MySQL JSON_EXTRACT 路径表达式错误

Question

语法对我来说是正确的，任何帮助将不胜感激！

mysql> select fieldnames from tablename limit 5;
+--------------------------------------------------------+
| fieldnames                                             |
+--------------------------------------------------------+
| {"example-field-1": "val2"}                            |
| {"example-field-2": "val1"}                            |
| {"example-field-1": "val1", "example-field-3": "val1"} |
| {"example-field-2": "val1"}                            |
| {"example-field-2": "val2"}                            |
+--------------------------------------------------------+
mysql> select JSON_EXTRACT(fieldnames, '$.example-field-1') from tablename;
ERROR 3143 (42000): Invalid JSON path expression. The error is around character position 17 in '$.example-field-1'.

MySQL 5.7.10

Answer 1

你可以试试https://dev.mysql.com/doc/refman/5.7/en/json.html的这个建议

如前所述，命名键的路径组件必须用引号引起来 如果不带引号的键名在路径表达式中不合法。让 $ 参考 到这个值。

select JSON_EXTRACT(fieldnames, '$."example-field-1"') from tablename;

注意

如果您有更多字段，则必须引用每个键而不是整个路径：

select JSON_EXTRACT(fieldnames, '$."field"."example-field-1"') from tablename;