这是一个表,要求累计金额 。表名t ,字段lsh,shangpinbh,jine。 流水号商品编号金额累计金额1100125002500210013000550031004400095004100610001050051007200012500 select lsh,shangpinbh,jine,(select sum(jine) from t where shangpinbh=t1.
思路:先将需要进行累积的值变成负值,再进行累加,即得到累计差 with x as (select rownum as seq,a.* from (select 编号,项目,金额 from detail order by 编号) a) select 编号, 项目, 金额, sum(case when seq=1 then 金额
import numpy as npimport scipy.io as ioimport matplotlib.pyplot as pltfrom scipy.interpolate import InterpolatedUnivariateSplinefrom scipy.optimize import curve_fitnp.random.seed(0)# 原始数据x = np.
查询结果 查询语句 select tDailyCount.ddate, tDailyCount.count, @dailycount := @dailycount + tDailyCount.count as scount from (select tTable.dd as ddate,count(*) as count from (SELECT jira_key, min(date_fo
不优雅的地方 对结果的获取需要额外引入CountDownLatch等待所有线程执行完毕 CompletableFuture Future接口天然可以通过回调获取结果,所以可以利用CompletableFuture实现并行,并调用CompletableFuture.join获取结果 private static void completableFuture(BigInteger n, int