string result = String.Join(delimiter, list);

是充分的。

我认为如果你在扩展方法中定义逻辑，代码将更易于阅读:

public static class EnumerableExtensions { 
  public static string Join<T>(this IEnumerable<T> self, string separator) {  
    return String.Join(separator, self.Select(e => e.ToString()).ToArray()); 
  } 
} 

public class Person {  
  public string FirstName { get; set; }  
  public string LastName { get; set; }  
  public override string ToString() {
    return string.Format("{0} {1}", FirstName, LastName);
  }
}  

// ...

List<Person> people = new List<Person>();
// ...
string fullNames = people.Join(", ");
string lastNames = people.Select(p => p.LastName).Join(", ");

我已经使用LINQ做到了这一点:

var oCSP = (from P in db.Products select new { P.ProductName });

string joinedString = string.Join(",", oCSP.Select(p => p.ProductName));

你可以简单地使用:

List<string> items = new List<string>() { "foo", "boo", "john", "doe" };

Console.WriteLine(string.Join(",", items));

编码快乐!

警告-严重性能问题

虽然这个答案确实产生了预期的结果，但与这里的其他答案相比，它的性能较差。在决定使用它时要非常谨慎

通过使用LINQ，这应该工作;

string delimiter = ",";
List<string> items = new List<string>() { "foo", "boo", "john", "doe" };
Console.WriteLine(items.Aggregate((i, j) => i + delimiter + j));

类描述:

public class Foo
{
    public string Boo { get; set; }
}

用法:

class Program
{
    static void Main(string[] args)
    {
        string delimiter = ",";
        List<Foo> items = new List<Foo>() { new Foo { Boo = "ABC" }, new Foo { Boo = "DEF" },
            new Foo { Boo = "GHI" }, new Foo { Boo = "JKL" } };

        Console.WriteLine(items.Aggregate((i, j) => new Foo{Boo = (i.Boo + delimiter + j.Boo)}).Boo);
        Console.ReadKey();

    }
}

这是我最好的:)

items.Select(i => i.Boo).Aggregate((i, j) => i + delimiter + j)

好问题。我一直在用

List<string> myStrings = new List<string>{ "ours", "mine", "yours"};
string joinedString = string.Join(", ", myStrings.ToArray());

它不是LINQ，但它可以工作。