LINQ member expression getting column name

谁说我不能喝 提交于 2019-11-30 07:39:47

If you need to also decompose simple (or nearly simple) expressions, you'll need some extra legwork to handle the different situations. Here is some starter code that handles some common cases:

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;
}

Which produces output like:

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"
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