M’agradaria assenyalar que un dels problemes és què passa si els mateixos recursos es troben en diversos arxius jar.Digamos que vostè vol llegir /org/node/foo.txt però no d’un arxiu, però des de tots i cadascun dels arxius jar.
M’he topat amb aquest mateix problema diverses vegades abans. jo estava esperant al JDK 7 que algú escrigui una ruta de classes de sistema de fitxers, però, per desgràcia, encara no. a la primavera té la classe de Recurs que li permet carregar classpath de recursos bastant bé.
Vaig escriure un petit prototip per resoldre aquest gran problema de la lectura dels recursos de la forma de múltiples arxius jar. El prototip no manejar cada cas extrem, però sí manejar a la recerca de recursos en els directoris que es troben en els arxius jar.
He utilitzat Stack Overflow durant força temps. Aquesta és la segona resposta que jo recordi contestar una pregunta així que em perdoni si vaig massa temps (és la meva naturalesa).
Aquest és un prototip dels recursos de l’lector. El prototip està desproveït d’un robust sistema de comprovació d’errors.
Tinc dos prototips d’arxius jar que tinc el programa d’instal·lació.
<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> Els arxius jar cada un té un arxiu a / org / node / trucada resource.txt.
Això és només un prototip del que un controlador es veuria amb classpath: // També tinc un resource.foo.txt en el meu local dels recursos per a aquest projecte.
Recull a tots ells i els imprimeix.
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 : //"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.split("!"); String zipFileName = split2; String sresource = split2; 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(); } } } } }
Es pot veure un exemple més complet aquí amb la sortida d’exemple.