C# 将 List<int> 转换为逗号分隔的字符串
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原文地址: http://stackoverflow.com/questions/1528724/
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Converting a List<int> to a comma separated string
提问by mrblah
Is there a way to take a List and convert it into a comma separated string?
有没有办法将 List 转换为逗号分隔的字符串？
I know I can just loop and build it, but somehow I think some of you guys a more cool way of doing it?
我知道我可以循环并构建它，但不知何故，我认为你们中的一些人有一种更酷的方式来做到这一点？
I really want to learn these types of 'tricks', so please explain or link to the docs on the method you use.
我真的很想学习这些类型的“技巧”，所以请解释或链接到有关您使用的方法的文档。
采纳答案by Pavel Minaev
List<int> list = ...;
string.Join(",", list.Select(n => n.ToString()).ToArray())
回答by Yuriy Faktorovich
List<int> list = new List<int> { 1, 2, 3 };
Console.WriteLine(String.Join(",", list.Select(i => i.ToString()).ToArray()));
回答by Larsenal
My "clever" entry:
我的“聪明”条目：
List<int> list = new List<int> { 1, 2, 3 };
StringBuilder sb = new StringBuilder();
var y = list.Skip(1).Aggregate(sb.Append(x.ToString()),
(sb1, x) => sb1.AppendFormat(",{0}",x));
// A lot of mess to remove initial comma
Console.WriteLine(y.ToString().Substring(1,y.Length - 1));
Just haven't figured how to conditionally add the comma.
只是还没想好如何有条件地添加逗号。
回答by cdiggins
For extra coolness I would make this an extension method on IEnumerable<T> so that it works on any IEnumerable:
为了更酷，我会将其作为 IEnumerable<T> 上的扩展方法，以便它适用于任何 IEnumerable：
public static class IEnumerableExtensions {
public static string BuildString<T>(this IEnumerable<T> self, string delim = ",") {
return string.Join(delim, self)
}
}
Use it as follows:
使用方法如下：
List<int> list = new List<int> { 1, 2, 3 };
Console.WriteLine(list.BuildString(", "));
回答by Eric Lippert
For approximately one gazillion solutions to a slightly more complicated version of this problem -- many of which are slow, buggy, or don't even compile -- see the comments to my article on this subject:
对于这个问题的稍微复杂一点的版本的大约一亿个解决方案——其中许多是缓慢的、有问题的，或者甚至不能编译——请参阅我关于这个主题的文章的评论：
http://blogs.msdn.com/ericlippert/archive/2009/04/15/comma-quibbling.aspx
http://blogs.msdn.com/ericlippert/archive/2009/04/15/comma-quibbling.aspx
and the StackOverflow commentary:
和 StackOverflow 评论：
回答by Gregory
Seems reasonablly fast.
似乎相当快。
IList<int> listItem = Enumerable.Range(0, 100000).ToList();
var result = listItem.Aggregate<int, StringBuilder, string>(new StringBuilder(), (strBuild, intVal) => { strBuild.Append(intVal); strBuild.Append(","); return strBuild; }, (strBuild) => strBuild.ToString(0, strBuild.Length - 1));
回答by Nitin Daware
Simple solution is
简单的解决办法是
List<int> list = new List<int>() {1,2,3};
string.Join<int>(",", list)
I used it just now in my code, working funtastic.
我刚刚在我的代码中使用了它，工作起来很有趣。
回答by Maurico Bello
you can use, the System.Linq library; It is more efficient:
您可以使用 System.Linq 库；它更有效：
using System.Linq;
string str =string.Join(",", MyList.Select(x => x.NombreAtributo));
