如何从jar文件读取文件？
我有一个JAR文件中的文件。 例如，它是1.txt 。
我怎样才能访问它？ 我的源代码是：
Double result=0.0; File file = new File("1.txt")); //how get this file from a jar file BufferedReader input = new BufferedReader(new FileReader(file)); String line; while ((line = input.readLine()) != null) { if(me==Integer.parseInt(line.split(":")[0])){ result= parseDouble(line.split(":")[1]); } } input.close(); return result;
你不能使用File，因为这个文件在文件系统上不是独立存在的。 相反，你需要getResourceAsStream（），如下所示：
... InputStream in = getClass().getResourceAsStream("/1.txt"); BufferedReader input = new BufferedReader(new InputStreamReader(in)); ...
如果你的jar包在类path中：
InputStream is = YourClass.class.getResourceAsStream("1.txt");
如果它不在类path上，那么可以通过以下方式访问它：
URL url = new URL("jar:file:/absolute/location/of/yourJar.jar!/1.txt"); InputStream is = url.openStream();
类似于这个答案是你需要的东西。
您需要以特殊方式将文件从存档中取出。
BufferedReader input = new BufferedReader(new InputStreamReader( this.getClass().getClassLoader().getResourceAsStream("1.txt")));
Jar文件是一个zip文件…..
所以要阅读一个jar文件，试试
ZipFile file = new ZipFile("whatever.jar"); if (file != null) { ZipEntries entries = file.entries(); //get entries from the zip file... if (entries != null) { while (entries.hasMoreElements()) { ZipEntry entry = entries.nextElement(); //use the entry to see if it's the file '1.txt' //Read from the byte using file.getInputStream(entry) } } }
希望这可以帮助。
这工作对我来说，复制一个txt文件从jar文件到另一个txt文件
public static void copyTextMethod() throws Exception{ String inputPath = "path/to/.jar"; String outputPath = "Desktop/CopyText.txt"; File resStreamOut = new File(outputPath); int readBytes; JarFile file = new JarFile(inputPath); FileWriter fw = new FileWriter(resStreamOut); try{ Enumeration<JarEntry> entries = file.entries(); while (entries.hasMoreElements()){ JarEntry entry = entries.nextElement(); if (entry.getName().equals("readMe/tempReadme.txt")) { System.out.println(entry +" : Entry"); InputStream is = file.getInputStream(entry); BufferedWriter output = new BufferedWriter(fw); while ((readBytes = is.read()) != -1) { output.write((char) readBytes); } System.out.println(outputPath); output.close(); } } } catch(Exception er){ er.printStackTrace(); } } }
private String loadResourceFileIntoString(String path) { //path = "/resources/html/custom.css" for example BufferedReader buffer = new BufferedReader(new InputStreamReader(getClass().getResourceAsStream(path))); return buffer.lines().collect(Collectors.joining(System.getProperty("line.separator"))); }
之前我已经遇到过几次同样的问题。 我希望在JDK 7中有人会写一个类path文件系统，但还没有。
Spring有Resource类，它可以很好地加载classpath资源。
答案是非常好的，但我想我可以添加到讨论与显示一个例子，使用类path资源的文件和目录。
我写了一个小原型来解决这个问题。 原型不处理每个边界情况，但它处理在jar文件中的目录中查找资源。
我已经使用堆栈溢出了相当长的一段时间。 这是我第一次记得回答一个问题，所以如果我长时间地原谅我（这是我的本性）。
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(); } } } } }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(); } } } } }
你可以在这里看到一个更完整的例子。
InputStream inputStreamLastName = RandomDataGeneratorUtils.class.getClassLoader().getResourceAsStream(Constants.LAST_NAMES_FILE + ".txt"); try { br = new BufferedReader(new InputStreamReader(inputStreamLastName, "UTF-8")); String sCurrentLine; ArrayList<String> lastNamesList = new ArrayList<String>(); while ((sCurrentLine = br.readLine()) != null) { if(sCurrentLine.length()>=min && sCurrentLine.length()<=max){ lastNamesList.add(sCurrentLine); } }
