Question

这就是我想要做的事情

select something1,something2,account_id,
(select u.organization_id
          from public.sfdc_contact sfdcc
          join public.users u on u.email=sfdcc.email
          where sfdcc.account_id=account_id
          group by u.organization_id
          order by count(*)
      limit 1
          )
from something

Redshift提示我错误，因为我试图按顺序计数。但我不能在子查询中有2列，任何提示？

Answer 1

不完全确定此结构是否完全符合您的需求，但使用＆＃34; window functions＆＃34;例如ROW_NUMBER（）OVER（）可用于从连接的派生表（子查询）提供单行。例如：

SELECT
      s.something1
    , s.something2
    , s.account_id
    , d2.organization_id
    , d2.cn
FROM something s
LEFT JOIN (
      SELECT
            organization_id
          , account_id
          , cn
          , ROW_NUMBER() OVER (PARTITION BY organization_id ORDER BY cn) rn
      FROM (
            SELECT
                  u.organization_id
                , sfdcc.account_id
                , COUNT(*) OVER (PARTITION BY u.organization_id, sfdcc.account_id) cn
            FROM public.sfdc_contact sfdcc
            JOIN public.users u ON u.email = sfdcc.email
            ) d1
      ) d2 ON s.account_id = d2.account_id and d2.rn = 1

使用COUNT（）OVER（）可能是不必要的，这可能更实用：

SELECT
      s.something1
    , s.something2
    , s.account_id
    , d2.organization_id
    , d2.cn
FROM something s
LEFT JOIN (
      SELECT
            organization_id
          , account_id
          , cn
          , ROW_NUMBER() OVER (PARTITION BY organization_id ORDER BY cn) rn
      FROM (
            SELECT
                  u.organization_id
                , sfdcc.account_id
                , COUNT(*) cn
            FROM public.sfdc_contact sfdcc
            JOIN public.users u ON u.email = sfdcc.email
            GROUP BY
                  u.organization_id
                , sfdcc.account_id
            ) d1
      ) d2 ON s.account_id = d2.account_id and d2.rn = 1

另请注意，如果您想要最高计数，请将row_number中使用的顺序更改为DESCending：

, ROW_NUMBER() OVER (PARTITION BY organization_id ORDER BY cn DESC) rn

避免的方法＆＃34;不支持这种类型的相关子查询模式＆＃34;

1 个答案: