我想从我的jar中读取资源,如下所示:
File file;
file = new File(getClass().getResource("/file.txt").toURI());
BufferredReader reader = new BufferedReader(new FileReader(file));
//Read the file
并且它在Eclipse中运行时工作正常,但是如果我将它导出到jar,那么运行它就会出现IllegalArgumentException:
Exception in thread "Thread-2"
java.lang.IllegalArgumentException: URI is not hierarchical
而且我真的不知道为什么,但经过一些测试我发现如果我改变了
file = new File(getClass().getResource("/file.txt").toURI());
至
file = new File(getClass().getResource("/folder/file.txt").toURI());
然后它的工作正好相反(它适用于jar而不是eclipse)。
我正在使用Eclipse,我的文件夹在一个类文件夹中。
而不是试图作为File解决资源只是要求ClassLoader返回InputStream资源,而不是通过getResourceAsStream:
InputStream in = getClass().getResourceAsStream("/file.txt");
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
只要file.txt
资源在类路径上可用,那么无论file.txt
资源是在classes/
目录中还是在jar
内,这种方法都将以相同的方式工作。
发生URI is not hierarchical
是因为jar文件中资源的URI看起来像这样:file:/example.jar!/file.txt
。您无法读取jar
(zip
文件)中的条目,就像它是一个普通的旧File。
这可以通过以下答案得到很好的解释:
如果您使用的是spring,那么您可以使用以下方法从src / main / resources中读取文件:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import org.springframework.core.io.ClassPathResource;
public String readFileToString(String path) throws IOException {
StringBuilder resultBuilder = new StringBuilder("");
ClassPathResource resource = new ClassPathResource(path);
try (
InputStream inputStream = resource.getInputStream();
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream))) {
String line;
while ((line = bufferedReader.readLine()) != null) {
resultBuilder.append(line);
}
}
return resultBuilder.toString();
}
由于某种原因,当我将Web应用程序部署到WildFly 14时,classLoader.getResource()
总是返回null。从getClass().getClassLoader()
或Thread.currentThread().getContextClassLoader()
获取classLoader将返回null。
getClass().getClassLoader()
API doc说,
“返回类的类加载器。一些实现可能使用null来表示引导类加载器。如果此类由引导类加载器加载,则此方法将在此类实现中返回null。”
如果你使用WildFly和你的web应用程序可能会尝试这个
request.getServletContext().getResource()
返回了资源网址。这里的请求是ServletRequest的一个对象。
问题是某些第三方库需要文件路径名而不是输入流。大多数答案都没有解决这个问题。
在这种情况下,一种解决方法是将资源内容复制到临时文件中。以下示例使用jUnit的TemporaryFolder
。
private List<String> decomposePath(String path){
List<String> reversed = Lists.newArrayList();
File currFile = new File(path);
while(currFile != null){
reversed.add(currFile.getName());
currFile = currFile.getParentFile();
}
return Lists.reverse(reversed);
}
private String writeResourceToFile(String resourceName) throws IOException {
ClassLoader loader = getClass().getClassLoader();
InputStream configStream = loader.getResourceAsStream(resourceName);
List<String> pathComponents = decomposePath(resourceName);
folder.newFolder(pathComponents.subList(0, pathComponents.size() - 1).toArray(new String[0]));
File tmpFile = folder.newFile(resourceName);
Files.copy(configStream, tmpFile.toPath(), REPLACE_EXISTING);
return tmpFile.getAbsolutePath();
}
下面的代码适用于Spring boot(kotlin):
val authReader = InputStreamReader(javaClass.getResourceAsStream("/file1.json"))
要访问jar中的文件,您有两个选择:
getClass().getResourceAsStream("file.txt")
访问它Thread.currentThread().getContextClassLoader().getResourceAsStream("file.txt")
访问它当jar用作插件时,第一个选项可能不起作用。
如果你想读作文件,我相信仍有类似的解决方案:
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("file/test.xml").getFile());
之前我遇到过这个问题,我做了后备加载方式。基本上第一种方式在.jar文件中工作,第二种方式在eclipse或其他IDE中工作。
public class MyClass {
public static InputStream accessFile() {
String resource = "my-file-located-in-resources.txt";
// this is the path within the jar file
InputStream input = MyClass.class.getResourceAsStream("/resources/" + resource);
if (input == null) {
// this is how we load file within editor (eg eclipse)
input = MyClass.class.getClassLoader().getResourceAsStream(resource);
}
return input;
}
}
确保使用正确的分隔符。我用/
替换了相对路径中的所有File.separator
。这在IDE中运行良好,但在构建JAR中不起作用。
到目前为止(2017年12月),这是我发现的唯一可在IDE内部和外部工作的解决方案。
使用PathMatchingResourcePatternResolver
注意:它也适用于spring-boot
在这个例子中,我正在读取位于src / main / resources / my_folder中的一些文件:
try {
// Get all the files under this inner resource folder: my_folder
String scannedPackage = "my_folder/*";
PathMatchingResourcePatternResolver scanner = new PathMatchingResourcePatternResolver();
Resource[] resources = scanner.getResources(scannedPackage);
if (resources == null || resources.length == 0)
log.warn("Warning: could not find any resources in this scanned package: " + scannedPackage);
else {
for (Resource resource : resources) {
log.info(resource.getFilename());
// Read the file content (I used BufferedReader, but there are other solutions for that):
BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(resource.getInputStream()));
String line = null;
while ((line = bufferedReader.readLine()) != null) {
// ...
// ...
}
bufferedReader.close();
}
}
} catch (Exception e) {
throw new Exception("Failed to read the resources folder: " + e.getMessage(), e);
}
经过大量的Java挖掘后,唯一对我有用的解决方案是手动读取jar文件本身,除非你在开发环境(IDE)中:
/** @return The root folder or jar file that the class loader loaded from */
public static final File getClasspathFile() {
return new File(YourMainClass.class.getProtectionDomain().getCodeSource().getLocation().getFile());
}
/** @param resource The path to the resource
* @return An InputStream containing the resource's contents, or
* <b><code>null</code></b> if the resource does not exist */
public static final InputStream getResourceAsStream(String resource) {
resource = resource.startsWith("/") ? resource : "/" + resource;
if(getClasspathFile().isDirectory()) {//Development environment:
return YourMainClass.class.getResourceAsStream(resource);
}
final String res = resource;//Jar or exe:
return AccessController.doPrivileged(new PrivilegedAction<InputStream>() {
@SuppressWarnings("resource")
@Override
public InputStream run() {
try {
final JarFile jar = new JarFile(getClasspathFile());
String resource = res.startsWith("/") ? res.substring(1) : res;
if(resource.endsWith("/")) {//Directory; list direct contents:(Mimics normal getResourceAsStream("someFolder/") behaviour)
ByteArrayOutputStream baos = new ByteArrayOutputStream();
Enumeration<JarEntry> entries = jar.entries();
while(entries.hasMoreElements()) {
JarEntry entry = entries.nextElement();
if(entry.getName().startsWith(resource) && entry.getName().length() > resource.length()) {
String name = entry.getName().substring(resource.length());
if(name.contains("/") ? (name.endsWith("/") && (name.indexOf("/") == name.lastIndexOf("/"))) : true) {//If it's a folder, we don't want the children's folders, only the parent folder's children!
name = name.endsWith("/") ? name.substring(0, name.length() - 1) : name;
baos.write(name.getBytes(StandardCharsets.UTF_8));
baos.write('\r');
baos.write('\n');
}
}
}
jar.close();
return new ByteArrayInputStream(baos.toByteArray());
}
JarEntry entry = jar.getJarEntry(resource);
InputStream in = entry != null ? jar.getInputStream(entry) : null;
if(in == null) {
jar.close();
return in;
}
final InputStream stream = in;//Don't manage 'jar' with try-with-resources or close jar until the
return new InputStream() {//returned stream is closed(closing the jar closes all associated InputStreams):
@Override
public int read() throws IOException {
return stream.read();
}
@Override
public int read(byte b[]) throws IOException {
return stream.read(b);
}
@Override
public int read(byte b[], int off, int len) throws IOException {
return stream.read(b, off, len);
}
@Override
public long skip(long n) throws IOException {
return stream.skip(n);
}
@Override
public int available() throws IOException {
return stream.available();
}
@Override
public void close() throws IOException {
try {
jar.close();
} catch(IOException ignored) {
}
stream.close();
}
@Override
public synchronized void mark(int readlimit) {
stream.mark(readlimit);
}
@Override
public synchronized void reset() throws IOException {
stream.reset();
}
@Override
public boolean markSupported() {
return stream.markSupported();
}
};
} catch(Throwable e) {
e.printStackTrace();
return null;
}
}
});
}
注意:上面的代码似乎只适用于jar文件,如果它在主类中。我不知道为什么。
你可以使用类加载器,它将从类路径读取为ROOT路径(开头没有“/”)
InputStream in = getClass().getClassLoader().getResourceAsStream("file.txt");
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
我认为这应该也适用于java。我使用的以下代码是使用kotlin。
val resource = Thread.currentThread().contextClassLoader.getResource('resources.txt')