Question

我做了一个这样的循环：

    int total;
    total = ((toVal - fromVal) + 1) * 2;
    RadProgressContext progress = RadProgressContext.Current;
    progress.Speed = "N/A";

    finYear = fromVal;

    for (int i = 0; i < total; i++)
    {
          decimal ratio = (i * 100 / total);

            progress.PrimaryTotal = total;
            progress.PrimaryValue = total;
            progress.PrimaryPercent = 100;

            progress.SecondaryTotal = 100; // total;
            progress.SecondaryValue = ratio;//i ;
            progress.SecondaryPercent = ratio; //i;


            progress.CurrentOperationText = "Step " + i.ToString();
            if (!Response.IsClientConnected)
            {
                //Cancel button was clicked or the browser was closed, so stop processing
                break;
            }

            progress.TimeEstimated = (total - i) * 100;
            //Stall the current thread for 0.1 seconds
            System.Threading.Thread.Sleep(100);


    }

现在我希望根据toVal & fromVal运行特定方法在前一个循环中但没有相同的循环数我想在这样的循环中运行它：

   for (fromVal; fromVal < toVal  ; fromVal++)
    {
        PrepareNewEmployees(calcYear, fromVal);
    }

例如：

fromVal =  2014 
toVal   = 2015

所以我想跑两次而不是4次！像这样：

PrepareNewEmployees(calcYear, 2014);
PrepareNewEmployees(calcYear, 2015);

但是在上一个循环for (int i = 0; i < total; i++)

中

Answer 1

您错过了进度条更新的重点。您不应该运行4次迭代并且每2次迭代执行一些工作，但是相反。做一个循环：

 for (int i = fromVal; i < toVal; i++)
{
    PrepareNewEmployees(...);
    decimal ratio = ((double)toVal-i)/(toVal-fromVal) *100;
    //Some other things, that need to be done twice in an iteration
}

Answer 2

因为您已经使用Thread，所以请考虑实施以下内容：

public void ResetProgress()
{
    SetProgress(0);
}

public SetProgress(int percents)
{
    // set progress bar to a given percents/ratio
    // you will have to use Invoke and blablabla
}

然后任何您的作业将如下所示

ResetProgress();
// note: you need to remember from which value you start to be able to calculate progress
for (int i = startVal; i < toVal  ; i++)
{
    PrepareNewEmployees(calcYear, i);
    SetProgress(100 * (i - startVal) / (toVal - startVal)); // in percents [0-100]
}
// optional, required if you exit loop or use suggestion below
SetProgress(100);

您也可以对其进行优化，不要在每个步骤后更新进度，而是在执行一定数量的步骤之后。例如，不要调用SetProgress，而是执行

if(i % 10 == 0)
    SetProgress();

这会使SetProgress的次数减少十倍。当然，有一些假设，例如：i从0开始，如果你想在结尾处有100％的条形，那么i应该可以被10分割。只是一个想法开始。 / p>

如何在某些条件下限制循环的循环次数？

2 个答案: