0

我有一个包含以下列和数据的表事务

id  transaction_date trans_type account_id agents_id transaction_date price miles
1   2012-02-08       Buy        1          1         2010-02-08       0.016 12000
2   2012-03-01       Sell       2          2         2012-03-10       0.256 -2000
3   2012-03-27       Buy        3          3         2012-03-27       0.256 10000
4   2012-03-28       Sell       4          4         2012-03-28       0.589 -11000
5   2012-03-29       Buy        5          5         2012-03-29       0.87  25000
6   2012-03-29       Sell       6          6         2012-02-29       0.879 -12000
7   2012-04-01       Sell       7          7         2012-04-01       0.058 -15000

  Account Table
  id    Program_id
  1     1
  2     1
  3     2

  Program table
  id      Abbreviation
  1       AA
  2       AC

  Agents table
  id      Name
  1       Bob
  2       Ben

我想获得第一个销售日期和第一个购买日期以获取交易售出前的平均天数，以获取交易在库存中的天数，所以应该是

  (Sell date)2012-03-01 - (Buy date)2012-02-08

我正在尝试这个

SELECT 
    case when t.trans_type ='Sell' then transaction_date end as SellDate
   ,case when t.trans_type ='Buy' then transaction_date end as BuyDate
   ,DATEDIFF(case when t.trans_type ='Sell' then transaction_date end
            ,case when t.trans_type ='Buy' then transaction_date end) as Date
   ,transaction_date
FROM transactions t
order by transaction_date

但总是在 Date 中获得 NULL

这是完整的查询

SELECT p.abbreviation,ag.name
  ,sum(-1.00 * t.miles * t.price - coalesce(t.fees,0) - coalesce(c.cost,0)) as profit
  ,sum(t.miles) 'Totakl Miles'
  ,avg(price / miles) 'Average'
  ,transaction_date
FROM transactions t
inner join accounts a on t.account_id = a.id
inner join programs p on a.program_id = p.id
inner join agents ag on t.agent_id = ag.id
LEFT JOIN (
           SELECT rp.sell_id, sum(rp.miles * t.price) as cost
           from report_profit rp
           inner join transactions t on rp.buy_id = t.id
           where t.miles > 50000
           group by rp.sell_id
           order by rp.sell_id
          ) c on t.id = c.sell_id
where t.transaction_date BETWEEN '2012-03-14' AND '2012-04-14'
Group by p.id , ag.id

编辑

我尝试了liquevicar 答案，但由于我添加的Group by，它给出了错误“子查询返回多个记录”

任何人都可以指导我吗？

提前致谢...

4

3 回答 3

1

尝试这样的子查询

SELECT 
    DATEDIFF(
      (
      SELECT MIN(date)
      FROM Transaction
      WHERE trans_type='Sell'
      ) AS first_sell_date
   ,
      (
      SELECT MIN(date)
      FROM Transaction
      WHERE trans_type='Buy'
      ) AS first_buy_date
   )

编辑：遵循 OP 评论并使用完整查询更新问题。

你能不能把 DATEDIFF 包裹在一个 MIN 电话上？

DATEDIFF(
    MIN(case when t.trans_type ='Sell' then transaction_date end),
    MIN(case when t.trans_type ='Buy' then transaction_date end)
) as Date
于 2012-04-13T11:55:45.803 回答
0

这个怎么样 -

SELECT *, DATEDIFF(sale.transaction_data, purchase.transaction_date)
FROM transactions purchase
INNER JOIN (
    SELECT *
    FROM transactions
    WHERE trans_type = 'Sell'
    ORDER BY transaction_date ASC
) sale
    ON purchase.transaction_date < sale.transaction_date
WHERE purchase.trans_type = 'Buy'
GROUP BY purchase.transaction_date

这将根据日期将所有买入交易链接到下一个卖出交易。

或者可能是这样的 -

SELECT DATEDIFF((SELECT MIN(transaction_date)
                 FROM transactions t
                 WHERE t.trans_type = 'Sell'
                 AND t.transaction_date > purchase.transaction_date),
            purchase.transaction_date)
FROM transactions purchase
WHERE trans_type = 'Buy'
于 2012-04-13T12:13:10.583 回答
0

首先感谢大家的每一个帮助

这是返回确切结果的查询

select p_id,ag_id,
     p_abb,ag_name
    ,sum(-1.00 * miles * price - coalesce(fees,0) - coalesce(cost,0)) as profit
    ,sum(miles) 'Total Miles',avg(price / miles) 'Average'
    ,DATEDIFF(min(buy_dt),min(sell_dt)) as 'Days'
     From
     (
         SELECT p.id 'p_id',ag.id 'ag_id',p.abbreviation 'p_abb',ag.name 'ag_name'
         ,miles
         ,price
         ,fees
         ,c.cost
         ,case when t.trans_type ='Sell' then transaction_date end 'sell_dt'
         ,case when t.trans_type ='Buy' then transaction_date end 'buy_dt'
         ,transaction_date
       FROM transactions t
       inner join accounts a on t.account_id = a.id
       inner join programs p on a.program_id = p.id
       inner join agents ag on t.agent_id = ag.id
       LEFT JOIN (
            SELECT rp.sell_id, sum(rp.miles * t.price) as cost
           from report_profit rp
           inner join transactions t on rp.buy_id = t.id
           where t.miles > 50000
           group by rp.sell_id
           order by rp.sell_id
          ) c on t.id = c.sell_id

  ) t1
  group by p_id, ag_id

再次感谢大家。。

于 2012-04-13T14:27:03.893 回答