Question

将IList集合转换为逗号分隔id的字符串的优雅方法是什么？

“1,234,2,324,324,2”

Answer 1

    IList<int> list = new List<int>( new int[] { 1, 2, 3 } );
    Console.WriteLine(string.Join(",", list));

Answer 2

你可以这样做：

// Given: IList<int> collection;

string commaSeparatedInts = string.Join(",",collection.Select(i => i.ToString()).ToArray());

Answer 3

这样做

IList<int> strings = new List<int>(new int[] { 1,2,3,4 });
string[] myStrings = strings.Select(s => s.ToString()).ToArray();
string joined = string.Join(",", myStrings);

或完全与Linq

string aggr = strings.Select(s=> s.ToString()).Aggregate((agg, item) => agg + "," + item);

Answer 4

// list = IList<MyObject>

var strBuilder = new System.Text.StringBuilder();

foreach(var obj in list)
{
  strBuilder.Append(obj.ToString());
  strBuilder.Append(",");
}

strBuilder = strBuilder.SubString(0, strBuilder.Length -1);
return strBuilder.ToString();

Answer 5

List<int> intList = new List<int>{1,234,2,324,324,2};
var str = intList.Select(i => i.ToString()).Aggregate( (i1,i2) => string.Format("{0},{1}",i1,i2));
Console.WriteLine(str);

Answer 6

mstrickland对使用字符串构建器有一个好主意，因为它具有较大列表的速度。但是，您不能将stringbuilder设置为字符串。试试这个。

    var strBuilder = new StringBuilder();

    foreach (var obj in list)
    {
        strBuilder.Append(obj.ToString());
        strBuilder.Append(",");
    }

    return strBuilder.ToString(0, strBuilder.Length - 1);

将IList <int>集合转换为逗号分隔列表</int>

6 个答案: