mysql-返回不同记录的条件最小最大值

adam78 • 4 年前 • 801 次点击

我有一个数据库转储 the geonames website for Great Britain . 它包含大约60000条记录。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

功能代码为ADM2的第一条记录表示它是管理级别2 具有特征码RGN的分段记录意味着它是一个区域。

我想按地名搜索记录以建立自动完成功能。如果记录具有相同的名称,如果其中一个记录是一个区域,即具有特征代码RGN,那么我只想返回否则,我想返回与具有最低id的名称匹配的记录。

我试过以下方法,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

它返回错误的记录:

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

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801 次点击

文章 [ 3 ] | 最新文章 4 年前

• 1 楼

Zaynul Abadin Tuhin 5 年前

一种方法可以存在并联合所有

select t1.* from location t1
where exists ( select 1 from location t2 where t2.name=t1.name and t2.feature_code='RGN'
             )
 and t1.feature_code='RGN'
union all

select t1.* from location t1
where not exists ( select 1 from location t2 where t2.name=t1.name and 
                t2.feature_code='RGN'
                  )
  and t1.id=(select min(id) from location t2 where t2.name=t1.name)

• 2 楼

Gordon Linoff 5 年前

在MySQL中,可以使用相关的子查询:

select l.*
from locations l
where l.id = (select l2.id
              from locations l2
              where l2.name = l.name
              order by (feature_code = 'RGN') desc,  -- put regions first
                       id asc
             );

在MySQL 8 +中,也可以使用 row_number() :

select l.*
from (select l.*,
             row_number() over (partition by name 
                                order by (feature_code = 'RGN') desc, id
                               ) as seqnum
      from locations l
     ) l
where seqnum = 1;

• 3 楼

GMB 5 年前

我认为 条件聚合 应该会成功的。可以通过以下方式过滤记录 name ,然后在聚合函数中应用逻辑。如果记录存在于 feature_code = 'RGN' 然后选择它,否则选择最小值 id 在匹配记录中。

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

Demo on DB Fiddle 搜索时 'Cornwall' :

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

注意:如果您想要整个匹配记录,一个解决方案是 JOIN 以上结果与原表一致:

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;

Demo :

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

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