MySQL JSON_EXTRACT路徑表達式錯誤

Question

語法看起來不錯，任何幫助將不勝感激！

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'. 

的MySQL 5.7.10

Answer 1

你可以嘗試從https://dev.mysql.com/doc/refman/5.7/en/json.html

這個意見正如前面提到的，該名稱密鑰必須用引號引起來 路徑組件如果未加引號的鍵名不合法路徑表達式。讓$參考 到這個值。

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