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

iam using mysql 5.7.17 (latest version)

iam使用mysql 5.7.17(最新版本)

this is my table samp1

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"]} |
+-----+----------------------------------------------------------------+

when i use below query it's executing properly

当我使用下面的查询它正在执行

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

but when i use this query with single number key. it throwing error,why?

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.

1 个解决方案

#1

This should work:

这应该工作:

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