获取两个查询返回的差值作为一个查询的返回答案

【问题标题】：Get the difference returned from two queries as the return of one query获取两个查询返回的差值作为一个查询的返回
【发布时间】：2014-03-27 19:50:17
【问题描述】：

使用以下查询：

Select Count(ID)
FROM Used
Where 
    ID = 54
    AND QTY = 1.875
    AND DateReceived = '2014-03-27 00:00:00'
    AND VendorID = 12400
    AND WithDrawn = 0;


Select Count(ID)
FROM Used
Where 
    ID = 54
    AND QTY = 1.875
    AND DateReceived = '2014-03-27 00:00:00'
    AND VendorID = 12400
    AND WithDrawn = 1;

如何将它们组合成一个返回计数差异的 wuery？

【问题讨论】：

标签： sql select sqlite difference

【解决方案1】：

你可以只减去这两个值：

SELECT (SELECT COUNT(ID)
        FROM Used
        WHERE ID = 54
          AND QTY = 1.875
          AND DateReceived = '2014-03-27 00:00:00'
          AND VendorID = 12400
          AND WithDrawn = 0) -
       (SELECT COUNT(ID)
        FROM Used
        WHERE ID = 54
          AND QTY = 1.875
          AND DateReceived = '2014-03-27 00:00:00'
          AND VendorID = 12400
          AND WithDrawn = 1);

或者，为每条记录构造一个 +1 或 -1 的值，然后取其总和：

SELECT SUM(CASE WithDrawn WHEN 0 THEN 1 ELSE -1 END)
FROM Used
WHERE ID = 54
  AND QTY = 1.875
  AND DateReceived = '2014-03-27 00:00:00'
  AND VendorID = 12400;

【讨论】：

【解决方案2】：

或者，您可以在 1 次扫描中完成此操作...

SELECT 
    (Count(CASE 
    WHEN WithDrawn = 0
        THEN ID END) 
        - 
    Count(CASE 
    WHEN WithDrawn = 1
        THEN ID
    END)) RESULT
FROM Used
WHERE ID = 54
    AND QTY = 1.875
    AND DateReceived = '2014-03-27 00:00:00'
    AND VendorID = 12400;

【讨论】：