尝试读取控制台应用程序的一些命令行参数

发布于 2024-07-26 22:00:18 字数 498 浏览 1 评论 0原文

我有一个非常简单的.NET 控制台应用程序。我希望传递一些命令行参数......没什么难的。

现在，最关键的是，我通常有一个我希望传递的项目列表，用于一个参数。在本例中，是文件列表。我不确定执行此操作的最佳方法。

eg. myapp.exe files=aaa.txt,bbb.txt,ccc.txt

但是带空格的文件名又如何呢？我应该用逗号分隔它吗？

为了让事情变得有趣，我有其他参数，它们接受单个值..所以我不确定是否应该将所有参数数据放在双引号中..

eg. myapp.exe files=aaa.txt,bbb.txt formatting=true foo=hello

或者

eg myapp.exe files="aaa.txt","bbb.txt" formatting="true" foo="hello"

原文

i've got a pretty simple .NET console application. I wish to pass in some command line arguments ... nothing tough there.

Now, the kicker is, i usually have a LIST of items i wish to pass in, for one argument. In this case, a list of files. I'm not sure of the best way to do this.

eg. myapp.exe files=aaa.txt,bbb.txt,ccc.txt

but what about file names with spaces? Should i comma delimate it?

to make things interesting, i've got other args, which accept single values .. so i'm not sure if i should put all the arg data in double quotes..

eg. myapp.exe files=aaa.txt,bbb.txt formatting=true foo=hello

eg myapp.exe files="aaa.txt","bbb.txt" formatting="true" foo="hello"

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

不及他 2024-08-02 22:00:18

对于您正在开发的程序来说可能有点晚了，但是您可以使用 Mono.Options

static void Main(string[] args)
{
    //let's try this mono.options thing shall we?
    string xapfile = null;
    var files = new List<string>();
    var p = new Mono.Options.OptionSet()
    {
        { "xap=", v => xapfile =v },
        { "file:", v => files.Add(v)}
    };
    List<string> extra = p.Parse(args);


    Console.WriteLine(@"
    xap = '{0}', 
    file(s)= {1}", 
         xapfile, 
         string.Join(",", files.ToArray())
    );
    Console.WriteLine(@"
    left overs: {0}", 
        string.Join(Environment.NewLine, extra.ToArray())
    );
    // rest of your Main here

调用语法为

myapp.exe --file="file 1" --file="file 2" --file="file 3"

你甚至可以放弃整个 files 变量而只处理剩下的作为文件，并将其命名为 myapp.exe "file 1" "file 2" "file 3" 并将文件从额外列表中取出。

Mono.Options 非常强大。你甚至不必使用 --，看来你可以使用 /file= 或 -file=

不幸的是，Windows 不会为你做通配符（除非你将选项传递给 powershell cmdlet），因此您必须自己执行此操作。不过这很简单，这里一些代码我过去曾使用过扩展 *.foo 或 "c:\temp\*.bar|c:\temp\*.txt"

另外，除非你将文件名包装在有声电影中，我可能不会用逗号分隔列表，因为逗号在文件名中是有效的。你知道有一天它会咬你:-) Pipe 是一个不错的选择，但前提是你将整个表达式包裹在有声电影中，否则 Windows 会认为你正在做 Pipeline。啊，命令行处理的乐趣:-)

Probably a bit late for the program you are working on, but here's how you could do it using Mono.Options

static void Main(string[] args)
{
    //let's try this mono.options thing shall we?
    string xapfile = null;
    var files = new List<string>();
    var p = new Mono.Options.OptionSet()
    {
        { "xap=", v => xapfile =v },
        { "file:", v => files.Add(v)}
    };
    List<string> extra = p.Parse(args);


    Console.WriteLine(@"
    xap = '{0}', 
    file(s)= {1}", 
         xapfile, 
         string.Join(",", files.ToArray())
    );
    Console.WriteLine(@"
    left overs: {0}", 
        string.Join(Environment.NewLine, extra.ToArray())
    );
    // rest of your Main here

Calling syntax would be

myapp.exe --file="file 1" --file="file 2" --file="file 3"

You could even forgo the whole files variable and just treat the left overs as the files, and call it like myapp.exe "file 1" "file 2" "file 3" and pluck the files out of the extra list.

Mono.Options is pretty powerful. You don't even have to use --, it appears you can use /file= or -file=

Unfortunately, Windows doesn't do globbing for you (unless you're passing the options to a powershell cmdlet), so you have to do it yourself. It's pretty easy though, here's a bit of code I've used in the past to expand *.foo or "c:\temp\*.bar|c:\temp\*.txt"

Also, unless you wrap the filenames in talkies, I probably wouldn't comma separate the list, as comma is valid in a filename. You know that one is going to bite you one day :-) Pipe makes a good choice, but only if you wrap the whole expression in talkies, otherwise windows thinks you're doing piping. Ah, the joys of command line processing :-)

回复收藏 0 原文