英文:
Get all files inside directory's in jar file in java
问题
假设我的项目位置是“some/random/guy”，在项目内部有一个路径为“some/random/guy/versions”的目录，我想获取该文件夹内的所有文件/目录，以便获取所有可用版本。
但是版本目录只有在项目编译一次后才会创建，所以它不在我的集成开发环境中。
在已经搜索了大约2小时后，我仍然没有找到答案。
目前我正在使用来自Oracle文档的代码，但它也不起作用：
try (DirectoryStream<Path> stream = Files.newDirectoryStream(Paths.get("some/random/guy/versions"))) {
for (Path file : stream) {
System.out.println(file.getFileName());
}
} catch (IOException | DirectoryIteratorException e) {
System.err.print(e);
}
我如何才能获取位于jar文件内部的任何文件夹中的所有文件/目录？
如果可能的话，我想避免解压缩jar文件，谢谢。
英文:
Lets say the location of my project is "some/random/guy", and inside the project there is a directory located with the path "some/random/guy/versions", I want to get all files/directory's inside that folder so I can get all the availible versions.
But the versions dir only gets created once the project is compiled, so its not in my IDE.
After searching like 2 hours already I couldn't find my answer.
Currently Im using the code from oracle docs but it also doesn't work:
try (DirectoryStream<Path> stream = Files.newDirectoryStream(Paths.get("some/random/guy/versions"))) {
for (Path file: stream) {
System.out.println(file.getFileName());
}
} catch (IOException | DirectoryIteratorException e) {
System.err.print(e);
}
How can I get all files/directory's inside any folder thats located inside a jar file?
I would like to avoid unzipping the jar file if possible, thanks.
答案1
得分: 1
以下是翻译好的内容：
这段代码将使用NIO FileSystem处理显示任何jar/zip存档的内容，您需要传入jar的位置和可选的jar内路径：
public static void main(String[] args) throws IOException {
Path jar = Path.of(args[0]);
String relPath = args.length > 1 ? args[1] : "/";
System.out.println("ZIP/JAR: " + jar + " isRegularFile()=" + Files.isRegularFile(jar));
try (FileSystem fs = FileSystems.newFileSystem(jar)) {
Path path = fs.getPath(relPath);
System.out.println("SCAN " + jar + " starting at path: " + path);
try(Stream<Path> str = Files.find(path, Integer.MAX_VALUE, (p,a) -> true)) {
str.forEach(System.out::println);
}
}
}
英文:
This code will show contents of any jar/zip archive using NIO FileSystem handling, you need to pass in a location of the jar and optional path under the jar:
public static void main(String[] args) throws IOException {
Path jar = Path.of(args[0]);
String relPath = args.length > 1 ? args[1] : "/";
System.out.println("ZIP/JAR: "+jar+" isRegularFile()="+Files.isRegularFile(jar));
try (FileSystem fs = FileSystems.newFileSystem(jar)) {
Path path = fs.getPath(relPath);
System.out.println("SCAN "+ jar+" starting at path: "+path);
try(Stream<Path> str = Files.find(path, Integer.MAX_VALUE, (p,a) -> true)) {
str.forEach(System.out::println);
}
}
}
答案2
得分: 0
jar文件就是普通的zip压缩文件。你不需要解压jar文件就可以获取所有条目的名称，因为zip文件包含了CentralDirectory。你只需要使用标准的ZipInputStream或者其中一个框架，比如zip4jvm：
import ru.olegcherednik.zip4jvm.ZipMisc;
public static List<String> getAllEntryNames(Path jar) throws IOException {
return ZipMisc.zip(jar).getEntries()
.map(ZipFile.Entry::getFileName)
.collect(Collectors.toList());
}
这段代码可以检索出所有已存在的条目名称。目录（如果它们作为单独的条目存在）以/结尾。
英文:
jar file is plain old zip archive. You do not neet unzip jar to get all entries' names, because zip files contains CentralDirectory. All you need is to use standard ZipInputStream or one of framework, e.g. zip4jvm:
import ru.olegcherednik.zip4jvm.ZipMisc;
public static List<String> getAllEntryNames(Path jar) throws IOException {
return ZipMisc.zip(jar).getEntries()
.map(ZipFile.Entry::getFileName)
.collect(Collectors.toList());
}
This snippet retrieves all existed entry names. Directory (if they exists as separate entry) ends with /.
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