Msg 512，Level 16錯誤

Question

我該如何解決以下問題？

Msg 512，Level 16，State 1，Line 1子查詢返回的值超過1。當子查詢遵循=，！=，<，< =，>，> =或當子查詢用作表達式時，這是不允許的。該語句已終止。

我試圖讓頻率，如：

| Frequency 
| 19 
| 23 

但不知何故，我不能將烘焙查詢得到比我更值：

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 

謝謝。

Answer 1

你內心的子選擇將返回不止一個結果。您將得到第1,2,3和4列的最小和最大PER組合。擺脫子選擇，它將解決問題。

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

Answer 2

您可以使用聚合函數：

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