如何在MySQL中加入第一个有序匹配？

Question

我有2张这样的桌子

Cities:
id__| country__
--------------
_1__| 1000
_2__| 1001
_3__| 1002

Translates:
id__ | city_id  |      name       | __lang__ | _order_
-------------------------------------------------------
1____|     1    | Dnipro          | en       | 2
2____|     1    | Dnipropetrovsk  | en       | 1
3____|     1    | Ekaterinoslav   | en       | 3

如何加入仅按城市ID命令排序的翻译的第一个city_id？

我想要下一个结果：

id__| country__ | city_id  |      name       | __lang__ | _order_
1   | 1000      |    1     |  Dnipropetrovsk |   en     |    1

我正在尝试这样的代码

SELECT * FROM `cities` 
INNER JOIN `translates` ON `cities`.`id` = `translates`.`city_id` 
ORDER BY `order`
GROUP BY `city_id`

但是GROUP BY忽略ORDER BY并且在ORDER BY和没有ORDER BY的情况下始终保持相同的行

Answer 1

有很多方法可以做到这一点，但这是一种方式！

SELECT
  cities.*,
  translates.*
FROM
  cities
INNER JOIN
  translates
ON 
  translates.city_id = cities.id
INNER JOIN
(
  SELECT
    city_id, 
    MIN(order) as order
  FROM 
    translates 
  GROUP BY
    city_id
) relevant_translates
ON
  relevant_translates.city_id = translates.city_id
AND
  relevant_translates.order = translates.order
;

relevant_translates子查询获取翻译表中每个city_id值的最小订单值。然后通过使用翻译进行内部联接，您只需将翻译值保留为最小订单值。祝好运！

Answer 2

如果你的每个order的city_id并不总是从1开始那么你可以使用相关的子查询来获得每个城市最小的order translate记录：

SELECT * 
FROM `cities` 
  INNER JOIN `translates` t1 ON `cities`.`id` = t1.`city_id` 
WHERE `order` = (SELECT min(`order`) FROM translates WHERE t1.city_id = translates.city_id);

如果它确实从1开始，那么只需要WHERE order=1并完成它。

Answer 3

这是一种方法：

SELECT c.id, c. 
FROM cities c
INNER JOIN translates t ON cities.id = t.city_id 
WHERE t.order = (SELECT MIN(order)
                     FROM translates t1 
                     WHERE t1.city_id = c.id)