如何使用 apache commons 从 TAR 解压缩特定文件? [英] How uncompress a specific file from a TAR using apache commons?
问题描述
我正在使用 Apache Commons 1.4.1 库来解压缩.tar"文件.
I'm using the Apache Commons 1.4.1 library to uncompress ".tar" files.
问题:我不必提取所有文件.我必须从 tar 存档中的特定位置提取特定文件.我只需要提取几个 .xml 文件,其中 TAR 文件的大小约为 300 MB &解压整个内容很浪费资源.
Problem: I don't have to extract all files. I have to extract specific files from specific location inside a tar archive. i have to extract only few .xml files where as the size of the TAR file is around 300 MB & it is waste of resource in uncompressing the entire content.
我被困住了&困惑我是否必须进行嵌套目录比较或者有什么办法吗?
I am stuck up & confused whether i have to do a nested directory compare or is there is any way around?
注意:.XML(必需文件)的位置始终相同.
Note: location of the .XML(required files) is always same.
TAR 的结构是:
directory:E:\Root\data
file:E:\Root\datasheet.txt
directory:E:\Root\map
file:E:\Root\mapers.txt
directory:E:\Root\ui
file:E:\Root\ui\capital.txt
file:E:\Root\ui\info.txt
directory:E:\Root\ui\sales
file:E:\Root\ui\sales\Reqest_01.xml
file:E:\Root\ui\sales\Reqest_02.xml
file:E:\Root\ui\sales\Reqest_03.xml
file:E:\Root\ui\sales\Reqest_04.xml
directory:E:\Root\ui\sales\stores
directory:E:\Root\ui\stores
directory:E:\Root\urls
directory:E:\Root\urls\fullfilment
file:E:\Root\urls\fullfilment\Cams_01.xml
file:E:\Root\urls\fullfilment\Cams_02.xml
file:E:\Root\urls\fullfilment\Cams_03.xml
file:E:\Root\urls\fullfilment\Cams_04.xml
directory:E:\Root\urls\fullfilment\profile
directory:E:\Root\urls\fullfilment\registration
file:E:\Root\urls\options.txt
directory:E:\Root\urls\profile
约束:我不能使用 JDK 7 &必须坚持使用 Apache 公共库.
Constraint: i cant use JDK 7 & have to stick with Apache commons library.
我目前的解决方案:
public static void untar(File[] files) throws Exception {
String path = files[0].toString();
File tarPath = new File(path);
TarEntry entry;
TarInputStream inputStream = null;
FileOutputStream outputStream = null;
try {
inputStream = new TarInputStream(new FileInputStream(tarPath));
while (null != (entry = inputStream.getNextEntry())) {
int bytesRead;
System.out.println("tarpath:" + tarPath.getName());
System.out.println("Entry:" + entry.getName());
String pathWithoutName = path.substring(0, path.indexOf(tarPath.getName()));
System.out.println("pathname:" + pathWithoutName);
if (entry.isDirectory()) {
File directory = new File(pathWithoutName + entry.getName());
directory.mkdir();
continue;
}
byte[] buffer = new byte[1024];
outputStream = new FileOutputStream(pathWithoutName + entry.getName());
while ((bytesRead = inputStream.read(buffer, 0, 1024)) > -1) {
outputStream.write(buffer, 0, bytesRead);
}
System.out.println("Extracted " + entry.getName());
}
}
推荐答案
TAR 文件格式设计为作为流写入或读取(即,从磁带驱动器写入或从磁带驱动器读取),并且没有集中的标头.所以不,没有办法通过读取整个文件来提取单个条目.
The TAR file format is designed to be written or read as a stream (ie, to/from a tape drive), and does not have a centralized header. So no, there's no way around reading the entire file to extract individual entries.
如果你想随机访问,你应该使用ZIP格式,并使用JDK的ZipFile
打开.假设您有足够的虚拟内存,该文件将进行内存映射,从而使随机访问非常快(我还没有查看如果无法进行内存映射,它是否会使用随机访问文件).
If you want random access, you should use the ZIP format, and open using the JDK's ZipFile
. Assuming that you have enough virtual memory, the file will be memory-mapped, making random access very fast (I haven't looked to see if it will use a random-access file if unable to memory-map).
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