【问题】
Let’s say I have given data as string:
I have two tables, one with the main data and a second table with historical values.
Tablestocks
+----------+-------+-------------+ | stock_id | symbol| name | +--------------------------------+ | 1| AAPL | Apple | | 2| GOOG | Google | | 3| MSFT | Microsoft | +----------+-------+-----------+
Tableprices
+----------+-------+---------------------+ | stock_id | price | date | +----------------------------------------+ | 1| 0.05| 2015-02-2401:00:00| | 2| 2.20| 2015-02-2401:00:00| | 1| 0.50| 2015-02-2323:00:00| | 2| 1.90| 2015-02-2323:00:00| | 3| 2.10| 2015-02-2323:00:00| | 1| 1.00| 2015-02-2319:00:00| | 2| 1.00| 2015-02-2319:00:00| +----------+-------+---------------------+
I need a query that returns:
+----------+-------+-----------+-------+ | stock_id | symbol| name | diff | +--------------------------------------+ | 1| AAPL | Apple | -0.45| | 2| GOOG | Google | 0.30| | 3| MSFT | Microsoft | NULL| +----------+-------+-----------+-------+
Where diff is the result of subtracting from the newest price of a stock the previous price. If one or less prices are present for a particular stock I should get NULL.
I have the following queries that return the last price and the previous price but I don’t know how to join everything
/\* last */ SELECTprice FROMprices WHEREstock_id = '1' ORDERBYdate DESC LIMIT 1 /\* previous */ SELECTprice FROMprices WHEREstock_id = '1' ORDERBYdate DESC LIMIT 1,1
【回答】
这类组内有序计算需要引用“第 1 条”和“第 2 条”,用 SQL 表达起来非常麻烦。这种情况用 SPL 实现很简单,只需 2 行代码:
| A | |
| 1 | $(db1)select s.stock_id stock_id,s.symbol symbol,s.name name,p.price price,p.date from stocks s,price p where s.stock_id=p.stock_id order by p.date desc |
| 2 | = A1.group(stock_id; symbol, name, if(p2=~.m(2).price,~.m(1).price-p2):diff) |
代码中 ~.m(i) 表示本组记录的第 i 条。
集算器还能用 m(-2)取倒数第 2 条,用 [1] 表示下一条,这种方式可以很容易进行有序计算和跨行计算,可参考:【集算器的序号思维及定位计算】
本文探讨了一种SQL问题,即如何在分组后计算组内数据的差值,特别是获取最新价格与前一个价格的差值。当股票的价格历史记录不足两条时,结果应为NULL。虽然SQL实现此操作复杂,但使用SPL语言可以简化为两行代码,利用其序号思维和定位计算功能实现高效处理。
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