blocks|key|3379595|text|你没有说这是一个桌面应用程序还是web应用程序。如果是桌面应用程序，我会从适当的ClassLoader使用getResourceAsStream()方法；如果是web应用程序，我会使用上下文方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3379596|entityMap^0|1H|L|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|$]]

You don't say if this is a desktop or web app. I would use the <code>getResourceAsStream()</code> method from an appropriate ClassLoader if it's a desktop or the Context if it's a web app.

blocks|key|943522|text|在您的技术之外，为什么不使用标准的Java+JarFile+class来获得您想要的引用呢？从那时起，你的大部分问题应该都会消失。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|943523|entityMap|0|LINK|mutability|MUTABLE|url|http://java.sun.com/javase/6/docs/api/java/util/jar/JarFile.html^0|M|D|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

Outside of your technique, why not use the standard Java <a href="http://java.sun.com/javase/6/docs/api/java/util/jar/JarFile.html" rel="nofollow noreferrer">JarFile class</a> to get the references you want? From there most of your problems should go away.

blocks|key|3379673|text|看起来您想要使用java.lang.Class.getResourceAsStream(String)，请参见|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3379674|https://docs.oracle.com/javase/8/docs/api/java/lang/Class.html#getResourceAsStream-java.lang.String-|3379675|entityMap|0|LINK|mutability|MUTABLE|url^0|8|17|0|0|2S|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|R|8|@]|D|@$9|S|A|T|1|U]]|E|$]]|$1|H|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|G]]]]

Looks like you want to use <code>java.lang.Class.getResourceAsStream(String)</code>, see
<a href="https://docs.oracle.com/javase/8/docs/api/java/lang/Class.html#getResourceAsStream-java.lang.String-" rel="nofollow noreferrer">https://docs.oracle.com/javase/8/docs/api/java/lang/Class.html#getResourceAsStream-java.lang.String-</a>

blocks|key|163096|text|看起来您似乎正在使用URL.toString结果作为FileReader构造函数的参数。URL.toString有点问题，通常你应该使用url.toURI().toString()。在任何情况下，该字符串都不是文件路径。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|163097|相反，您应该：|163098|163099|将URL传递给ServicesLoader，并让它调用openStream或类似的方法。|unordered-list-item|163100|使用Class.getResourceAsStream，只是传递流，可能是在InputSource中。(记住检查null值，因为API有点混乱。)|163101|163102|entityMap^0|A|C|Q|A|18|C|1W|M|0|0|0|1|3|7|E|R|A|0|2|P|12|B|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|10|8|@]|D|@]|E|$]]|$1|I|3|J|5|K|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]|$9|16|A|17|B|C]]|D|@]|E|$]]|$1|L|3|M|5|K|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|N|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]]|P|$]]

It looks as if you are using the <code>URL.toString</code> result as the argument to the <code>FileReader</code> constructor. <code>URL.toString</code> is a bit broken, and instead you should generally use <code>url.toURI().toString()</code>. In any case, the string is not a file path.

Instead, you should either:

<ul>
<li>Pass the <code>URL</code> to <code>ServicesLoader</code> and let it call <code>openStream</code> or similar.</li>
<li>Use <code>Class.getResourceAsStream</code> and just pass the stream over, possibly inside an <code>InputSource</code>. (Remember to check for nulls as the API is a bit messy.)</li>
</ul>

blocks|key|3379754|text|问题是我在调用XMLReader的解析方法上走得太远了。parse方法接受InputSource，因此甚至没有理由使用FileReader。将上面代码的最后一行改为|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3379755|xr.parse(+new+InputSource(+filename+));|code-block|syntax|javascript|3379756|工作得很好。|3379757|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

The problem was that I was going a step too far in calling the parse method of XMLReader. The parse method accepts an InputSource, so there was no reason to even use a FileReader. Changing the last line of the code above to

<pre><code>xr.parse( new InputSource( filename ));
</code></pre>

works just fine.

