6

你好，

我在 C# 4.0 中使用 LINQ 和 EF。我已将基本的 ELMAH 表拖到 EF 中（多次构建和保存）。一切都按预期工作。

但是试图过于雄心勃勃并且需要一点帮助 - 我试图从作为变量传递的表达式中获取列名。

我想要的是这样的：

传入：x=>x.ErrorId

并得到：“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！担。

4

1 回答 1

6

如果您还需要分解简单（或几乎简单）的表达式，则需要一些额外的工作来处理不同的情况。下面是一些处理一些常见情况的起始代码：

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;
}

产生如下输出：

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"
于 2011-06-07T18:42:45.703 回答