SQL的最高銷售的產品

訂購樣品數據：SQL的最高銷售的產品

ORDER_DAY ORDER_ID PRODUCT_ID QUANTITY PRICE 
---------- --------- ----------- ---------- --------- 
01-JUL-11 O1  P1   5    5 
01-JUL-11 O2  P2   2    10 
01-JUL-11 O3  P3   10    25 
01-JUL-11 O4  P1   20    5 
02-JUL-11 O5  P3   5    25 
02-JUL-11 O6  P4   6    20 
02-JUL-11 O7  P1   2    5 
02-JUL-11 O8  P5   1    50 
02-JUL-11 O9  P6   2    50 
02-JUL-11 O10  P2   4    10

問：我拿最高銷售的產品（數量*價格）這兩天

所需的輸出：

DATE  PRODUCT_ID SOLD_AMOUNT 
01-JUL-11  P3  250 
02-JUL-11  P3  125

來源

2013-10-31 Rakesh Bharadwaj

拉克什嗨，你似乎是新來的。在尋求幫助之前，一定要自己去解決問題，向我們展示你的嘗試。你有沒有嘗試過使用谷歌搜索嗎？你發現了什麼？爲什麼它不能幫助你？ –

-3

select order_day as orderdate,product_id,max(quantity * price) as sold_amount 
from order 
group by order_day 
order by product_id

來源

2013-10-31 05:33:33

爲什麼downvote？任何原因？ –

那麼，不知道爲什麼downvote，但分組是錯誤的肯定。 –

因爲這只是錯誤的。 –

-1

select t.ORDER_DAY as date1 ,t.PRODUCT_ID,max(t.quantity * t.price) 
as sold_amount from table t group by t.ORDER_DAY,t.PRODUCT_ID

來源

2013-10-31 05:39:13

不，您甚至沒有使用group by。這應該如何工作呢？ –

@ThorstenKettner：更新了它 –

好的，現在在語法上很好。但是，您選擇每天和產品的最大利潤，而不是每天的最大利潤:-)所需的查詢其實並不那麼簡單，正如您從我自己的答案中看到的那樣。 –

-1

select date，product_id，max（product_id * price）as sold_amount form order group by order_day order by order_day

來源

2013-10-31 05:45:02

不可以。您按order_day分組。那麼你想要顯示哪個product_id？ –

您將從獲取每天和產品的銷售金額開始。有了這些數據，你首先選擇每天最大的利潤，然後選擇符合此的條目：

WITH PRODUCT_PER_DAY AS 
(
    SELECT ORDER_DAY, PRODUCT_ID, SUM(QUANTITY * PRICE) AS SOLD_AMOUNT 
    FROM MYTABLE 
    GROUP BY ORDER_DAY, PRODUCT_ID 
) 
SELECT ORDER_DAY, PRODUCT_ID, SOLD_AMOUNT 
FROM PRODUCT_PER_DAY 
WHERE (ORDER_DAY, SOLD_AMOUNT) IN 
(
    SELECT ORDER_DAY, MAX(SOLD_AMOUNT) 
    FROM PRODUCT_PER_DAY 
    GROUP BY ORDER_DAY 
) 
ORDER BY ORDER_DAY, PRODUCT_ID;

來源

2013-10-31 08:48:12

-2

請使用這個查詢，你會期望的output.I在MySQL都試過了。

SELECT 
order_date, 
Product_id, 
sold_amount 
FROM 
(
    SELECT 
     date_format(order_day,'%d-%b-%y') AS order_date, 
     Product_id, 
     SUM(price*quantity) AS sold_amount 
    FROM 
     ecommerce 
    GROUP BY 
     date_format(order_day,'%d-%b-%y'), 
     product_id) a 
WHERE 
(
    order_date,sold_amount) IN 
(
    SELECT 
     DATE, 
     MAX(sold_amt) AS sold_amount 
    FROM 
     (
      SELECT 
       date_format(order_day,'%d-%b-%y') AS DATE, 
       product_id, 
       SUM(price*quantity) AS sold_amt 
      FROM 
       ecommerce 
      GROUP BY 
       date_format(order_day,'%d-%b-%y'), 
       product_id)a 
    GROUP BY 
     DATE)

來源

2016-07-11 07:01:51

嘗試以下查詢：

select order_day, product_id, totalsale 
from (select order_day, 
      product_id, 
      nvl(QUANTITY, 0) * PRICE as totalsale, 
      dense_rank() over(partition by ORDER_DAY order by(nvl(QUANTITY, 0) * PRICE) desc) as maxsum 
     from orders) 
where maxsum = 1;

來源

2017-02-27 18:14:10 Arjun

select b.product_id,a.order_day,a.total_price from(select order_day, max(quantity*price)as total_price from order group by order_day)a 
join (select product_id , quantity * price as total_amount from order)b on 
a.total_price= b.total_price

來源

2017-04-25 13:32:02 mano

嘗試了這一點（修改表和列名稱按您的需要）：

Select a.order_day, b.product_id, a.sales 
    from 
      (select order_day, max(quantity*price) as sales 
      from ordr 
      group by order_day) a 
    inner join 
     (select order_day, product_id, quantity*price as sales 
      from ordr) b 
    on a.order_day = b.order_day and a.sales = b.sales;

來源

2017-07-01 20:10:40 riyaB

SQL的最高銷售的產品

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