blocks|key|4226281|text|如果您广泛使用资源，则可以考虑使用Commons+VFS。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4226282|还支持：*本地文件*+FTP、SFTP+*+HTTP和HTTPS+*临时文件“普通文件系统支持”*+Zip、Jar和Tar+(未压缩、tgz或tbz2)+*+gzip和bzip2+*资源*内存-“内存驱动器”*+mime|4226283|还有JBoss+VFS+--但没有太多的文档记录。|4226284|entityMap|0|LINK|mutability|MUTABLE|url|http://commons.apache.org/vfs/filesystems.html|1|http://java.dzone.com/news/jboss-virtual-file-system^0|H|B|0|0|0|2|9|1|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@$A|S|B|T|1|U]]|C|$]]|$1|D|3|E|5|6|7|V|8|@]|9|@]|C|$]]|$1|F|3|G|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|H|3|-4|5|6|7|10|8|@]|9|@]|C|$]]]|I|$J|$5|K|L|M|C|$N|O]]|P|$5|K|L|M|C|$N|Q]]]]

If you use resources extensively, you might consider using 
 <a href="http://commons.apache.org/vfs/filesystems.html" rel="nofollow">Commons VFS</a>.

Also supports:
 * Local Files
 * FTP, SFTP
 * HTTP and HTTPS
 * Temporary Files "normal FS backed)
 * Zip, Jar and Tar (uncompressed, tgz or tbz2)
 * gzip and bzip2
 * resources
 * ram - "ramdrive"
 * mime

There's also <a href="http://java.dzone.com/news/jboss-virtual-file-system" rel="nofollow">JBoss VFS</a> - but it's not much documented.

blocks|key|943633|text|我有两个用于读取数据的CSV文件。java程序被导出为可运行的jar文件。当你导出它的时候，你会发现它不会和你的资源一起导出。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|943634|我在eclipse的项目下添加了一个名为data的文件夹。在该文件夹中，我存储了我的csv文件。|943635|当我需要引用这些文件时，我会这样做...|943636|private+static+final+String+ZIP_FILE_LOCATION_PRIMARY+=+"free-zipcode-database-Primary.csv";
private+static+final+String+ZIP_FILE_LOCATION+=+"free-zipcode-database.csv";

private+static+String+getFileLocation(){
++++String+loc+=+new+File("").getAbsolutePath()+%2B+File.separatorChar+%2B
++++++++"data"+%2B+File.separatorChar;
++++if+(usePrimaryZipCodesOnly()){++++++++++++++
++++++++loc+=+loc.concat(ZIP_FILE_LOCATION_PRIMARY);
++++}+else+{
++++++++loc+=+loc.concat(ZIP_FILE_LOCATION);
++++}
++++return+loc;
}|code-block|syntax|javascript|943637|然后，当您将jar放在一个位置以便可以通过命令行运行时，请确保将包含资源的数据文件夹添加到与jar文件相同的位置。|943638|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Q|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|R|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I have 2 CSV files that I use to read data. The java program is exported as a runnable jar file. When you export it, you will figure out it doesn't export your resources with it. 

I added a folder under project called data in eclipse. In that folder i stored my csv files.

When I need to reference those files I do it like this...

<pre><code>private static final String ZIP_FILE_LOCATION_PRIMARY = "free-zipcode-database-Primary.csv";
private static final String ZIP_FILE_LOCATION = "free-zipcode-database.csv";

private static String getFileLocation(){
 String loc = new File("").getAbsolutePath() + File.separatorChar +
 "data" + File.separatorChar;
 if (usePrimaryZipCodesOnly()){ 
 loc = loc.concat(ZIP_FILE_LOCATION_PRIMARY);
 } else {
 loc = loc.concat(ZIP_FILE_LOCATION);
 }
 return loc;
}
</code></pre>

Then when you put the jar in a location so it can be ran via commandline, make sure that you add the data folder with the resources into the same location as the jar file.

