Question

我目前有以下代码，它允许我调用我对象的EmailAddress属性所需的任何方法，并且效果很好：

public static Expression<Func<T, bool>> BuildEmailAddressLambda(string method, params object[] args) {
    var e = Expression.Parameter(typeof(T), "e");
    var propertyInfo = typeof(T).GetProperty("EmailAddress");
    var m = Expression.MakeMemberAccess(e, propertyInfo);
    var mi = m.Type.GetMethod(method, args.Select(a => a.GetType()).ToArray());
    var c = args.Select(a => Expression.Constant(a, a.GetType())).ToArray();

    Expression<Func<T, bool>> lambda = Expression.Lambda<Func<T, bool>>(Expression.Call(m, mi, c), e);
    return lambda;
}

// called:
lambda = LambdaExpressionHelper<MailingListMember>.BuildEmailAddressLambda("StartsWith", "r", StringComparison.OrdinalIgnoreCase);

但是，我现在需要修改代码，使其适用于EmailAddress以外的对象成员。具体来说，我想创建一个树来覆盖以下表达式，其中使用了多个方法调用：

e.GetStringValue(12).StartsWith("r", StringComparison.OrdinalIgnoreCase);

我做了几次尝试，所有这些都会导致各种错误。我觉得我在如何创建ExpressionTrees的逻辑中缺少一些东西，并且非常感谢对此的一些帮助。

由于

Answer 1

这对你有好处吗？

public static Expression<Func<T, bool>> BuildEmailAddressLambda<T>(
    string member, IEnumerable<object> memberArgs, string method, params object[] args)
{
    var e = Expression.Parameter(typeof(T), "e");
    var memberInfo =
        (MemberInfo) typeof(T).GetField(member) ??
        (MemberInfo) typeof(T).GetProperty(member) ??
        (MemberInfo) typeof(T).GetMethod(member, (memberArgs ?? Enumerable.Empty<object>()).Select(p => p.GetType()).ToArray());
    Expression m;
    if (memberInfo.MemberType == MemberTypes.Method)
    {
        var a = memberArgs.Select(p => Expression.Constant(p));
        m = Expression.Call(e, (MethodInfo) memberInfo, a);
    }
    else
    {
        m = Expression.MakeMemberAccess(e, memberInfo);
    }
    var mi = m.Type.GetMethod(method, args.Select(a => a.GetType()).ToArray());
    var c = args.Select(a => Expression.Constant(a, a.GetType()));

    return Expression.Lambda<Func<T, bool>>(Expression.Call(m, mi, c), e);
}

// called:
lambda = LambdaExpressionHelper<MailingListMember>.BuildEmailAddressLambda("EmailAddress", null, "StartsWith", "r", StringComparison.OrdinalIgnoreCase);
// or
lambda = LambdaExpressionHelper<MailingListMember>.BuildEmailAddressLambda("GetStringValue", new object[] { 12 }, "StartsWith", "r", StringComparison.OrdinalIgnoreCase);

如何使用多个方法调用创建ExpressionTree

1 个答案: