blocks|key|2537213|text|我已经在Windows+XP、Vista、Win7、MacOS10.3-+10.6和各种Linux发行版上使用了您的方案，没有出现任何问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2537214|++++if+(System.getProperty("os.name").startsWith("Windows"))+{
++++++++//+includes:+Windows+2000,++Windows+95,+Windows+98,+Windows+NT,+Windows+Vista,+Windows+XP
++++}+else+{
++++++++//+everything+else
++++}+|code-block|syntax|javascript|2537215|本质上，检测类Unix是通过而不是检测Windows。|offset|length|style|BOLD|2537216|entityMap^0|0|0|E|3|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I've used your scheme in production code on Windows XP, Vista, Win7, Mac OS 10.3 - 10.6 and a variety of Linux distros without an issue: 

<pre><code> if (System.getProperty("os.name").startsWith("Windows")) {
 // includes: Windows 2000, Windows 95, Windows 98, Windows NT, Windows Vista, Windows XP
 } else {
 // everything else
 } 
</code></pre>

Essentially, detect Unix-like by not detecting Windows.

blocks|key|52290|text|Javadoc说:在UNIX系统上，这个*字段的值是'/'；在Microsoft+Windows系统上，它是'\'。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|52291|System.out.println(+File.separatorChar+==+'/'+?+"Unix"+:+"Windows"+);|code-block|syntax|javascript|52292|entityMap^0|Q|3|1I|3|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Javadoc says: On UNIX systems the value of this
 * field is <code>'/'</code>; on Microsoft Windows systems it is <code>'\'</code>.

<pre><code>System.out.println( File.separatorChar == '/' ? "Unix" : "Windows" );
</code></pre>

blocks|key|2537189|text|System.getProperty("os.name");大概是你能得到的最好的了。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2537190|entityMap^0|0|U|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

<code>System.getProperty("os.name");</code> is about the best you are going to get.

blocks|key|2747093|text|我同意@Fuzzy的观点，因为我认为Java让您能够获得这些信息的唯一方法是通过os.name属性。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2747094|我唯一能想到的其他事情是：|2747095|Java有一个外壳脚本或批处理文件包装器来启动应用程序，该应用程序使用-D参数将操作系统信息传递给|2747096|。虽然给出了你的描述，但这听起来并不可行。|ordered-list-item|2747097|你可以尝试检查特定于操作系统的目录是否存在。例如，您可以假设目录"/"将始终存在于类Unix系统上，但不存在于Windows上，并执行以下操作：|2747098|2747099|if((new+File("/")).exists())+{+System.out.println("I'm+on+a+Unix+system!")；}|2747100|尝试启动特定于Unix的命令行命令，如ls，并检查返回代码。如果它能工作，你就是在一个类Unix的系统上，如果不能工作，你就在Windows上。|unordered-list-item|2747101|2747102|不过，所有这些解决方案都只是一些技巧，坦率地说，我对它们中的任何一个都不是很满意。不幸的是，你最好还是带着你原来的想法。很有趣，是吧？|2747103|entityMap^0|14|7|0|0|Z|2|0|0|W|3|0|0|0|J|2|0|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|12|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|J|3|K|5|L|7|16|8|@]|D|@]|E|$]]|$1|M|3|N|5|L|7|17|8|@$9|18|A|19|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]|$1|P|3|Q|5|6|7|1B|8|@]|D|@]|E|$]]|$1|R|3|S|5|T|7|1C|8|@$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|U|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]|$1|V|3|W|5|6|7|1G|8|@]|D|@]|E|$]]|$1|X|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|Y|$]]

I agree with @Fuzzy in that I think the only way that Java intended you to be able to get that information was through the <code>os.name</code> property.

The only other things I can think of are:

<ol>
<li>Have a shell script or batch file wrapper to launch your Java app that passes in OS information using the <code>-D</code> argument to the JVM. Though given your description, this doesn't sound doable.</li>
<li>You could try to check for the existence of an OS-specific directory. For instance, you could assume the directory <code>"/"</code> will always exist on a Unix-like system, but not on Windows and do something like this:

if((new File("/")).exists())
{
 System.out.println("I'm on a Unix system!");
}</li>
<li>Try to kick off a Unix-specific command line command like <code>ls</code> and check the return code. If it worked, you're on a Unix-like system, if not you're on Windows.</li>
</ol>

All of those solutions are really just hacks though and frankly I don't really feel all that great about any of them. You're unfortunately probably best off with your original thought. Fun, eh?

blocks|key|52810|text|使用File.pathSeparator或File.separator。第一个将在Windows中返回";“，在Unix中返回":”。第二个将在Windows中返回"\“，在Unix中返回"/”。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|52811|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Use File.pathSeparator or File.separator. The first will return ";" in Windows and ":" in Unix. The second will return "\" in Windows and "/" in Unix.

blocks|key|2747070|text|您可以尝试执行uname命令-应该在所有unixoid系统上都可用。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2747071|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Uname^0|7|5|7|5|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@$9|Q|A|R|1|S]]|E|$]]|$1|F|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

