Question

请考虑以下代码：

string[] s = new[] { "Rob", "Jane", "Freddy" };

string joined = string.Join(", ", s);

// joined equals "Rob, Jane, Freddy"

出于UI原因，我可能希望显示字符串"Rob, Jane or Freddy"。

有关最简洁方法的任何建议吗？

修改

我正在寻找一些简洁的东西。由于我只连接小数字（＆lt; 10）字符串，所以我不担心运行时性能。

Answer 1

简明扼要的意思？还是跑？运行最快的将是StringBuilder手摇。但要打字，可能是（编辑处理0/1等）：

string joined;
switch (s.Length) {
    case 0: joined = ""; break;
    case 1: joined = s[0]; break;
    default:
        joined = string.Join(", ", s, 0, s.Length - 1)
               + " or " + s[s.Length - 1];
        break;
}

StringBuilder方法可能类似于：

static string JoinOr(string[] values) {
    switch (values.Length) {
        case 0: return "";
        case 1: return values[0];
    }
    StringBuilder sb = new StringBuilder();
    for (int i = 0; i < values.Length - 2; i++) {
        sb.Append(values[i]).Append(", ");
    }
    return sb.Append(values[values.Length-2]).Append(" or ")
        .Append(values[values.Length-1]).ToString();
}

Answer 2

连接除最后一个之外的所有内容。手动完成最后一个。

Answer 3

在string []上创建一个扩展方法，它实现与string.Join相同的逻辑，但最后一项将附加“或”。

string[] s = new[] { "Rob", "Jane", "Freddy" };

Console.WriteLine(s.BetterJoin(", ", " or "));

// --- 8＆lt; ----

namespace ExtensionMethods
{
    public static class MyExtensions
    {
        public static string BetterJoin(this string[] items, string separator, string lastSeparator)
        {
            StringBuilder sb = new StringBuilder();

            int length = items.Length - 2;
            int i = 0;

            while (i < length)
            {
                sb.AppendFormat("{0}{1}", items[i++], separator);
            }

            sb.AppendFormat("{0}{1}", items[i++], lastSeparator);
            sb.AppendFormat("{0}", items[i]);

            return sb.ToString();
        }
    }
}

Answer 4

怎么样：

if (s.Length > 1)
{
    uiText = string.Format("{0} and {1}", string.Join(", ", s, 0, s.Length - 1), s[s.Length - 1]);
}
else
{
    uiText = s.Length > 0 ? s[0] : "";
}

Answer 5

最高内存效率和可伸缩性将使用StringBuilder并预先计算最终字符串的长度以消除缓冲区重新分配。（这类似于String.Concat方法的工作方式。）

public static string Join(string[] items, string separator, string lastSeparator) {
    int len = separator.Length * (items.Length - 2) + lastSeparator.Length;
    foreach (string s in items) len += s.Length;
    StringBuilder builder = new StringBuilder(len);
    for (int i = 0; i < items.Length; i++) {
        builder.Append(items[i]);
        switch (items.Length - i) {
            case 1: break;
            case 2: builder.Append(lastSeparator); break;
            default: builder.Append(separator); break;
        }
    }
    return builder.ToString();
}

用法：

string joined = Join(s, ", ", " or ");

一个有趣的解决方案是使用递归算法。它适用于相当少量的字符串，但它不能很好地扩展。

public static string Join(string[] items, int index , string separator, string lastSeparator) {
    return items[index++]  + (index == items.Length-1 ? lastSeparator + items[index] : separator + Join(items, index, separator, lastSeparator));
}

用法：

string joined = Join(s, 0, ", ", " or ");

Answer 6

任何类型T的通用解决方案。

static class IEnumerableExtensions
{
  public static string Join<T>(this IEnumerable<T> items,
                               string seperator, string lastSeperator)
  {
    var sep = "";
    return items.Aggregate("", (current, item) =>
      {
        var result = String.Concat(current,
          // not first  OR not last
          current == "" || !items.Last().Equals(item) ? sep : lastSeperator,
          item.ToString());
        sep = seperator;
        return result;
      });
  }
}

用法：

var three = new string[] { "Rob", "Jane", "Freddy" };
var two = new string[] { "Rob", "Jane" };
var one = new string[] { "Rob" };
var threeResult = three.Join(", ", " or "); // = "Rob, Jane or Freddy"
var twoResult = two.Join(", ", " or "); // = "Rob or Jane"
var oneResult = one.Join(", ", " or "); // = "Rob"

Answer 7

string[] name_storage = new[] { "emre" , "balc" };                              
name_storage[name_storage.Count() - 1] += "ı";
string name = name_storage[0];
string sur_name = name_storage[1];
divElement.InnerHtml += name + " - " + sur_name;
//result = emre - balcı

将字符串数组连接到“string1，string2或string3”

修改

7 个答案: