我有来自the geonames website for Great Britain的数据库转储。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
具有特征代码ADM2的第一条记录表示它是管理级别2具有特征代码RGN的secord记录表示它是一个区域。
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,并且如果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,否则我想返回与该ID最低的名称匹配的记录。
我尝试了以下但它不起作用:
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
我认为条件聚合应该可以解决问题。您可以通过name
过滤记录,然后在聚合函数中应用逻辑。如果feature_code = 'RGN'
存在记录,那么你想选择它,否则你会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;
Demo on DB Fiddle在搜索'Cornwall'
时:
| id_found |
| -------- |
| 11609029 |
注意:如果你想要整个匹配记录,一个解决方案是简单地用原始表格JOIN
上面的结果集:
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 |
在MySQL中,您可以使用相关子查询:
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,您还可以使用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;
一种方法可以存在并且联合所有
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)