好.所以我有一个非常简单的问题:我希望能够从正在运行的.jar文件中加载资源(整个文件夹),但我无法让它工作.这是我尝试过的(如果类名是“myClass”,文件夹被称为“myFolder”),但它总是抛出NullPointerException:
URL folderURL = myClass.class.getClassLoader().getResource("myFolder/"); String folderPath = folderURL.getPath(); File myFolder = new File(folderPath);
在创建“myFolder”之前,总是抛出NullPointerException.
更多信息:我必须从静态上下文访问该文件夹.正在访问该文件夹的类与文件夹本身所在的目录不在同一目录中.(该文件夹位于jar内的根目录中,该类是几个子包.)
解决方法
这没办法.您正在尝试从JAR内的资源创建File对象.这不会发生.加载资源的最佳方法是将一个包文件夹作为资源文件夹,然后在其中创建一个Resources.jar,将资源转储到同一目录中,然后在其他目录中使用Resources.class.getResourceAsStream(resFileName) Java类文件.
如果你需要“暴力破解”由getResource(..)给出的URL指向的JAR目录中的子文件,请使用以下内容(尽管这有点像黑客!).它也适用于普通的文件系统:
/** * List directory contents for a resource folder. Not recursive. * This is basically a brute-force implementation. * Works for regular files and also JARs. * * @author Greg Briggs * @param clazz Any java class that lives in the same place as the resources you want. * @param path Should end with "/",but not start with one. * @return Just the name of each member item,not the full paths. * @throws URISyntaxException * @throws IOException */ String[] getResourceListing(Class clazz,String path) throws URISyntaxException,IOException { URL dirURL = clazz.getClassLoader().getResource(path); if (dirURL != null && dirURL.getProtocol().equals("file")) { /* A file path: easy enough */ return new File(dirURL.toURI()).list(); } if (dirURL == null) { /* * In case of a jar file,we can't actually find a directory. * Have to assume the same jar as clazz. */ String me = clazz.getName().replace(".","/")+".class"; dirURL = clazz.getClassLoader().getResource(me); } if (dirURL.getProtocol().equals("jar")) { /* A JAR path */ String jarPath = dirURL.getPath().substring(5,dirURL.getPath().indexOf("!")); //strip out only the JAR file JarFile jar = new JarFile(URLDecoder.decode(jarPath,"UTF-8")); Enumeration<JarEntry> entries = jar.entries(); //gives ALL entries in jar Set<String> result = new HashSet<String>(); //avoid duplicates in case it is a subdirectory while(entries.hasMoreElements()) { String name = entries.nextElement().getName(); if (name.startsWith(path)) { //filter according to the path String entry = name.substring(path.length()); int checkSubdir = entry.indexOf("/"); if (checkSubdir >= 0) { // if it is a subdirectory,we just return the directory name entry = entry.substring(0,checkSubdir); } result.add(entry); } } return result.toArray(new String[result.size()]); } throw new UnsupportedOperationException("Cannot list files for URL "+dirURL); }
然后,您可以修改getResource(..)给出的URL并将文件附加到末尾,并将这些URL传递给getResourceAsStream(..),以备加载.如果您不理解这一点,则需要阅读类加载.