blocks|key|4226737|text|下面是关于如何正确读取jar文件(在本例中是当前正在执行的jar文件)中的文件的示例代码。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4226738|如果jar文件不是当前运行的文件，只需将可执行文件更改为jar文件的路径即可。|4226739|然后将filePath更改为您希望在jar文件中使用的文件的路径。例如，如果你的文件在|4226740|4226741|+someJar.jar\img\test.gif|blockquote|4226742|4226743|4226744|。将filePath设置为"img\test.gif“|4226745|File+executable+=+new+File(BrowserViewControl.class.getProtectionDomain().getCodeSource().getLocation().toURI());
JarFile+jar+=+new+JarFile(executable);
InputStream+fileInputStreamReader+=+jar.getInputStream(jar.getJarEntry(filePath));
byte[]+bytes+=+new+byte[fileInputStreamReader.available()];

int+sizeOrig+=+fileInputStreamReader.available();
int+size+=+fileInputStreamReader.available();
int+offset+=+0;
while+(size+!=+0){
++++fileInputStreamReader.read(bytes,+offset,+size);
++++offset+=+sizeOrig+-+fileInputStreamReader.available();
++++size+=+fileInputStreamReader.available();
}|code-block|syntax|javascript|4226746|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|X|8|@]|9|@]|A|$]]|$1|G|3|H|5|I|7|Y|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|10|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|11|8|@]|9|@]|A|$]]|$1|N|3|O|5|P|7|12|8|@]|9|@]|A|$Q|R]]|$1|S|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|T|$]]

Here's a sample code on how to read a file properly inside a jar file (in this case, the current executing jar file)

Just change executable with the path of your jar file if it is not the current running one.

Then change the filePath to the path of the file you want to use inside the jar file. I.E. if your file is in 

<blockquote>
 someJar.jar\img\test.gif
</blockquote>

. Set the filePath to "img\test.gif"

<pre><code>File executable = new File(BrowserViewControl.class.getProtectionDomain().getCodeSource().getLocation().toURI());
JarFile jar = new JarFile(executable);
InputStream fileInputStreamReader = jar.getInputStream(jar.getJarEntry(filePath));
byte[] bytes = new byte[fileInputStreamReader.available()];

int sizeOrig = fileInputStreamReader.available();
int size = fileInputStreamReader.available();
int offset = 0;
while (size != 0){
 fileInputStreamReader.read(bytes, offset, size);
 offset = sizeOrig - fileInputStreamReader.available();
 size = fileInputStreamReader.available();
}
</code></pre>

I'm trying to access an XML file within a jar file, from a separate jar that's running as a desktop application. I can get the URL to the file I need, but when I pass that to a FileReader (as a String) I get a FileNotFoundException saying "The file name, directory name, or volume label syntax is incorrect."

As a point of reference, I have no trouble reading image resources from the same jar, passing the URL to an ImageIcon constructor. This seems to indicate that the method I'm using to get the URL is correct.

<pre><code>URL url = getClass().getResource("/xxx/xxx/xxx/services.xml");
ServicesLoader jsl = new ServicesLoader( url.toString() );
</code></pre>

Inside the ServicesLoader class I have 

<pre><code>XMLReader xr = XMLReaderFactory.createXMLReader();
xr.setContentHandler( this );
xr.setErrorHandler( this );
xr.parse( new InputSource( new FileReader( filename )));
</code></pre>

What's wrong with using this technique to read the XML file?

How do I read a resource file from a Java jar file?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我试图从作为桌面应用程序运行的单独jar中访问jar文件中的XML文件。我可以获得所需文件的URL，但是当我将其作为字符串传递给FileReader时，会得到一个FileNotFoundException，提示“文件名、目录名或卷标语法不正确”。作为参考，我可以毫不费力地从同一个jar中读取图像资源，并将URL传递给I...

问如何从Java jar文件中读取资源文件？
EN

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问如何从Java jar文件中读取资源文件？EN