java ZipEntry 到文件
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/36890005/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
ZipEntry to File
提问by JF Meier
Is there a direct way to unpack a java.util.zip.ZipEntryto a File?
有没有直接的方法将 a 解包java.util.zip.ZipEntry到 a File？
I want to specify a location (like "C:\temp\myfile.java") and unpack the Entry to that location.
我想指定一个位置（如“C:\temp\myfile.java”）并将 Entry 解压到该位置。
There is some code with streams on the net, but I would prefer a tested library function.
网上有一些带有流的代码，但我更喜欢经过测试的库函数。
回答by Evgeniy Dorofeev
Use ZipFile class
使用 ZipFile 类
ZipFile zf = new ZipFile("zipfile");
Get entry
获取条目
ZipEntry e = zf.getEntry("name");
Get inpustream
获取输入流
InputStream is = zf.getInputStream(e);
Save bytes
节省字节
Files.copy(is, Paths.get("C:\temp\myfile.java"));
回答by Coffee Monkey
Use ZipInputStreamto move to the desired ZipEntryby iterating using the getNextEntry()method. Then use the ZipInputStream.read(...)method to read the bytes for the current ZipEntry. Output those bytes to a FileOutputStreampointing to a file of your choice.
用于通过使用该方法进行迭代ZipInputStream来移动到所需ZipEntry的位置getNextEntry()。然后使用该ZipInputStream.read(...)方法读取当前ZipEntry. 将这些字节输出到FileOutputStream指向您选择的文件的指针。
回答by Shakthifuture
I have extractor the zip file into File and added it into the list using ZipEntry. Hopefully, this will help you.
我已将 zip 文件提取到 File 中，并使用 ZipEntry 将其添加到列表中。希望这会对您有所帮助。
private List<File> unzip(Resource resource) {
List<File> files = new ArrayList<>();
try {
ZipInputStream zin = new ZipInputStream(resource.getInputStream());
ZipEntry entry = null;
while((entry = zin.getNextEntry()) != null) {
File file = new File(entry.getName());
FileOutputStream os = new FileOutputStream(file);
for (int c = zin.read(); c != -1; c = zin.read()) {
os.write(c);
}
os.close();
files.add(file);
}
} catch (IOException e) {
log.error("Error while extract the zip: "+e);
}
return files;
}
