Question

如何解决以下问题？

Msg 512，Level 16，State 1，Line 1 Subquery返回的值超过1。当子查询遵循=，！=，＆lt;，＆lt; =，＆gt;，＆gt; =或子查询用作表达式时，不允许这样做。声明已经终止。

我尝试将频率设为：

| Frequency
| 19
| 23

但不知何故，通过触发此查询，我无法获得超过我的价值：

SELECT ts.TimeDifference/ ts.EntryAmount as Frequency
FROM
(
    SELECT COUNT(*) as EntryAmount,
    (
        SELECT DATEDIFF(d, t.minimum, t.maximum)
        FROM (SELECT max(Date) AS maximum, min(Date) AS minimum FROM Times GROUP BY column1, column2, column3, column4 HAVING COUNT(*) > 1) t
    ) AS TimeDifference
    FROM Times
) ts

谢谢。

Answer 1

您最内层的子选择返回多个结果。您将获得第1,2,3和4列的最小和最大PER组合。摆脱子选择，它将解决问题。

{{1}}

Answer 2

您可以使用聚合函数：

SELECT ts.TimeDifference/ ts.EntryAmount as Frequency
FROM
(
    SELECT COUNT(*) as EntryAmount,
    (
        SELECT Max(DATEDIFF(d, t.minimum, t.maximum))
        FROM (SELECT max(Date) AS maximum, min(Date) AS minimum FROM Times GROUP BY column1, column2, column3, column4 HAVING COUNT(*) > 1) t
    ) AS TimeDifference
    FROM Times
) ts

消息512，级别16错误

2 个答案: