Question

我有两个扩展功能，可在List<int>和List<string>上运行。

public static string StringTokenizr(this List<int> list, NpStringTokenizrType type, string splitter = ",")
{
    return string.Join(",", list.ConvertAll(name => string.Format(type == NpStringTokenizrType.StringLike ? "'{0}'" : "{0}", name.ToString().Replace("'", "\\'"))));
}

public static string StringTokenizr(this List<string> list, NpStringTokenizrType type, string splitter = ",")
{
    return string.Join(",", list.ConvertAll(name => string.Format(type == NpStringTokenizrType.StringLike ? "'{0}'" : "{0}", name.ToString().Replace("'", "\\'"))));
}

有没有办法让它更通用，以便它可以只是一个功能？

Answer 1

使您的方法通用：

public static string StringTokenizr<T>(
       this List<T> list, 
       NpStringTokenizrType type, 
       string splitter = ",")
{
   return string.Join(",", list.ConvertAll(name => string.Format(type == NpStringTokenizrType.StringLike ? "'{0}'" : "{0}", name.ToString().Replace("'", "\\'"))));
}

Answer 2

通过这种方式你可以使它一般，但这个函数不会局限于int和字符串它将接受其他类型

public static string StringTokenizr<T>(this List<T> list, NpStringTokenizrType type, string splitter = ",")
{
    return string.Join(",", list.ConvertAll(name => string.Format(type == NpStringTokenizrType.StringLike ? "'{0}'" : "{0}", name.ToString().Replace("'", "\\'"))));
}

Answer 3

使用像List<T>这样的泛型类型参数，并检查函数体内的类型，或者将List作为string投射到函数中。

将两个扩展功能合二为一

3 个答案: