可以表达式>和destination.x=source.x 被重构吗？

Question

我正在生成具体实现复印机的接口。后面的步骤是确定我是否可以根据请求的复制类型（直接复制、尝试或 try-catch-addToValidationDictionary）轻松添加不同的行为。这是我需要/正在处理的主要问题是 try-catch-addToValidationDictionary。如果复制语句本身（result.AssetTag = asset.AssetTag）能够以列表形式为不需要 try/catch/validation 功能的另一个消费者重用，那就太好了。

一般形式是这样的：

public static AssetService
{
public static ModelAsset CreateAssetDomain(IAmAnAsset asset, IValidationDictionary validationDictionary)
 {

 var result=new ModelAsset();
 var hasExceptions=false;
  try
        {
            result.AssetTag = asset.AssetTag;
        }
        catch (System.Exception exception)
        {
            validationDictionary.AddError(Member.Name<IAmAnAsset>(lIAmAnAsset => lIAmAnAsset.AssetTag), exception.Message);
            hasExceptions = true;
        }
 try
        {
            result.LocationIdentifer = asset.LocationIdentifer;
        }
        catch (System.Exception exception)
        {
            validationDictionary.AddError(Member.Name<IAmAnAsset>(lIAmAnAsset => lIAmAnAsset.LocationIdentifer), exception.Message);
            hasExceptions = true;
        }
  ...
  if (hasExceptions)
    throw new ArgumentException("Failed validation");

 return result;
 }
}

我试图用 lambda 来分解一些重复，但来自 这篇帖子似乎只需要一个Expression> 并且我不确定您将如何使用表达式>超载。

一次尝试如下：

 Action<Action, Expression<Func<IAmAnAsset, object>>> CopyActions = (copyAction, expression) =>
            {
                try
                {
                    copyAction();
                }
                catch (Exception exception)
                {

                    validationDictionary.AddError(Member.Name<IAmAnAsset>(expression), exception.Message);
                    hasExceptions = true;
                }
            };

  var copyActions = new Dictionary<string,Action>()
    {
    Member.Name<IAmAnAsset>(z=>z.AddedBy),()=>result.AddedBy=asset.AddedBy},
    Member.Name<IAmAnAsset>(z=>z.AssetTag),()=>result.AssetTag=asset.AssetTag},
    ...
    }
 foreach (var item in copyActions)
        {
            tryCopyAction(item.Value, item.Key);
        }
 if (hasExceptions)
throw new ArgumentException("Failed validation");

 return result;

我希望有一种解决方案可以减少固有的重复 Member.Name(z=>z.AddedBy),()=>result.AddedBy=asset.AddedBy}, 满足以下任何条件：

• 每行上有 2 个位置需要 IAmAnAsset.AddedBy
• 同一行上需要 .AddedBy 3 次
• Member.Name 在每行上
  • 是否可以使用此表达式来检索字符串名称或评估它的值？

Answer 1

一个简单的函数怎么样？

public static class Member
{
  private static string GetMemberName(Expression expression)
  {
    switch (expression.NodeType)
    {
      case ExpressionType.MemberAccess: var memberExpression = (MemberExpression)expression; var supername = GetMemberName(memberExpression.Expression); if (String.IsNullOrEmpty(supername)) return memberExpression.Member.Name; return String.Concat(supername, '.', memberExpression.Member.Name);
      case ExpressionType.Call: var callExpression = (MethodCallExpression)expression; return callExpression.Method.Name;
      case ExpressionType.Convert: var unaryExpression = (UnaryExpression)expression; return GetMemberName(unaryExpression.Operand);
      case ExpressionType.Parameter: return String.Empty;
      default: throw new ArgumentException("The expression is not a member access or method call expression");
    }
  }
  public static string Name<T,V>(Expression<Func<T, V>> expression)
  {
    return GetMemberName(expression.Body);
  }
  public static string Name<T>(Expression<Action<T>> expression)
  {
    return GetMemberName(expression.Body);
  }
}

void Copy<D, S, V>(D dest, S source, Expression<Func<S, V>> getVal, Action<D, V> setVal, IDictionary validationDictionary)
{
  Func<S, V> doGetVal = getVal.Compile();
  try { setVal(dest, (V)doGetVal(source)); }
  catch (System.Exception exception)
  {
    validationDictionary.Add(Member.Name<S,V>(getVal), exception.Message);
  }
}

class TestAsset { public string AssetTag { get; set; } public string LocationIdentifier { get; set; } }
TestAsset Test()
{
  Dictionary<string, string> validationDictionary = new Dictionary<string, string>();
  var result = new TestAsset{ AssetTag = "a", LocationIdentifier = "b" };
  var asset = new TestAsset{ AssetTag = "A", LocationIdentifier = "B" };
  var validationCount = validationDictionary.Count(); 
  Copy(result, asset, x => asset.AssetTag, (x, v) => x.AssetTag = v, validationDictionary); 
  Copy(result, asset, x => asset.LocationIdentifier, (x, v) => x.LocationIdentifier = v, validationDictionary); 
  if (validationCount < validationDictionary.Count) throw new ArgumentException("Failed validation"); 
  return result;
}