Question

我有一个这样的交易表：数量是基于不同单价的库存总量。我们称之为T

id | transaction_time | item | unit_price | quantity | subtotal
1      2012-5-15          A         1.00         15         15.00
2      2012-5-15          A         3.00         15         45.00
3      2012-5-15          B         1.00         10         10.00
4      2012-6-10          A         2.00         15         30.00
5      2012-6-15          A         2.00         10         20.00

我需要随着时间的推移得到每个商品的总价值......但是，相同的商品是基于不同的单价。例如A的结果是：

transaction_time | item |  quantity | subtotal
    2012-5-15          A     30         60.00
    2012-6-10          A     45         90.00
    2012-6-15          A     40         80.00

2012-5-15，我们有15件商品A价格1.00,15件商品A价格3.00，所以总数量是30，小计是15 * 1 + 15 * 3 = 60。

2012-6-10我们还有15个项目A，价格2，所以总量变为30 + 15 = 45，小计变为60 + 15 * 2 = 90

2012-6-15我们有10个项目A，价格2，所以价格为2的项目A从15下降到10.总数量变为40，小计下降-2 * 5，变为80。

我试过

select transaction_time,sum(quantity),sum(subtotal)
where id in(select max(id) from T group by unit_price,item)
group by item
having item=A

这只给我最后一行

2012-6-15          A     40         80.00

Answer 1

在查询（某种复杂，可能很慢，需要优化）之后，请检查DEMO

SELECT tr_sub.cur_tt, tr_sub.item, sum(tr.quantity), sum(tr.quantity*tr.unit_price) 
FROM 
  (SELECT tr1.transaction_time as cur_tt, max(tr2.transaction_time) as prev_tt, tr1.item as item, 
     IF (tr1.unit_price=tr2.unit_price, tr1.unit_price, tr2.unit_price) as t_p 
     FROM transactions tr1 LEFT JOIN transactions tr2 ON 
     tr1.transaction_time>=tr2.transaction_time AND tr1.item=tr2.item
    GROUP BY tr1.item, tr1.transaction_time,  t_p
   ) as tr_sub INNER JOIN transactions tr ON 
   tr_sub.prev_tt=tr.transaction_time 
   AND tr_sub.item=tr.item 
   AND tr_sub.t_p=tr.unit_price
GROUP BY tr_sub.item, tr_sub.cur_tt
ORDER BY tr_sub.cur_tt, tr_sub.item

Answer 2

首先需要确定特定项目的所有unit_price值：

SELECT DISTINCT unit_price
FROM t
WHERE item = 'A'

<强>输出：

unit_price
----------
1
3
2

您还需要识别所有可能的transaction_times：

SELECT DISTINCT transaction_time
FROM t
WHERE item = 'A';

<强>输出：

transaction_time
----------------
2012-05-15
2012-06-10
2012-06-15

现在在上述两组之间执行CROSS JOIN

SELECT *
FROM (
   SELECT DISTINCT transaction_time
   FROM t
   WHERE item = 'A') AS times
CROSS JOIN (
   SELECT DISTINCT unit_price
   FROM t
   WHERE item = 'A') AS up
ORDER BY times.transaction_time

得到：

transaction_time unit_price
----------------------------
2012-05-15       3
2012-05-15       2
2012-05-15       1
2012-06-10       3
2012-06-10       2
2012-06-10       1
2012-06-15       1
2012-06-15       3
2012-06-15       2

现在使用上述内容并执行相关子查询，从unit_price项transaction_time获取'A'：

SELECT transaction_time, unit_price,
       (SELECT quantity
        FROM t
        WHERE t.item = 'A' 
              AND t.unit_price = up.unit_price
              AND t.transaction_time <= times.transaction_time
        ORDER BY transaction_time DESC LIMIT 1) AS quantity
FROM (
   SELECT DISTINCT transaction_time
   FROM t
   WHERE item = 'A') AS times
CROSS JOIN (
   SELECT DISTINCT unit_price
   FROM t
   WHERE item = 'A') AS up
ORDER BY times.transaction_time

<强>输出：

transaction_time     unit_price  quantity
----------------------------------------
15.05.2012 00:00:00  1           15
15.05.2012 00:00:00  3           15
15.05.2012 00:00:00  2           NULL
10.06.2012 00:00:00  1           15
10.06.2012 00:00:00  3           15
10.06.2012 00:00:00  2           15
15.06.2012 00:00:00  1           15
15.06.2012 00:00:00  3           15
15.06.2012 00:00:00  2           10

最终结果只是在上面执行GROUP BY：

SELECT transaction_time, 
       'A' AS item,
       SUM(quantity) AS quantity,
       SUM(quantity*unit_price) AS subtotal
FROM (           
    SELECT transaction_time, unit_price,
           (SELECT quantity
            FROM t
            WHERE t.item = 'A' 
                  AND t.unit_price = up.unit_price
                  AND t.transaction_time <= times.transaction_time
            ORDER BY transaction_time DESC LIMIT 1) AS quantity
    FROM (
       SELECT DISTINCT transaction_time
       FROM t
       WHERE item = 'A') AS times
    CROSS JOIN (
       SELECT DISTINCT unit_price
       FROM t
       WHERE item = 'A') AS up) AS x
GROUP BY transaction_time

<强>输出：

transaction_time  item  quantity  subtotal
----------------------------------------------
15.05.2012        A     30        60
10.06.2012        A     45        90
15.06.2012        A     40        80

Demo here

SQL Query根据不同时间的不同单价，数量获取总价值

2 个答案: