如何从JavaJAR文件中读取资源文件？
我试图从一个单独的jar（作为桌面应用程序运行）中访问jar文件中的XML文件。我可以获取所需文件的URL，但当我将其作为字符串传递给FileReader时，会得到一个FileNotFoundException，表示文件名、目录名或卷标语法不正确 作为参考，我可以从同一个jar读取图像资源，并将URL传递给ImageIcon构造函数。这似乎表明我用来获取URL的方法是正确的如何从JavaJAR文件中读取资源文件？,java,jar,resources,Java,Jar,Resources,我试图从一个单独的jar（作为桌面应用程序运行）中访问jar文件中的XML文件。我可以获取所需文件的URL，但当我将其作为字符串传递给FileReader时，会得到一个FileNotFoundException，表示文件名、目录名或卷标语法不正确 作为参考，我可以从同一个jar读取图像资源，并将URL传递给ImageIcon构造函数。这似乎表明我用来获取URL的方法是正确的 URL url = getClass().getResource("/xxx/xxx/xxx/services.xml")
URL url = getClass().getResource("/xxx/xxx/xxx/services.xml");
ServicesLoader jsl = new ServicesLoader( url.toString() );
XMLReader xr = XMLReaderFactory.createXMLReader();
xr.setContentHandler( this );
xr.setErrorHandler( this );
xr.parse( new InputSource( new FileReader( filename )));
使用此技术读取XML文件有什么问题？您不能说这是桌面应用还是web应用。如果是桌面，我会使用相应类加载器中的getResourceAsStream方法；如果是web应用，我会使用上下文方法。在您的技术之外，为什么不使用标准Java来获取您想要的引用？从那以后，您的大多数问题都应该消失。看起来您想使用java.lang.Class.getResourceAsStreamString，请参阅
看起来您正在使用URL.toString结果作为FileReader构造函数的参数。URL.toString有点不完整，通常应该使用URL.toURI.toString。在任何情况下，字符串都不是文件路径 相反，您应该： 将URL传递给ServicesLoader，并让它调用openStream或类似程序。 使用Class.getResourceAsStream并只传递流，可能在InputSource内部。记住检查空值，因为API有点混乱。
问题是我调用XMLReader的解析方法时走得太远了。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更改为
xr.parse( new InputSource( filename ));
工作很好。
< P>如果你广泛地使用资源，你可以考虑使用。 还支持： *本地文件 *FTP，SFTP *HTTP和HTTPS *备份的临时文件 *Zip、Jar和Tar未压缩、tgz或tbz2 *gzip和bzip2 *资源 *冲压驱动 *默剧还有-但没有太多的文档。我有两个用于读取数据的CSV文件。java程序作为可运行的jar文件导出。当你导出它时，你会发现它并没有导出你的资源 我在eclipse项目下添加了一个名为data的文件夹。在那个文件夹中，我存储了我的csv文件 当我需要引用这些文件时，我会这样做
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;
}
然后，当您将jar放在一个可以通过命令行运行的位置时，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。我想指出的一个问题是，如果相同的资源位于多个jar文件中会发生什么。 假设您想读取/org/node/foo.txt，但不是从一个文件中读取，而是从每个jar文件中读取 我以前曾多次遇到过同样的问题。 我希望在JDK7中有人会编写一个类路径文件系统，但遗憾的是现在还没有 Spring有一个资源类，它允许您很好地加载类路径资源 我编写了一个小原型来解决从多个jar文件中读取资源的问题。原型并不处理每一种边缘情况，但它可以处理在jar文件中的目录中查找资源的问题 我使用堆栈溢出已有相当一段时间了。这是我记得回答一个问题的第二个答案，所以如果我回答得太久，请原谅，这是我的天性 这是一个原型资源读取器。原型缺乏健壮的错误检查 我已经设置了两个原型jar文件
<pre>
<dependency>
<groupId>invoke</groupId>
<artifactId>invoke</artifactId>
<version>1.0-SNAPSHOT</version>
</dependency>
<dependency>
<groupId>node</groupId>
<artifactId>node</artifactId>
<version>1.0-SNAPSHOT</version>
</dependency>
package com.foo;
import java.io.File;
import java.io.FileReader;
import java.io.InputStreamReader;
import java.io.Reader;
import java.net.URI;
import java.net.URL;
import java.util.Enumeration;
import java.util.zip.ZipEntry;
import java.util.zip.ZipFile;
/**
* Prototype resource reader.
* This prototype is devoid of error checking.
*
*
* I have two prototype jar files that I have setup.
* <pre>
* <dependency>
* <groupId>invoke</groupId>
* <artifactId>invoke</artifactId>
* <version>1.0-SNAPSHOT</version>
* </dependency>
*
* <dependency>
* <groupId>node</groupId>
* <artifactId>node</artifactId>
* <version>1.0-SNAPSHOT</version>
* </dependency>
* </pre>
* The jar files each have a file under /org/node/ called resource.txt.
* <br />
* This is just a prototype of what a handler would look like with classpath://
* I also have a resource.foo.txt in my local resources for this project.
* <br />
*/
public class ClasspathReader {
public static void main(String[] args) throws Exception {
/* This project includes two jar files that each have a resource located
in /org/node/ called resource.txt.
*/
/*
Name space is just a device I am using to see if a file in a dir
starts with a name space. Think of namespace like a file extension
but it is the start of the file not the end.
*/
String namespace = "resource";
//someResource is classpath.
