MySQL - 条件 MIN MAX 返回不同的记录

Question

我有一个来自 the geonames website for Great Britain 的数据库转储。它由大约 60000 条记录组成。示例数据如下:

id       |     name    |   admin1   |   admin2   |  admin3  |  feature_class  |  feature_code
-------------------------------------------------------------------------------------------
2652355  |   Cornwall  |   ENG      |     C6     |          |      A          |    ADM2
11609029 |   Cornwall  |   ENG      |            |          |      L          |    RGN
6269131  |   England   |   ENG      |            |          |      A          |    ADM1

第一条记录的特征代码为ADM2，表示管理级别为2第二条记录的要素代码为RGN，表示它是一个区域。

我想按地名搜索记录以构建自动完成功能。如果记录具有相同的名称，并且其中一个记录是一个区域，即具有 feature_code RGN，那么我只想返回该记录记录，否则我想返回与具有最低 id 的名称匹配的记录。

我已尝试以下方法，但不起作用:

   SELECT IF(t0.feature_code = 'RGN', MAX(t0.id), MIN(t0.id)) as id
       , CONCAT_WS(', ', t0.name,
                  IF(t3.name != t0.name, t3.name, NULL),
                  IF(t2.name != t0.name, t2.name, NULL),
                  IF(t1.name != t0.name, t1.name, NULL)) AS name
     FROM locations t0
  LEFT JOIN locations t1 ON t1.admin1 = t0.admin1 AND t1.feature_code = 'ADM1'
  LEFT JOIN locations t2 ON t2.admin2 = t0.admin2 AND t2.feature_code = 'ADM2'
  LEFT JOIN locations t3 ON t3.admin3 = t0.admin3 AND t3.feature_code = 'ADM3'
  WHERE 
      (t0.feature_class IN ('P', 'A') OR (t0.feature_class = 'L' AND t0.feature_code = 'RGN' ) )
      AND t0.name like 'Cornwall%' 
  GROUP BY CONCAT_WS(', ', t0.name,
                     IF(t3.name != t0.name, t3.name, NULL),
                     IF(t2.name != t0.name, t2.name, NULL),
                     IF(t1.name != t0.name, t1.name, NULL))
  ORDER BY t0.name 

它返回不正确的记录:

id      | name
---------------------------
2652355 | Cornwall, England

Answer 1

我认为条件聚合应该可以解决问题。您可以按名称过滤记录，然后应用聚合函数中的逻辑。如果存在 feature_code = 'RGN' 的记录，那么您需要选择它，否则您将选择匹配记录中的最小 id。

SELECT IFNULL(MAX(CASE WHEN feature_code = 'RGN' THEN id END), MIN(id)) id_found
FROM mytable
WHERE name = @name;

<强> Demo on DB Fiddle 搜索'Cornwall'时:

| id_found |
| -------- |
| 11609029 |

注意:如果您想要整个匹配记录，一种解决方案是简单地将上述结果集与原始表JOIN:

SELECT t.*
FROM mytable t
INNER JOIN (
    SELECT IFNULL(MAX(CASE WHEN feature_code = 'RGN' THEN id END), MIN(id)) id_found
    FROM mytable
    WHERE name = @name
) x ON x.id_found = t.id;

<强> Demo :

| id       | name     | admin1 | admin2 | admin3 | feature_class | feature_code |
| -------- | -------- | ------ | ------ | ------ | ------------- | ------------ |
| 11609029 | Cornwall | ENG    |        |        | L             | RGN          |