2016-07-15 44 views
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branch_id  bill_amount_total  billdate 
6    1000     05-06-12 
6    1000     06-06-12 
7    2000     05-06-12 
7    500     06-06-12 
8    700     05-06-12 
8    800     06-06-12 
6    200     05-06-12 
7    200     05-06-12 
8    200     05-06-12 



SELECT branch_id, sum(bill_amount_total) as max_amount,bill_date 
    FROM pos_master 
    group by bill_date order by max_amount desc 

我所需要的最大纸币金额合计值之和为特定日期总和最大账单金额合计值的日期

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首先,使用日期数据类型划分日期。其次,一旦数据集得到纠正,期望的结果集是什么样子? – Strawberry

回答

0

你可以试试这个:

SELECT branch_id,bill_date,max(max_amount,bill_date) FROM (
SELECT 
    branch_id, 
    bill_date, 
    sum(bill_amount_total) as max_amount,bill_date 
FROM pos_master group by bill_date order by max_amount desc) as x 
GROUP BY bill_date having bill_date between '2012-06-05' and '2012-06-07' ; 
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这将返回一个任意的分支ID – Strawberry

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不支持不支持branch_id的..group。 –

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SELECT branch_id,bill_date,最大值(MAX_AMOUNT)FROM(SELECT branch_id, bill_date, 总和(bill_amount_total)作为MAX_AMOUNT FROM pos_master组由bill_date为了通过MAX_AMOUNT降序)为x –

-1

U可以尝试下面方法

我添加样品的日期和表

SELECT SUM(Total) FROM Orders 

WHERE OrderDate BETWEEN ‘3/1'2014' AND ‘3/31/2014' 
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SELECT 2 BETWEEN 3 AND 1; – Strawberry