我查看了JDK和Apache压缩库附带的默认Zip库,我对它们不满意,原因有3个:
They are bloated and have bad API design. I have to write 50 lines of boiler plate byte array output, zip input, file out streams and close relevant streams and catch exceptions and move byte buffers on my own? Why can't I have a simple API that looks like this Zipper.unzip(InputStream zipFile, File targetDirectory, String password = null) and Zipper.zip(File targetDirectory, String password = null) that just works?
It seems zipping unzipping destroys file meta-data and password handling is broken.
Also, all the libraries I tried were 2-3x slow compared to the command line zip tools I get with UNIX?
对我来说(2)和(3)是次要的点,但我真的想要一个良好的测试库与一行接口。
你看了http://commons.apache.org/vfs/吗?它声称可以为你简化很多事情。但我从未在项目中使用过。
除了JDK或Apache压缩之外,我也不知道Java-Native压缩库。
我记得有一次我们从Apache Ant中剥离了一些特性——它们有很多内置的压缩/解压utils。
VFS的示例代码如下所示:
File zipFile = ...;
File outputDir = ...;
FileSystemManager fsm = VFS.getManager();
URI zip = zipFile.toURI();
FileObject packFileObject = fsm.resolveFile(packLocation.toString());
FileObject to = fsm.toFileObject(destDir);
FileObject zipFS;
try {
zipFS = fsm.createFileSystem(packFileObject);
fsm.toFileObject(outputDir).copyFrom(zipFS, new AllFileSelector());
} finally {
zipFS.close();
}
解压zip文件及其所有子文件夹,只使用JDK:
private void extractFolder(String zipFile,String extractFolder)
{
try
{
int BUFFER = 2048;
File file = new File(zipFile);
ZipFile zip = new ZipFile(file);
String newPath = extractFolder;
new File(newPath).mkdir();
Enumeration zipFileEntries = zip.entries();
// Process each entry
while (zipFileEntries.hasMoreElements())
{
// grab a zip file entry
ZipEntry entry = (ZipEntry) zipFileEntries.nextElement();
String currentEntry = entry.getName();
File destFile = new File(newPath, currentEntry);
//destFile = new File(newPath, destFile.getName());
File destinationParent = destFile.getParentFile();
// create the parent directory structure if needed
destinationParent.mkdirs();
if (!entry.isDirectory())
{
BufferedInputStream is = new BufferedInputStream(zip
.getInputStream(entry));
int currentByte;
// establish buffer for writing file
byte data[] = new byte[BUFFER];
// write the current file to disk
FileOutputStream fos = new FileOutputStream(destFile);
BufferedOutputStream dest = new BufferedOutputStream(fos,
BUFFER);
// read and write until last byte is encountered
while ((currentByte = is.read(data, 0, BUFFER)) != -1) {
dest.write(data, 0, currentByte);
}
dest.flush();
dest.close();
is.close();
}
}
}
catch (Exception e)
{
Log("ERROR: "+e.getMessage());
}
}
Zip文件及其所有子文件夹:
private void addFolderToZip(File folder, ZipOutputStream zip, String baseName) throws IOException {
File[] files = folder.listFiles();
for (File file : files) {
if (file.isDirectory()) {
addFolderToZip(file, zip, baseName);
} else {
String name = file.getAbsolutePath().substring(baseName.length());
ZipEntry zipEntry = new ZipEntry(name);
zip.putNextEntry(zipEntry);
IOUtils.copy(new FileInputStream(file), zip);
zip.closeEntry();
}
}
}
我知道有点晚了,有很多答案,但这个zip4j是我使用过的最好的压缩库之一。它很简单(没有锅炉代码),可以很容易地处理密码保护文件。
import net.lingala.zip4j.exception.ZipException;
import net.lingala.zip4j.core.ZipFile;
public static void unzip(){
String source = "some/compressed/file.zip";
String destination = "some/destination/folder";
String password = "password";
try {
ZipFile zipFile = new ZipFile(source);
if (zipFile.isEncrypted()) {
zipFile.setPassword(password);
}
zipFile.extractAll(destination);
} catch (ZipException e) {
e.printStackTrace();
}
}
Maven依赖项是:
<dependency>
<groupId>net.lingala.zip4j</groupId>
<artifactId>zip4j</artifactId>
<version>1.3.2</version>
</dependency>