Mysql 5.7 ERROR 3143(42000):无效的JSON路径表达式.错误在字符位置3附近

Question

iam使用mysql 5.7.17(最新版本)

这是我的表samp1

mysql> select *from samp1;  

+-----+----------------------------------------------------------------+
| id  | jdoc                                                           |
+-----+----------------------------------------------------------------+
| 101 | {"k1": 12, "k2": 34}                                           |
| 111 | {"k1": 12, "k2": 34, "k3": {"L1": "value1", "L2": "value2"}}   |
| 112 | {"k1": 12, "k2": 34, "k3": {"L1": "value1", "L2": "value2"}}   |
| 125 | {"7": "123", "8": 10, "9": "hey", "10": ["dar", "sne", "swo"]} |
+-----+----------------------------------------------------------------+

当我使用下面的查询它正在执行

select id,jdoc -> "$.k3" as f from samp1  ;

但是当我使用单个数字键的查询时.它抛出错误,为什么？

select id,jdoc -> "$.7" as f from samp1  ;

 ERROR 3143 (42000): Invalid JSON path expression. The error is around character position 3.

Answer 1

这应该工作:

select id,jdoc -> '$."7"' as f from samp1