mysql - 从表中选择一行或另一行，具体取决于其他行是否存在

Question

我想选择给定语言的产品翻译，但也包括翻译不存在时的后备。这是数据集:

+----+------------+---------------+-----------------+
| id | product_id | language_code |      name       |
+----+------------+---------------+-----------------+
|  1 |          1 | en            | Name in english |
|  2 |          1 | fr            | Name in french  |
|  3 |          1 | es            | Name in spanish |
|  4 |          2 | en            | Another product |
+----+------------+---------------+-----------------+

我正在处理一个查询，该查询选择 language_code='fr' 并在 'fr' 不存在时回退到 'en'。结果应该类似于:

+----+------------+---------------+-----------------+
| id | product_id | language_code |      name       |
+----+------------+---------------+-----------------+
|  2 |          1 | fr            | Name in french  |
|  4 |          2 | en            | Another product |
+----+------------+---------------+-----------------+

我尝试了多种方法:

SELECT translations.id, translations.language_code, translations.name
FROM translations
WHERE language_code = 'fr' 
OR language_code = 'en';

这将限制为给定语言，但不会删除重复项 (id=1)。

我见过多种使用 GROUP BY 的解决方案，但我一直遇到 only_full_group_by 问题，而且我不知道如何解决。

SELECT *
FROM translations
WHERE language_code = 'fr' 
OR language_code = 'en'
GROUP BY product_id;

Answer 1

您可以使用不存在和或:

select t.id, t.language_code, t.name
from translations t
where t.language_code = 'fr' OR
      (t.language_code = 'en' and
       not exists (select 1 from translations t2 where t2.product_id = t.product_id and t2.language_code = 'fr'
      ));