String someResource = args.length > 0 ? args[0] :
//"classpath:///org/node/resource.txt"; It works with files
"classpath:///org/node/"; //It also works with directories
URI someResourceURI = URI.create(someResource);
System.out.println("URI of resource = " + someResourceURI);
someResource = someResourceURI.getPath();
System.out.println("PATH of resource =" + someResource);
boolean isDir = !someResource.endsWith(".txt");
/** Classpath resource can never really start with a starting slash.
* Logically they do, but in reality you have to strip it.
* This is a known behavior of classpath resources.
* It works with a slash unless the resource is in a jar file.
* Bottom line, by stripping it, it always works.
*/
if (someResource.startsWith("/")) {
someResource = someResource.substring(1);
}
/* Use the ClassLoader to lookup all resources that have this name.
Look for all resources that match the location we are looking for. */
Enumeration resources = null;
/* Check the context classloader first. Always use this if available. */
try {
resources =
Thread.currentThread().getContextClassLoader().getResources(someResource);
} catch (Exception ex) {
ex.printStackTrace();
}
if (resources == null || !resources.hasMoreElements()) {
resources = ClasspathReader.class.getClassLoader().getResources(someResource);
}
//Now iterate over the URLs of the resources from the classpath
while (resources.hasMoreElements()) {
URL resource = resources.nextElement();
/* if the resource is a file, it just means that we can use normal mechanism
to scan the directory.
*/
if (resource.getProtocol().equals("file")) {
//if it is a file then we can handle it the normal way.
handleFile(resource, namespace);
continue;
}
System.out.println("Resource " + resource);
/*
Split up the string that looks like this:
jar:file:/Users/rick/.m2/repository/invoke/invoke/1.0-SNAPSHOT/invoke-1.0-SNAPSHOT.jar!/org/node/
into
this /Users/rick/.m2/repository/invoke/invoke/1.0-SNAPSHOT/invoke-1.0-SNAPSHOT.jar
and this
/org/node/
*/
String[] split = resource.toString().split(":");
String[] split2 = split[2].split("!");
String zipFileName = split2[0];
String sresource = split2[1];
System.out.printf("After split zip file name = %s," +
" \nresource in zip %s \n", zipFileName, sresource);
/* Open up the zip file. */
ZipFile zipFile = new ZipFile(zipFileName);
/* Iterate through the entries. */
Enumeration entries = zipFile.entries();
while (entries.hasMoreElements()) {
ZipEntry entry = entries.nextElement();
/* If it is a directory, then skip it. */
if (entry.isDirectory()) {
continue;
}
String entryName = entry.getName();
System.out.printf("zip entry name %s \n", entryName);
/* If it does not start with our someResource String
then it is not our resource so continue.
*/
if (!entryName.startsWith(someResource)) {
continue;
}
/* the fileName part from the entry name.
* where /foo/bar/foo/bee/bar.txt, bar.txt is the file
*/
String fileName = entryName.substring(entryName.lastIndexOf("/") + 1);
System.out.printf("fileName %s \n", fileName);
/* See if the file starts with our namespace and ends with our extension.
*/
if (fileName.startsWith(namespace) && fileName.endsWith(".txt")) {
/* If you found the file, print out
the contents fo the file to System.out.*/
try (Reader reader = new InputStreamReader(zipFile.getInputStream(entry))) {
StringBuilder builder = new StringBuilder();
int ch = 0;
while ((ch = reader.read()) != -1) {
builder.append((char) ch);
}
System.out.printf("zip fileName = %s\n\n####\n contents of file %s\n###\n", entryName, builder);
} catch (Exception ex) {
ex.printStackTrace();
}
}
//use the entry to see if it's the file '1.txt'
//Read from the byte using file.getInputStream(entry)
}
}
}
/**
* The file was on the file system not a zip file,
* this is here for completeness for this example.
* otherwise.
*
* @param resource
* @param namespace
* @throws Exception
*/
private static void handleFile(URL resource, String namespace) throws Exception {
System.out.println("Handle this resource as a file " + resource);
URI uri = resource.toURI();
File file = new File(uri.getPath());
if (file.isDirectory()) {
for (File childFile : file.listFiles()) {
if (childFile.isDirectory()) {
continue;
}
String fileName = childFile.getName();
if (fileName.startsWith(namespace) && fileName.endsWith("txt")) {
try (FileReader reader = new FileReader(childFile)) {
StringBuilder builder = new StringBuilder();
int ch = 0;
while ((ch = reader.read()) != -1) {
builder.append((char) ch);
}
System.out.printf("fileName = %s\n\n####\n contents of file %s\n###\n", childFile, builder);
} catch (Exception ex) {
ex.printStackTrace();
}
}
}
} else {
String fileName = file.getName();
if (fileName.startsWith(namespace) && fileName.endsWith("txt")) {
try (FileReader reader = new FileReader(file)) {
StringBuilder builder = new StringBuilder();
int ch = 0;
while ((ch = reader.read()) != -1) {
builder.append((char) ch);
}
System.out.printf("fileName = %s\n\n####\n contents of file %s\n###\n", fileName, builder);
} catch (Exception ex) {
ex.printStackTrace();
}
}
}
}
}
下面是一个示例代码，介绍如何正确读取jar文件中的文件（在本例中为当前正在执行的jar文件） 只要用jar文件的路径更改可执行文件，如果它不是当前正在运行的文件 然后将filePath更改为要在jar文件中使用的文件的路径。例如，如果您的文件在 someJar.jar\img\test.gif 。将文件路径设置为img\test.gif
@JVerstry您的回答对我很有帮助，遗憾的是，它被完全低估了。您可以发布工作代码吗？请确保在使用文件API时，使用unix文件路径分隔符。
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();
}