有没有一种优雅的方法将IList集合转换为逗号分隔的ID字符串?
"1,234,2,324,324,2"
"1,234,2,324,324,2"
IList<int> list = new List<int>( new int[] { 1, 2, 3 } );
Console.WriteLine(string.Join(",", list));
ToArray
。所使用的 string.Join
重载操作的是 IEnumerable<string>
。 - Stefan GlienkeSelect
,只需传入一个IEnumerable<T>
(在这种情况下是int),它会为您获取每个值的字符串值。(该重载也同时添加)。 - Servy你可以做:
// Given: IList<int> collection;
string commaSeparatedInts = string.Join(",",collection.Select(i => i.ToString()).ToArray());
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);
或者完全使用Linq
string aggr = strings.Select(s=> s.ToString()).Aggregate((agg, item) => agg + "," + item);
// 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();
var strBuilder = new StringBuilder();
foreach (var obj in list)
{
strBuilder.Append(obj.ToString());
strBuilder.Append(",");
}
return strBuilder.ToString(0, strBuilder.Length - 1);
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);