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我有一個包含正數和負數的表。我必須使用子查詢找出正負數之和 
您的問題不是很清楚(沒有給出表名或列名),也不清楚為什麼需要子查詢(如果可以避免的話,這不是一個好主意)。您可以使用'case'語句獲取所需的值
以下計算正負值的數量
選擇總和(當列> = 0時為1,否則0結束時為正),
總和(當列<0時為1,否則為0結束)為負
從表
而以下是正值和負值的總數
選擇總和(當列> = 0時,則列為0,否則結束)為正,
總和(當acolumn <0時,acolumn否則為0結束)為負
從表
|
對於負數的總和:
SELECT SUM(numberColumn)FROM tableFoo WHERE numberColumn <0
對於正數的總和:
從tableFoo的SELECT SUM(numberColumn)那裡numberColumn> = 0
結合兩者(QUERY1和QUERY2是前兩個查詢):
選擇(QUERY1),(QUERY2)
|
選擇總和(當a> = 0時為else 0結束的情況)為正,
總和(當a <0時,否則else 0結束)為負
從一個
|
通過使用CTE(公用表表達式),我們可以獲得輸出。
;與Psum_CTE
如
(SELECT SUM(num)AS PositiveSum
從樣品
num> = 0
)
,Nsum_CTE
如
(
SELECT SUM(num)AS NegativeSum
從樣品
num <0
)
SELECT PositiveSum,NegativeSum
從Psum_CTE,Nsum_CTE
|
選擇 (
(SELECT SUM(numberColumn)FROM tableFoo WHERE numberColumn <0)-
(SELECT SUM(numberColumn)FROM tableFoo WHERE numberColumn> 0)
)AS總計算
|
您可以使用符號分隔值:
選擇Sum((Sign(n)+ 1)/ 2 * n)作為PositiveSum,
Sum(-(Sign(n)-1)/ 2 * n)為負數
來自YourTableOData;
根據輸入值的符號,符號返回1、0或-1。可以根據符號進行一些算術運算,將其轉換為1或0:(Sign(n)+ 1)/ 2對於所有正值均為1,否則為0。請注意檢查負值((Sign(n)-1 )/ 2)返回-1或0,因此使用負號(-)可以避免對要求和的值的符號進行翻轉。
|
你的答案
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