Question

您好

我正在使用LINQ和EF和C＃4.0。我已将基本的ELMAH表拖入EF（已构建并保存了很多次）。一切都按照人们的预期运作。

但是他们试图过于雄心勃勃并需要一些帮助 - 我试图从一个作为变量传入的表达式中获取Column名称。

我想要的是：

传入：x =＆gt; x.ErrorId

并获取：“ErrorId”

public void GetColumnName(Expression<Func<T, object>> property)
{
  // The parameter passed in x=>x.Message
  // Message works fine (probably because its a simple string) using:
  string columnName = (property.Body as MemberExpression).Member.Name;

  // But if I attempt to use the Guid or the date field then it
  // is passed in as x => Convert(x.TimeUtc)
  // As a result the above code generates a NullReference exception
  // i.e. {"Object reference not set to an instance of an object."}

  // What is the correct code here to extract the column name generically?
  // Ideally in a way that won't bite me again in the future.

}

感谢您的帮助！丹。

Answer 1

如果你还需要分解简单（或接近简单）的表达式，你需要一些额外的工作来处理不同的情况。以下是一些处理一些常见情况的入门代码：

string GetColumnName<T,TResult>(Expression<Func<T,TResult>> property)
{
    var member = GetMemberExpression(property.Body);
    if (member == null)
        throw new ArgumentException("Not reducible to a Member Access", 
                                    "property");

    return member.Member.Name;
}

MemberExpression GetMemberExpression(Expression body)
{
    var candidates = new Queue<Expression>();
    candidates.Enqueue(body);
    while (candidates.Count > 0)
    {
        var expr = candidates.Dequeue();
        if (expr is MemberExpression)
        {
            return ((MemberExpression)expr);
        }
        else if (expr is UnaryExpression)
        {
            candidates.Enqueue(((UnaryExpression)expr).Operand);
        }
        else if (expr is BinaryExpression)
        {
            var binary = expr as BinaryExpression;
            candidates.Enqueue(binary.Left);
            candidates.Enqueue(binary.Right);
        }
        else if (expr is MethodCallExpression)
        {
            var method = expr as MethodCallExpression;
            foreach (var argument in method.Arguments)
            {
                candidates.Enqueue(argument);
            }
        }
        else if (expr is LambdaExpression)
        {
            candidates.Enqueue(((LambdaExpression)expr).Body);
        }
    }

    return null;
}

产生的输出如下：

GetColumnName((x) => x.X): "X"
GetColumnName((x) => x.X + 2): "X"
GetColumnName((x) => 2 + x.X): "X"
GetColumnName((x) => -x.X): "X"
GetColumnName((x) => Math.Sqrt(x.Y)): "Y"
GetColumnName((x) => Math.Sqrt(Math.Abs(x.Y))): "Y"

LINQ成员表达式获取列名

1 个答案: