blocks|key|2052633|text|在纯Java中没有可靠的方法可以做到这一点。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2052634|如果您将目录路径与文件路径分开构建，则File(String+parent,+String+child)构造函数可以提供帮助，从而使交换更容易。|offset|length|style|CODE|2052635|另一种方法是设置一个脚本从不同的目录运行Java，或者使用JNI本机代码as+suggested+below。|2052636|The+relevant+Sun+bug在2008年被关闭，因为“不会修复”。|2052637|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/8/docs/api/java/io/File.html#File-java.lang.String-java.lang.String-|1|/questions/840190#8204584|2|http://bugs.sun.com/bugdatabase/view_bug.do?bug_id=4045688^0|0|J|X|J|X|0|0|10|I|1|0|0|K|2|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Y|8|@$D|Z|E|10|F|G]]|9|@$D|11|E|12|1|13]]|A|$]]|$1|H|3|I|5|6|7|14|8|@]|9|@$D|15|E|16|1|17]]|A|$]]|$1|J|3|K|5|6|7|18|8|@]|9|@$D|19|E|1A|1|1B]]|A|$]]|$1|L|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]]|M|$N|$5|O|P|Q|A|$R|S]]|T|$5|O|P|Q|A|$R|U]]|V|$5|O|P|Q|A|$R|W]]]]

There is no reliable way to do this in pure Java. Setting the <code>user.dir</code> property via <code>System.setProperty()</code> or <code>java -Duser.dir=...</code> does seem to affect subsequent creations of <code>Files</code>, but not e.g. <code>FileOutputStreams</code>.

The <a href="https://docs.oracle.com/javase/8/docs/api/java/io/File.html#File-java.lang.String-java.lang.String-" rel="noreferrer"><code>File(String parent, String child)</code></a> constructor can help if you build up your directory path separately from your file path, allowing easier swapping.

An alternative is to set up a script to run Java from a different directory, or use JNI native code <a href="/questions/840190#8204584">as suggested below</a>.

<a href="http://bugs.sun.com/bugdatabase/view_bug.do?bug_id=4045688" rel="noreferrer">The relevant Sun bug</a> was closed in 2008 as "will not fix".

blocks|key|1838637|text|如果您使用ProcessBuilder运行您的遗留程序，您将能够指定它的工作directory。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1838638|entityMap|0|LINK|mutability|MUTABLE|url|http://java.sun.com/javase/6/docs/api/java/lang/ProcessBuilder.html|1|http://java.sun.com/javase/6/docs/api/java/lang/ProcessBuilder.html#directory^0|5|E|0|12|9|1|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@$A|O|B|P|1|Q]|$A|R|B|S|1|T]]|C|$]]|$1|D|3|-4|5|6|7|U|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]|L|$5|G|H|I|C|$J|M]]]]

If you run your legacy program with <a href="http://java.sun.com/javase/6/docs/api/java/lang/ProcessBuilder.html" rel="noreferrer" title="ProcessBuilder">ProcessBuilder</a>, you will be able to specify its working <a href="http://java.sun.com/javase/6/docs/api/java/lang/ProcessBuilder.html#directory" rel="noreferrer">directory</a>.

blocks|key|2052907|text|可以更改PWD，使用JNA/JNI调用libc。JRuby人员有一个方便的java库，用于调用POSIX，名为jna-posix，这里是the+maven+info|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2052908|您可以看到它使用here的示例(Clojure代码，对不起)。查看函数chdirToRoot|2052909|entityMap|0|LINK|mutability|MUTABLE|url|http://kenai.com/projects/jna-posix|1|http://mvnrepository.com/artifact/org.jruby.ext.posix/jna-posix|2|https://github.com/arohner/lein-daemon/blob/master/daemon-runtime/src/leiningen/daemon/runtime.clj^0|1J|9|0|1W|E|1|0|8|4|2|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@$A|S|B|T|1|U]|$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|F|3|-4|5|6|7|12|8|@]|9|@]|C|$]]]|G|$H|$5|I|J|K|C|$L|M]]|N|$5|I|J|K|C|$L|O]]|P|$5|I|J|K|C|$L|Q]]]]

It is possible to change the PWD, using JNA/JNI to make calls to libc. The JRuby guys have a handy java library for making POSIX calls called <a href="http://kenai.com/projects/jna-posix" rel="noreferrer">jna-posix</a> Here's <a href="http://mvnrepository.com/artifact/org.jruby.ext.posix/jna-posix" rel="noreferrer">the maven info</a>

You can see an example of its use <a href="https://github.com/arohner/lein-daemon/blob/master/daemon-runtime/src/leiningen/daemon/runtime.clj" rel="noreferrer">here</a> (Clojure code, sorry). Look at the function chdirToRoot

blocks|key|1839090|text|如前所述，您不能更改JVM的CWD，但是如果要使用Runtime.exec()启动另一个进程，则可以使用重载方法来指定工作目录。这并不是真正用于在另一个目录中运行Java程序，但在许多情况下，当需要启动另一个程序(例如Perl脚本)时，您可以指定该脚本的工作目录，同时保持JVM的工作目录不变。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1839091|请参阅Runtime.exec+javadocs|offset|length|1839092|具体来说，|1839093|public+Process+exec(String[]+cmdarray,String[]+envp,+File+dir)+throws+IOException|code-block|syntax|javascript|1839094|其中dir是运行子进程的工作目录|style|CODE|1839095|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/6/docs/api/java/lang/Runtime.html#exec%2528java.lang.String%255B%255D,%2520java.lang.String%255B%255D,%2520java.io.File%2529^0|0|3|C|0|0|0|0|2|3|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@$D|10|E|11|1|12]]|A|$]]|$1|F|3|G|5|6|7|13|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|14|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|15|8|@$D|16|E|17|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

As mentioned you can't change the CWD of the JVM but if you were to launch another process using Runtime.exec() you can use the overloaded method that lets you specify the working directory. This is not really for running your Java program in another directory but for many cases when one needs to launch another program like a Perl script for example, you can specify the working directory of that script while leaving the working dir of the JVM unchanged.

See <a href="http://docs.oracle.com/javase/6/docs/api/java/lang/Runtime.html#exec%28java.lang.String%5B%5D,%20java.lang.String%5B%5D,%20java.io.File%29">Runtime.exec</a> javadocs

Specifically,

<pre><code>public Process exec(String[] cmdarray,String[] envp, File dir) throws IOException
</code></pre>

where <code>dir</code> is the working directory to run the subprocess in

blocks|key|2052245|text|如果我理解正确的话，Java程序是从当前环境变量的副本开始的。通过System.setProperty(String,+String)进行的任何更改都会修改副本，而不是原始环境变量。这并不是说这提供了Sun为什么选择这种行为的完整原因，但也许它提供了一点启示……|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2052246|entityMap^0|X|Y|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|$]]

If I understand correctly, a Java program starts with a copy of the current environment variables. Any changes via <code>System.setProperty(String, String)</code> are modifying the copy, not the original environment variables. Not that this provides a thorough reason as to why Sun chose this behavior, but perhaps it sheds a little light...

blocks|key|1838153|text|工作目录是操作系统的一个特性(在进程启动时设置)。为什么不直接传递自己的系统属性(-Dsomeprop=/my/path)，并在代码中将其用作文件的父对象：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1838154|File+f+=+new+File+(+System.getProperty("someprop"),+myFilename)|code-block|syntax|javascript|1838155|entityMap^0|15|J|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

The working directory is a operating system feature (set when the process starts). 
Why don't you just pass your own System property (<code>-Dsomeprop=/my/path</code>) and use that in your code as the parent of your File:

<pre><code>File f = new File ( System.getProperty("someprop"), myFilename)
</code></pre>

blocks|key|1838583|text|更聪明/更简单的做法是修改代码，而不是假设文件存在于当前工作目录中(我假设您正在执行类似new+File("blah.txt")的操作)，而是自己构建文件的路径。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1838584|让用户传入基目录，从配置文件中读取它，如果找不到其他属性，则回退到user.dir，等等。但是改进程序中的逻辑要比更改环境变量的工作方式容易得多。|1838585|entityMap^0|18|K|0|X|8|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|P|8|@]|D|@]|E|$]]]|I|$]]

The smarter/easier thing to do here is to just change your code so that instead of opening the file assuming that it exists in the current working directory (I assume you are doing something like <code>new File("blah.txt")</code>, just build the path to the file yourself.