You could try to execute the <a href="http://en.wikipedia.org/wiki/Uname" rel="nofollow noreferrer"><code>uname</code></a> command - should be available on all unixoid systems.

blocks|key|52813|text|package+com.appspot.x19290;

import+java.io.File;
import+java.io.FileInputStream;
import+java.io.IOException;

public+class+UnixCheck+{
++++public+static+void+main(String[]+args)+{
++++++++UnixCheck+s+=+UnixCheck.S;
++++++++String+isUnix+=+s.unix+?+"is+Unix"+:+"not+Unix";
++++++++try+{
++++++++++++System.out.println(isUnix+%2B+",+devnull:+"+%2B+s.devnull.getPath());
++++++++}+catch+(NullPointerException+e)+{
++++++++++++System.out.println(isUnix+%2B+",+devnull:+unknown");
++++++++}
++++}

++++public+static+final+UnixCheck+S+=+new+UnixCheck();
++++public+static+final+UnixCheck+TEST+=+new+UnixCheck(true);

++++public+final+boolean+unix;
++++public+final+File+devnull;

++++private+UnixCheck()+{
++++++++this(false);
++++}

++++private+UnixCheck(boolean+testing)+{
++++++++String+path;
++++++++path+=+testing+?+"/<dev>/<no><such><null><device>"+:+"/dev/null";
++++++++File+devnull+=+devnullOrNone(path);
++++++++if+(devnull+==+null)+{
++++++++++++this.unix+=+false;
++++++++++++path+=+testing+?+"<no><such><null><device>"+:+"nul";
++++++++++++this.devnull+=+devnullOrNone(path);
++++++++}+else+{
++++++++++++this.unix+=+true;
++++++++++++this.devnull+=+devnull;
++++++++}
++++}

++++private+static+File+devnullOrNone(String+name)+{
++++++++File+file+=+new+File(name);
++++++++if+(file.isFile())
++++++++++++return+null;
++++++++if+(file.isDirectory())
++++++++++++return+null;
++++++++try+{
++++++++++++FileInputStream+i+=+new+FileInputStream(file);
++++++++++++try+{
++++++++++++++++i.read();
++++++++++++}+finally+{
++++++++++++++++i.close();
++++++++++++}
++++++++}+catch+(IOException+e)+{
++++++++++++return+null;
++++++++}
++++++++return+file;
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|52814|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>package com.appspot.x19290;

import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;

public class UnixCheck {
 public static void main(String[] args) {
 UnixCheck s = UnixCheck.S;
 String isUnix = s.unix ? "is Unix" : "not Unix";
 try {
 System.out.println(isUnix + ", devnull: " + s.devnull.getPath());
 } catch (NullPointerException e) {
 System.out.println(isUnix + ", devnull: unknown");
 }
 }

 public static final UnixCheck S = new UnixCheck();
 public static final UnixCheck TEST = new UnixCheck(true);

 public final boolean unix;
 public final File devnull;

 private UnixCheck() {
 this(false);
 }

 private UnixCheck(boolean testing) {
 String path;
 path = testing ? "/&lt;dev&gt;/&lt;no&gt;&lt;such&gt;&lt;null&gt;&lt;device&gt;" : "/dev/null";
 File devnull = devnullOrNone(path);
 if (devnull == null) {
 this.unix = false;
 path = testing ? "&lt;no&gt;&lt;such&gt;&lt;null&gt;&lt;device&gt;" : "nul";
 this.devnull = devnullOrNone(path);
 } else {
 this.unix = true;
 this.devnull = devnull;
 }
 }

 private static File devnullOrNone(String name) {
 File file = new File(name);
 if (file.isFile())
 return null;
 if (file.isDirectory())
 return null;
 try {
 FileInputStream i = new FileInputStream(file);
 try {
 i.read();
 } finally {
 i.close();
 }
 } catch (IOException e) {
 return null;
 }
 return file;
 }
}
</code></pre>

Ok, I know that <code>System.getProperty("os.name")</code> will give me the name of the OS I'm running under, but that's not a lot of help. What I need to know is if the OS I'm running on is a 'Unix-like' OS, I don't care if it's HP-UX, AIX, Mac OS X or whatever.

From the <a href="http://lopica.sourceforge.net/os.html" rel="noreferrer">list of possible os.name values</a> it seems like a quick and dirty way of detecting a 'Unix-like' OS is checking if <code>os.name</code> does not contain "Windows". The false positives that will give me are OSes my code is very unlikely to encounter! Still, I'd love to know a better way if there is one.

How can I detect a Unix-like OS in Java?

好吧，我知道System.getProperty("os.name")会给我提供我正在运行的操作系统的名称，但这不会有太多帮助。我需要知道的是，如果我运行的操作系统是“类Unix”操作系统，我并不关心它是HP-UX、AIX、Mac还是其他什么。从上看，检测“类Unix”操作系统的一种快捷方法是检查os.name是否包含“”而不是“”。给我的误报是OSes，我的代码不太可能遇到！尽管如此，如果有更好

问如何在Java中检测类Unix操作系统？
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