正確的地方加入的try-catch C＃

Question

我想用try-catch檢查由0

((double)o.Close - (double)subList.Skip(idx - t + 1).First().Close)/
(double)subList.Skip(idx - t + 1).First().Close 

被劃分...但我不知道在哪裏添加try-catch。我嘗試添加包括整個var newList =，但不允許，因爲變量newList以後不再存在。那麼添加try-catch的適當位置在哪裏？

var newList = data.GroupBy(o => o.Date).Where(o => o.Key <= Beginday).OrderByDescending(o => o.Key).Take(Y).SelectMany(o => o).GroupBy(x => new { x.Symbol }) 
       .Select 
       (
        x => 
        { 
         var subList = x.OrderBy(y => y.Date).ToList(); 
         return subList.Select((y, idx) => 
         { 
          return new 
          { 
           Symbol = y.Symbol, 
           Close = y.Close, 
           Date = y.Date, 
           Vol = (idx < t - 1) ? 0 : new DescriptiveStatistics(subList.Skip(idx - t + 1).Take(t) 
           .Select(o => 
           ((double)o.Close - (double)subList.Skip(idx - t + 1).First().Close)/(double)subList.Skip(idx - t + 1).First().Close).ToList()).StandardDeviation,        
         }; 
         }); 
        } 
       ) 
       .SelectMany(x => x) 
       .ToList();  

Answer 1

取決於你要在你釣到的魚做什麼 - 如果你想newList有一個有效的值，即使除數爲0，那麼你應該有你的0檢查集成到主碼流而比使用例外流量。

但是，要直接以書面形式回答您的問題，請先將newList聲明爲默認值，然後在try/catch塊內重新分配（將您編寫的代碼的全部內容放入try中，並使用顯式的default嘗試之前的聲明）。然後，當您需要稍後使用它時，newList將存在。

Answer 2

如果你想趕上一切，然後

 try 
      { 
      var newList = data.GroupBy(o => o.Date).Where(o => o.Key <=      Beginday).OrderByDescending(o => o.Key).Take(Y).SelectMany(o => o).GroupBy(x => new { x.Symbol }) 
      .Select 
      (
       x => 
       { 
        var subList = x.OrderBy(y => y.Date).ToList(); 
        return subList.Select((y, idx) => 
        { 
         return new 
         { 
          Symbol = y.Symbol, 
          Close = y.Close, 
          Date = y.Date, 
          Vol = (idx < t - 1) ? 0 : new DescriptiveStatistics(subList.Skip(idx - t + 1).Take(t) 
          .Select(o => 
          ((double)o.Close - (double)subList.Skip(idx - t + 1).First().Close)/(double)subList.Skip(idx - t + 1).First().Close).ToList()).StandardDeviation,        
        }; 
        }); 
       } 
      ) 
      .SelectMany(x => x) 
      .ToList();  

      } 

    catch(Exception ex) 
    { 
    //your error 
     } 

Answer 3

你當然可以這樣做：

var newList = new[] 
{ 
    new { Symbol = "", Close = 0.0, Date = DateTime.Now, Vol = 0 } 
}.Take(0).ToList(); 

try 
{ 
    var newList = data.GroupBy(o => o.Date).Where(o => o.Key <= Beginday).OrderByDescending(o => o.Key).Take(Y).SelectMany(o => o).GroupBy(x => new { x.Symbol }) 
     .Select 
     (
      x => 
      { 
       var subList = x.OrderBy(y => y.Date).ToList(); 
       return subList.Select((y, idx) => 
       { 
        return new 
        { 
         Symbol = y.Symbol, 
         Close = y.Close, 
         Date = y.Date, 
         Vol = (idx < t - 1) ? 0 : new DescriptiveStatistics(subList.Skip(idx - t + 1).Take(t) 
         .Select(o => 
         ((double)o.Close - (double)subList.Skip(idx - t + 1).First().Close)/(double)subList.Skip(idx - t + 1).First().Close).ToList()).StandardDeviation,        
       }; 
       }); 
      } 
     ) 
     .SelectMany(x => x) 
     .ToList();  
} 
catch (Exception ex) 
{ 
} 

只要名稱和類型是兩個匿名類型相同，則這工作正常。