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sql计算每日总数除前一天的总数

发布于 2024-08-20 16:22:42 字数 733 浏览 2 评论 0原文

我有一个包含日期、项目和数量的表。

我需要一个sql查询来返回每天的总数，但总数是数量减去前一天的总数。数量随着月份的推移而累积。所以第一个可能有 5 个，第二个有 12 个，第三个有 20 个。

所以第一个加了 5 第二次加 7 得到 12 3rd 加 8 等于 20。

我过去做过类似的事情，但找不到或不记得了。我知道我需要一个相关的子查询。

TIA

—

编辑 1

我正在使用 Microsoft Access。日期是日期时间字段，项目是文本，并且数量是数字

编辑 2

好的，这就是我所拥有的，

SELECT oos.report_date, oos.tech, oos.total_cpe, oos_2.total_cpe
   FROM oos INNER JOIN (
     SELECT oos_2.tech, Sum(oos_2.total_cpe) AS total_cpe
     FROM oos_2
     WHERE (((oos_2.report_date)<#10/10/2010#))
     GROUP BY oos_2.tech
   ) oos_2 ON oos.tech = oos_2.tech;

如何将 oos.report_date 放入我所说的 #10/10/2010# 中。我以为我可以像mysql一样把它放在那里，但没有运气。我要继续研究。

原文

I have a table that has a date, item, and quantity.

I need a sql query to return the totals per day, but the total is the quantity minus the previous day totals. The quantity accumulates as the month goes on. So the 1st could have 5 the 2nd have 12 and the 3rd has 20.

So the 1st adds 5
2nd adds 7 to make 12
3rd adds 8 to make 20.

I've done something like this in the past, but can not find it or remember. I know i'll need a correlated sub-query.

TIA

Edit 1

I'm using Microsoft Access.
Date is a datetime field,
item is a text, and
quantity is number

Edit 2

Ok this is what i have

SELECT oos.report_date, oos.tech, oos.total_cpe, oos_2.total_cpe
   FROM oos INNER JOIN (
     SELECT oos_2.tech, Sum(oos_2.total_cpe) AS total_cpe
     FROM oos_2
     WHERE (((oos_2.report_date)<#10/10/2010#))
     GROUP BY oos_2.tech
   ) oos_2 ON oos.tech = oos_2.tech;

How do i get the oos.report_date into where i says #10/10/2010#. I thought I could just stick it in there like mysql, but no luck. I'm gonna continue researching.

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预谋 2024-08-27 16:22:42

通过在日期上加 1 并使值变为负数来对它们求和，从而从今天中取出昨天的总数：

SELECT report_date, tech, Sum(total_cpe) AS total_cpe 
FROM (
    SELECT oos.report_date, oos.tech, oos.total_cpe
    FROM oos 
    UNION ALL
    SELECT oos.report_date+1, oos.tech, 0-oos.total_cpe 
    FROM oos 
)
WHERE (report_date < #10/10/2010#) 
GROUP BY report_date, tech 
ORDER BY report_date, tech

Sum them by adding one to the date and making the value negative, thus taking yesterday's total from today's:

SELECT report_date, tech, Sum(total_cpe) AS total_cpe 
FROM (
    SELECT oos.report_date, oos.tech, oos.total_cpe
    FROM oos 
    UNION ALL
    SELECT oos.report_date+1, oos.tech, 0-oos.total_cpe 
    FROM oos 
)
WHERE (report_date < #10/10/2010#) 
GROUP BY report_date, tech 
ORDER BY report_date, tech

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征棹 2024-08-27 16:22:42

好吧，我明白了。

SELECT o.report_date, o.tech, o.total_cpe, 
o.total_cpe -  (
    SELECT  IIf(Sum(oos.total_cpe) is null, 0,Sum(oos.total_cpe)) AS total_cpe 
    FROM oos 
    WHERE (((oos.tech)=o.tech) AND ((oos.report_date)<o.report_date))
) AS total
FROM oos o;

Ok, I figured it out.

SELECT o.report_date, o.tech, o.total_cpe, 
o.total_cpe -  (
    SELECT  IIf(Sum(oos.total_cpe) is null, 0,Sum(oos.total_cpe)) AS total_cpe 
    FROM oos 
    WHERE (((oos.tech)=o.tech) AND ((oos.report_date)<o.report_date))
) AS total
FROM oos o;

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