Let the user pass in the base directory, read it from a config file, fall back to <code>user.dir</code> if the other properties can't be found, etc. But it's a whole lot easier to improve the logic in your program than it is to change how environment variables work.

blocks|key|2052848|text|此问题的其他可能答案可能取决于您打开文件的原因。这是一个属性文件，还是一个具有与应用程序相关的配置的文件？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2052849|如果是这种情况，您可以考虑尝试通过类路径加载器加载文件，这样就可以加载Java可以访问的任何文件。|2052850|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

The other possible answer to this question may depend on the reason you are opening the file. Is this a property file or a file that has some configuration related to your application?

If this is the case you may consider trying to load the file through the classpath loader, this way you can load any file Java has access to.

blocks|key|2053065|text|如果你在外壳程序中运行你的命令，你可以写类似"java+-cp“的代码，然后添加任何你想要的目录，用":”分隔。如果java在一个目录中找不到东西，它会尝试在其他目录中找到它们，这就是我所做的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2053066|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|$]]

If you run your commands in a shell you can write something like "java -cp" and add any directories you want separated by ":" if java doesnt find something in one directory it will go try and find them in the other directories, that is what I do.

blocks|key|2053248|text|使用FileSystemView|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2053249|private+FileSystemView+fileSystemView;
fileSystemView+=+FileSystemView.getFileSystemView();
currentDirectory+=+new+File(".");
//listing+currentDirectory
File[]+filesAndDirs+=+fileSystemView.getFiles(currentDirectory,+false);
fileList+=+new+ArrayList<File>();
dirList+=+new+ArrayList<File>();
for+(File+file+:+filesAndDirs)+{
if+(file.isDirectory())
++++dirList.add(file);
else
++++fileList.add(file);
}
Collections.sort(dirList);
if+(!fileSystemView.isFileSystemRoot(currentDirectory))
++++dirList.add(0,+new+File(".."));
Collections.sort(fileList);
//change
currentDirectory+=+fileSystemView.getParentDirectory(currentDirectory);|code-block|syntax|javascript|2053250|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Use FileSystemView 

<pre><code>private FileSystemView fileSystemView;
fileSystemView = FileSystemView.getFileSystemView();
currentDirectory = new File(".");
//listing currentDirectory
File[] filesAndDirs = fileSystemView.getFiles(currentDirectory, false);
fileList = new ArrayList&lt;File&gt;();
dirList = new ArrayList&lt;File&gt;();
for (File file : filesAndDirs) {
if (file.isDirectory())
 dirList.add(file);
else
 fileList.add(file);
}
Collections.sort(dirList);
if (!fileSystemView.isFileSystemRoot(currentDirectory))
 dirList.add(0, new File(".."));
Collections.sort(fileList);
//change
currentDirectory = fileSystemView.getParentDirectory(currentDirectory);
</code></pre>

How can I change the current working directory from within a Java program? Everything I've been able to find about the issue claims that you simply can't do it, but I can't believe that that's really the case.

I have a piece of code that opens a file using a hard-coded relative file path from the directory it's normally started in, and I just want to be able to use that code from within a different Java program without having to start it from within a particular directory. It seems like you should just be able to call <code>System.setProperty( "user.dir", "/path/to/dir" )</code>, but as far as I can figure out, calling that line just silently fails and does nothing.

I would understand if Java didn't allow you to do this, if it weren't for the fact that it allows you to get the current working directory, and even allows you to open files using relative file paths....

Changing the current working directory in Java?

Java

Perl

如何在Java程序中更改当前工作目录？我所能找到的关于这个问题的所有东西都声称你根本不能这样做，但我不敢相信这是真的。我有一段代码，它使用通常启动文件的目录中的硬编码相对文件路径打开文件，我只希望能够在不同的Java程序中使用该代码，而不必从特定目录中启动它。看起来您应该能够调用System.setProperty( "user.dir", "/path/to/dir" )，但是据我所知，调用这条

问是否在Java中更改当前工作目录？
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