Question

我使用以下代码按表达式转换顺序，因此也可以对可以为空的列进行排序。

protected virtual Expression<Func<T, object>> GetSorting(string ordering)
{
        Expression<Func<T, object>> expression = default(Expression<Func<T, object>>);
        IEnumerable<Order> sortObjects = string.IsNullOrEmpty(ordering) ? null : JsonConvert.DeserializeObject<IEnumerable<Order>>(ordering);
        if (sortObjects != null)
        {
            foreach (Order sortObject in sortObjects)
            {
                Expression<Func<T, object>> currentExpression = this.GetExpression(sortObject.Property);
                expression = this.CombineExpressions(expression, currentExpression);
            }
        }

        return expression;
}

private Expression<Func<T, object>> GetExpression(string propertyName)
{
        Type type = typeof(T);
        ParameterExpression parameter = Expression.Parameter(type, "x");
        MemberExpression propertyReference = Expression.Property(parameter, propertyName);
        Expression conversion = Expression.Convert(propertyReference, typeof(object));
        Expression<Func<T, object>> currentExpression = Expression.Lambda<Func<T, object>>(conversion, new[] { parameter });
        return currentExpression;

}

private Expression<Func<T, object>> CombineExpressions(Expression<Func<T, object>> expression, Expression<Func<T, object>> currentExpression)
{
        if (expression == default(Expression<Func<T, object>>))
        {
            expression = currentExpression;
        }
        else
        {
            // Combine the two expressions' body together
            BinaryExpression body = Expression.AndAlso(expression.Body, currentExpression.Body);
            ParameterExpression[] parameters = new ParameterExpression[1] { Expression.Parameter(typeof(T), expression.Parameters.First().Name) };

            // Convert the BinaryExpression to the requested type
            Expression<Func<T, object>> lambda = Expression.Lambda<Func<T, object>>(body, parameters);

            expression = lambda;
        }

        return expression;
}

此代码适用于所有非导航属性，但似乎不再查询导航属性。我使用Select表达式来加载导航属性，如下所示：

protected override Expression<Func<Resource, ResourceViewModel>> Selector
{
        get
        {
            return (x) => new ResourceViewModel()
            {
                ResourceId = x.ResourceId,
                DisplayName = x.DisplayName,                  
                ResourceType = x.ResourceType != null ? x.ResourceType.Name : string.Empty,
            }
        }
}

如果我没有订购的东西，则会加载导航属性。但是只要有任何要排序的东西，导航属性就是null。如果我跳过三元操作并直接转到ResourceType.Name属性，我会得到一个异常，告诉我lambda_method抛出了NullReference异常。

我知道订购导航属性不会起作用，但这不是问题。订购常规＆＃39;属性导致问题。

对此有何想法？

Answer 1

事实证明这个问题并不像我想象的那么复杂。问题是我创建了错误的表达式树。您可以嵌套表达式，以便导航到属性的属性。

下面的解决方案应该解释这一点（这不是我如何解决问题，但它应该说清楚）：

private Expression<Func<T, object>> GetExpression(string parentClass, string propertyName)
{
    Type type = typeof(T);
    ParameterExpression parameter = Expression.Parameter(type, "x");

    // Get parent class expression
    // Will result in (x) => x.MyNavigationPropertyWhichIsAClass
    MemberExpression propertyReference1 = Expression.Property(parameter, parentclass);

    // Navigate to the property of the navigation property class
    // Will result in (x) => x.MyNavigationPropertyWhichIsAClass.MyPropertyIWantToSort
    MemberExpression propertyReference2 = Expression.Property(propertyRefernce1, propertyName);

    Expression conversion = Expression.Convert(propertyReference2, typeof(object));
    Expression<Func<T, object>> currentExpression = Expression.Lambda<Func<T, object>>(conversion, new[] { parameter });
    return currentExpression;

}

排序后，实体框架的导航属性为空

1 个答案: