Aș dori să subliniez că una dintre probleme este, ceea ce se întâmplă dacă aceleași resurse sunt Ei găsesc în mai multe fișiere de jar. Să vrem să citim /org/node/foo.txt, dar nu dintr-un fișier, ci de la fiecare dintre fișierele de jar.
Am terminat cu aceeași problemă ori mai înainte Scrieți un mic prototip pentru a rezolva această mare problemă a citirii resurselor sub forma mai multor fișiere JAR. Prototipul nu se ocupă de fiecare caz extrem, dar se ocupă de căutarea resurselor din directoarele care se află în fișierele JAR.
Am folosit o depășire a stivei de ceva timp. Acesta este al doilea răspuns pe care îmi amintesc să răspund la o întrebare, așa că mă iert dacă mă duc prea mult (este natura mea).
Acesta este un prototip al resurselor cititorului. Prototipul este lipsit de un sistem robust de verificare a erorilor. Fișier în / org / nod / numit Resource.txt.
Acesta este doar un prototip al ceea ce un controler ar arăta cu CLASSPATH: // am, de asemenea, o resursă.foo.txt în resursele mele locale pentru acest proiect.
Strângeți toate și imprimați-le.
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(); } } } } }
Puteți vedea un exemplu mai complet aici cu ieșirea de exemplu.
.