query OK, 1 row affected (0.01 sec)
mysql> SELECT * FROM json_test;
+----+--------------------------------------------------------+
| id | jv |
+----+--------------------------------------------------------+
| 1 | {"value": "212765.700000000010000"} |
| 2 | {"whatever": "foo"} |
| 3 | {"value": "212765.700000000010000", "whatever": "bar"} |
+----+--------------------------------------------------------+
3 rows in set (0.01 sec)
...然后更新行以将值转换为 DECIMAL:
mysql> UPDATE json_test SET jv = JSON_REPLACE(jv, '$.value', CAST(JSON_EXTRACT(jv, '$.value') AS DECIMAL(21,15))) WHERE JSON_TYPE(JSON_EXTRACT(jv, '$.value')) = 'STRING';
Query OK, 2 rows affected (0.04 sec)
Rows matched: 2 Changed: 2 Warnings: 0
mysql> SELECT * FROM json_test;
+----+------------------------------------------------------+
| id | jv |
+----+------------------------------------------------------+
| 1 | {"value": 212765.700000000010000} |
| 2 | {"whatever": "foo"} |
| 3 | {"value": 212765.700000000010000, "whatever": "bar"} |
+----+------------------------------------------------------+
3 rows in set (0.01 sec)
mysql> SELECT JSON_TYPE(JSON_EXTRACT(jv, '$.value')) AS jt FROM json_test WHERE id=1;
+---------+
| jt |
+---------+
| DECIMAL |
+---------+
1 row in set (0.01 sec)