Question

我需要计算销售数量，并按日期总计销售总额，这很容易。但曲线间距是 - 我需要＆＃34;切断＆＃34;到下午6点而不是午夜。

前一天下午6点至下午6点。

扔我的是＆＃34;分组＆＃34;。我的计数只是拉动真实的日期而不是＆＃34; 6 pm＆＃34;资讯

Sort30   Day30      Total                 Counter
-------- ---------- --------------------- -----------
20120810 08/10/2012 675.32                9
20120809 08/09/2012 1314.68               16

查询：

SELECT top 30 CONVERT(VARCHAR(8), chickendate, 112) AS varSort30, 
CONVERT(VARCHAR(10), chickendate, 101) AS varDay30,
SUM(CAST(transAmount AS money)) AS varTotal, 
Count(chickendate) AS varCounter
FROM CHICKEN
WHERE 
    (chickendate >= dateadd(hour, 18, dateadd(day, datediff(day, 0, chickendate), -1))  AND 
    chickendate < dateadd(hour, 18, dateadd(day, datediff(day, 0, chickendate), 0))) 
GROUP BY CONVERT(VARCHAR(8), chickendate, 112), 
    CONVERT(VARCHAR(10), chickendate, 101)
ORDER BY CONVERT(VARCHAR(8), chickendate, 112) DESC

四处转转，我觉得它的某些东西在我脸上盯着我。感谢。

Answer 1

如果我的逻辑是正确的，这应该会给你正确的结果：

SELECT TOP 30
    CONVERT(VARCHAR(8), modifiedChickenDate, 112) AS varSort30, 
    CONVERT(VARCHAR(10), modifiedChickenDate, 101) AS varDay30,
    SUM(CAST(transAmount AS money)) AS varTotal, 
    COUNT(modifiedChickenDate) AS varCounter
FROM (
    SELECT
        transAmount,
        DATEADD(HOUR, 6, chickendate) AS modifiedChickenDate
    FROM CHICKEN
) sub
GROUP BY
    CONVERT(VARCHAR(8), modifiedChickenDate, 112) AS varSort30, 
    CONVERT(VARCHAR(10), modifiedChickenDate, 101) AS varDay30,
ORDER BY
    CONVERT(VARCHAR(8), modifiedChickenDate, 112) AS varSort30

Answer 2

如果chickendate是DATETIME而不是DATE，那么您可以使用

GROUP BY CONVERT(VARCHAR(10), DATEADD(hh, 6, chickendate), 101)

将日期提前6小时（截止时间比午夜提前6小时，或下午6点），然后在一年中分组。这仅适用于您存储时间信息的情况，我不确定您是谁。为必要的表发布一些模式。但我认为你正在寻找......

SELECT TOP 30 CONVERT(VARCHAR(10), DATEADD(hh, 6, chickendate), 101) as Date
        , SUM(CAST(transAmount as money)) AS Total
        , Count(*) as Counter
    FROM Chicken
    GROUP BY CONVERT(VARCHAR(10), DATEADD(hh, 6, chickendate), 101)
    ORDER BY CONVERT(VARCHAR(10), DATEADD(hh, 6, chickendate), 101) DESC

Answer 3

怎么样：

GROUP BY CONVERT(VARCHAR(8), dateadd(hour, 6, chickendate), 112)

所以'2012-08-09 18:00:00'按20120810分组，'2012-08-09 17:59:59'按20120809分组。

T-SQL选择每日总计但每天截止不是午夜

3 个答案: