Question

考虑我有两个表tab1和tab2：

+-----------+-----------+
| ID_t1     |  col2     |
+-----------+-----------+
| 1         |  1        |
+-----------+-----------+
| 2         |  1        |
+-----------+-----------+

tab2 ：

+-----------+-----------+
| ID_t2     |  ID_t1    |
+-----------+-----------+
| 1         |  1        |
+-----------+-----------+
| 2         |  1        |
+-----------+-----------+
| 3         |  1        |
+-----------+-----------+
| 4         |  2        |
+-----------+-----------+

我尝试在表格tab2中找到具有tab1.col1 = 1和最大行的对象，因此结果应该像对象1

+-----------+-----------+
| ID_t1     |  col2     |
+-----------+-----------+
| 1         |  1        |
+-----------+-----------+

我试试这个：

SELECT ID_t1 FROM tab1 
     WHERE tab1.col2 = 1 and 
           ID_t1 = (SELECT ID_t1 FROM tab2 WHERE -- MAX ONE REPEATED IN THIS TABLE)

我真的堆叠如何与这个查询竞争，你能帮我吗。

Answer 1

你在找这个：

SELECT ID_t1 
FROM tab1 
WHERE tab1.col2 = 1 AND
      ID_t1 = (SELECT ID_t1 
               FROM tab2 
               GROUP BY ID_t1 
               ORDER BY COUNT(*) DESC 
               LIMIT 1)

Answer 2

您可以在子查询中使用LIMIT：

select
from tabl1
where id_t1 in (
    select id_t1
    from tab2
    group by id_t1
    order by count(*) desc
    limit 1
) and col2 = 1;

Answer 3

试试这个

 select ID_t1, col2 from tab1 
 where ID_t1=1 and 
 col2=
 (
 select ID_t1 from tab2 
 group by ID_t1
 order by count(*) DESC
 limit 1
 )

Answer 4

我找到了一个问题的答案：

SELECT 
  tab1 as id, 
  COUNT(*) as nombre
FROM 
  tab1, tab2 
WHERE 
  tab2.ID_t1 = tab1.ID_t1 and
  tab1.col2 = 1
GROUP BY tab2.ID_t1
ORDER BY nombre desc LIMIT 1

PostgreSql - SELECT Max从另一个表重复

4 个答案: