Question

在MSAccess中，我有以下查询。当我尝试运行查询时，出现错误Your query does not include the specified expression "ID" as part of aggregate function，但找不到原因。我的查询出了什么问题？

SELECT
   Count(t2.subjectid) AS CountOfsubjectid,
   t2.pname,
   (
      select
         max(outcometime) 
      from
         table1 t1 
      where
         t1.id = t2.id 
   )
   AS showntime 
FROM
   table2 AS t2 
WHERE
   t2.outcome = "accepted" 
GROUP BY
   t2.pname,
   t2.showntime;

更新（示例数据）：

表1：

ID  outcometime pname  outcome  subjectid
1   20181111    USB    shown    Ux1ku
1   20181113    USB    shown    Ux1ku
2   20181115    USB    shown    Tsn2f
3   20181116    USB    shown    O93nf
2   20181114    USB    shown    Tsn2f
2   20181112    USB    shown    Tsn2f

表2：

ID  outcometime pname       outcome     subjectid
1   20181118    USB         accepted    Ux1ku
2   20181119    USB         accepted    Tsn2f
3   20181117    USB         accepted    O93nf

所需结果：

pname     showntime     countofsubjectid
USB       20181113         1
USB       20181115         1
USB       20181116         1

还更新了示例数据。那是错的。谢谢。

Answer 1

如果我理解正确，则需要这样写：

SELECT Count(t2.subjectid) AS CountOfsubjectid,
       t2.name,
       (select max(outcometime) 
        from table1 as t1 inner join
             table2 as tt2
             on t1.id = tt2.id
        where tt2.name = t2.name
      ) as showntime 
FROM table2 AS t2 
WHERE t2.outcome = "accepted" 
GROUP BY t2.name;

Answer 2

当前，您正在尝试在封装在聚合查询中的SELECT子句中运行相关子查询，然后在GROUP BY子句中通过别名引用此子查询。

考虑使用派生表首先通过子查询运行单元级别，然后在外部主查询中运行聚合。

SELECT 
    dt.pname,
    COUNT(subjectid) AS CountOfsubjectid,
    dt.showntime
FROM 
  (SELECT
         t2.subjectid
         t2.pname,
         (
            select
               max(outcometime) 
            from
               table1 t1 
            where
               t1.id = t2.id 
         )
         AS showntime 
   FROM
        table2 AS t2 
   WHERE
        t2.outcome = 'accepted'
  ) AS dt
GROUP BY
   dt.pname,
   dt.showntime;

但是，请考虑避免对表中的每一行运行低效的相关子查询，以加入对MAX计算的一次的汇总查询，然后再次对{{1} }。

COUNT

您的查询不包含指定表达式“ ID”作为聚合函数的一部分

2 个答案: