C＃传递参数与λ

Question

在下面，我需要通过nextDB的Lambda表达式中重试：

Retry.Do(() => 
{ 
    string nextDB = dbList.Next(); 
    using (DataBaseProxy repo = new DataBaseProxy(nextDB)) 
    { 
     return repo.DoSomething(); 
    } 
}); 

我该怎么办呢？这里是我Retry类：

public static class Retry 
{ 
    public static void Do(
     Action action, 
     int retryCount = 3) 
    { 
     Do<object>(() => 
     { 
      action(); 
      return null; 
     }, retryCount); 
    } 

    public static T Do<T>(
     Func<T> action, 
     int retryCount = 3) 
    { 
     var exceptions = new List<Exception>(); 

     for (int retry = 0; retry < retryCount; retry++) 
     { 
      try 
      { 
       return action(); 
      } 
      catch (Exception ex) 
      { 
       exceptions.Add(ex); 
      } 
     } 

     throw new AggregateException(exceptions); 
    } 
} 

Answer 1

我想你想在这里使用Action<T>。例如：

public static void Do<T>(
    Action<T> action, 
    T param, 
    int retryCount = 3) 
{ 
    var exceptions = new List<Exception>(); 

    for (int retry = 0; retry < retryCount; retry++) 
    { 
     try 
     { 
      action(param); 
      return; 
     } 
     catch (Exception ex) 
     { 
      exceptions.Add(ex); 
     } 
    } 

    throw new AggregateException(exceptions); 
} 

你会调用该函数是这样的：

Do(s => { 
    Console.WriteLine(s); 
}, "test", 3); 

---

根据您的意见，似乎要在多个数据库通过和每一个尝试连续直到你找到一个可行的。一个简单的选择是删除到retryCount，而不是传入你的数组。

public static void Do<T>(
    Action<T> action, 
    IEnumerable<T> items) 
{ 
    var exceptions = new List<Exception>(); 

    foreach(var item in items) 
    { 
     try 
     { 
      action(item); 
      return; 
     } 
     catch (Exception ex) 
     { 
      exceptions.Add(ex); 
     }   
    } 

    throw new AggregateException(exceptions); 
} 

现在你怎么称呼它是这样的：

Do(s => { 
    Console.WriteLine(s); 
}, new[] { "db1", "db2", "db3" });