Question

我希望我能得到一些帮助，因为我在SQL编程方面还很陌生。

以下代码有效，但大约需要花费时间。运行21分钟。能否以相同的结果更快地执行该查询？

表EOD大约有4600万条记录和6列（数据类型跟随名称）：

Symbol varchar(15)      
Date Date   
Open Float  
High Float  
Low Float   
Close Float  
Volume varchar(15)

该符号将根据我的分析而有所不同，因此未在CTE“ SMA”中编写where子句。

谢谢，让我知道是否可以说得更清楚。明智的。

with SMA as 
(
    select
        *,
        -- column names: symbol, open, low, high, close, volume(varchar(15))
        ((lag ([close],10)over (order by symbol, date)              --#varSMANUM must also change number for field 'WM'!
            +lag ([close],9)over (order by symbol, date) 
            +lag ([close],8)over (order by symbol, date) 
            +lag ([close],7)over (order by symbol, date) 
            +lag ([close],6)over (order by symbol, date) 
            +lag ([close],5)over (order by symbol, date) 
            +lag ([close],4)over (order by symbol, date) 
            +lag ([close],3)over (order by symbol, date) 
            +lag ([close],2)over (order by symbol, date) 
            +lag ([close],1)over (order by symbol, date) 
            )/ 10) as test,
    (cast(2 as float)/(10    --#varSMANUM must add lag statements above!!!!
            +1)) as WM
    from 
        qwid.dbo.eod 
)
select  
    date,
    [close],
    test 
from 
    SMA 
where 
    symbol = 'aapl' 
order by 
    date;

Answer 1

我想我先将查询重写为：

select date, [close],
       avg([close]) over (order by symbol, date rows between 10 preceding and 1 preceding) as test
    test 
from qwid.dbo.eod  
where symbol = 'aapl' 
order by date;

然后，我要在qwid.dbo.eod(symbol, date, [close])上添加索引。

请注意，这与您对前几个值的查询稍有不同。当您的版本返回NULL时，这将返回一个计算值。

SQL CTE该查询如何运行超过20分钟？

1 个答案: