3

我有两张桌子：

主要：id_main，field1，过滤器

main_logs（5000 万行）：auto inc、id_main、path

我正在寻找以下结果：id_main, field1, most common path

我尝试了以下查询：

select id_main, 
  field1, 
  (select path, count(*) as cpt 
   from main_log 
   where main_log.id_main=main.id_main group by path order by cpt desc limit 1) 
from main 
where filter in (1,3,5);

Mysql 返回：操作数应包含 1 列

如果我删除路径，结果是正确的，但我错过了路径值。

select id_main, 
  field1, 
  (select path, count(*) as cpt 
   from main_log 
   where main_log.id_main=main.id_main group by path order by cpt desc limit 1) 
from main 
where filter in (1,3,5);

我不需要 count(*) 的结果，但我需要它用于“order by”

如何编写此查询以获取结果？谢谢

主要的

id_main     | field1    | filter
1       | red       | 1
2       | blue      | 3
3       | pink      | 1

主日志

autoinc     | id_main   | path
1       | 1         | home1
2       | 1         | home2
3       | 1         | home2
4       | 2         | house2
5       | 2         | house7
6       | 2         | house7
7       | 3         | casee

预期结果

id_main     | fields1   | most common path
1       | red       | home2
2       | blue      | house7
3       | pink      | casee
4

4 回答 4

3

您需要使用：

SELECT id_main, field, 
    (SELECT path 
    FROM main_logs 
    WHERE id_main=main.id_main 
    GROUP BY path 
    ORDER BY count(path) DESC 
    LIMIT 1) AS most 
FROM main 
WHERE filter IN (1,3,5);

经测试，可以正常工作。

于 2013-01-04T12:17:24.750 回答
2

尝试这个：

SELECT m.id_main, m.field1, A.path 
FROM main m 
INNER JOIN (SELECT * 
            FROM (SELECT id_main, path, COUNT(*) cnt
                  FROM main_log ml  
                  WHERE EXISTS (SELECT * FROM main m WHERE ml.id_main = m.id_main AND filter IN (1,3,5))
                  GROUP BY id_main, path 
                  ORDER BY cnt DESC
                  ) AS A 
            GROUP BY id_main
            ) AS A ON m.id_main = A.id_main;

旧代码忽略

SELECT m.id_main, m.field1, A.path 
FROM main m 
INNER JOIN (SELECT * FROM (SELECT id_main, path, count(*) cnt
            FROM main_log 
            GROUP BY id_main, path 
            ORDER BY cnt DESC) GROUP BY id_main) as A on m.id_main = A.id_main 
WHERE filter IN (1,3,5);
于 2012-12-25T16:47:19.647 回答
0

您在子查询中返回两列。你只需要退回一个。

样本数据：

ID_MAIN     FIELD1  FILTER
1   h   1
2   x   2
3   y   3


AUTOINC     ID_MAIN     PATH
11  1   abc
12  2   abd
13  1   xyz
14  1   ghf
15  2   xyz

试试看：SQLFIDDLE

询问：

select id_main, 
  field1, 
  (select count(id_main) as cpt 
   from main_logs 
   where main_logs.id_main=main.id_main 
   group by path 
   order by cpt desc limit 1) as CPTs
from main 
where filter in (1,3,5);

结果：

ID_MAIN     FIELD1  CPTs
1       h   1
3       y   (null)

编辑为每个 ID 提供每个路径的最大计数

绝对不是最优雅的查询。连接和子查询也可能导致相当荒谬的性能滞后。

遵循您的示例数据，除了 table main， id = 3，pink filter = 5。因此它符合您的filter标准。但是，即使没有该条件，以下查询似乎也适用于逻辑。

询问：

select a.id, b.path, a.mx
from
(select x.id, x.path, max(x.ct) as mx
from (
select m.id_main as id, ml.path, 
count(ml.id_main) as ct
from main m
left join 
main_logs ml
on ml.id_main = m.id_main
group by ml.path) as x
group by x.id) as a
inner join 
(select m.id_main as id, m.filter, ml.path, 
count(ml.id_main) as ct
from main m
left join 
main_logs ml
on ml.id_main = m.id_main
group by ml.path) as b
on a.id = b.id
and a.mx = b.ct
where b.filter in (1,2,3)
order by a.mx desc
;

结果：

ID  PATH    MX
1   home2   2
2   house7  2
3   casee   1
于 2012-12-25T16:47:57.400 回答
0 
SELECT 
    m.id_main,
    m.field1,
    ml.path,
    IFNULL(ml.Count,0)
FROM main as m
LEFT JOIN (
           SELECT
                 id_main, 
                 path,
                 COUNT(path) as Count
           FROM main_logs
           GROUP BY id_main
          ) as ml on ml.id_main = m.id_main
于 2012-12-25T16:52:11.307 回答