我收到错误消息:
扩展方法必须在非通用静态类中定义
在线上:
public class LinqHelper
这是基于Mark Gavells代码的帮助程序类。我对这个错误的含义感到非常困惑,因为我确信当我在星期五离开它时,它可以正常工作!
using System; using System.Collections.Generic; using System.Linq; using System.Web; using System.Linq.Expressions; using System.Reflection; /// <summary> /// Helper methods for link /// </summary> public class LinqHelper { public static IOrderedQueryable<T> OrderBy<T>(this IQueryable<T> source, string property) { return ApplyOrder<T>(source, property, "OrderBy"); } public static IOrderedQueryable<T> OrderByDescending<T>(this IQueryable<T> source, string property) { return ApplyOrder<T>(source, property, "OrderByDescending"); } public static IOrderedQueryable<T> ThenBy<T>(this IOrderedQueryable<T> source, string property) { return ApplyOrder<T>(source, property, "ThenBy"); } public static IOrderedQueryable<T> ThenByDescending<T>(this IOrderedQueryable<T> source, string property) { return ApplyOrder<T>(source, property, "ThenByDescending"); } static IOrderedQueryable<T> ApplyOrder<T>(IQueryable<T> source, string property, string methodName) { string[] props = property.Split('.'); Type type = typeof(T); ParameterExpression arg = Expression.Parameter(type, "x"); Expression expr = arg; foreach (string prop in props) { // use reflection (not ComponentModel) to mirror LINQ PropertyInfo pi = type.GetProperty(prop); expr = Expression.Property(expr, pi); type = pi.PropertyType; } Type delegateType = typeof(Func<,>).MakeGenericType(typeof(T), type); LambdaExpression lambda = Expression.Lambda(delegateType, expr, arg); object result = typeof(Queryable).GetMethods().Single( method => method.Name == methodName && method.IsGenericMethodDefinition && method.GetGenericArguments().Length == 2 && method.GetParameters().Length == 2) .MakeGenericMethod(typeof(T), type) .Invoke(null, new object[] { source, lambda }); return (IOrderedQueryable<T>)result; } }
更改
至
public static class LinqHelper
创建扩展方法时,需要考虑以下几点:
non-generic
static
non-nested
this