java 如何在java中提取zip文件中的特定文件
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How to extract specific file in a zip file in java
提问by Mr rain
I need to provide a view of zip file to customer in system, and allow customers download choosed files.
我需要在系统中向客户提供 zip 文件的视图，并允许客户下载所选文件。
- parse the zip file and show on the web page. and remember every zipentry location(for example file1 is starting from byte 100 width length 1024bytes) in backend.
- download the specified file when customer click the download button.
- 解析 zip 文件并显示在网页上。并记住后端的每个 zipentry 位置（例如 file1 从字节 100 宽度长度 1024bytes 开始）。
- 当客户点击下载按钮时，下载指定的文件。
now I have rememberred all zipentry locations, but is there java zip tools to unzip the specified location of zip files?? API just like unzip(file, long entryStart, long entryLength);
现在我已经记住了所有的 zipentry 位置，但是有没有 java zip 工具来解压 zip 文件的指定位置？API 就像 unzip(file, long entryStart, long entryLength);
回答by Thunderforge
This can be done without messing with byte arrays or input streams using Java 7's NIO2:
这可以使用 Java 7 的 NIO2 来完成，而不会弄乱字节数组或输入流：
public void extractFile(Path zipFile, String fileName, Path outputFile) throws IOException {
// Wrap the file system in a try-with-resources statement
// to auto-close it when finished and prevent a memory leak
try (FileSystem fileSystem = FileSystems.newFileSystem(zipFile, null)) {
Path fileToExtract = fileSystem.getPath(fileName);
Files.copy(fileToExtract, outputFile);
}
}
回答by Amit Bhati
You can use the below code to extract a particular file from zip:-
您可以使用以下代码从 zip 中提取特定文件：-
public static void main(String[] args) throws Exception{
String fileToBeExtracted="fileName";
String zipPackage="zip_name_with_full_path";
OutputStream out = new FileOutputStream(fileToBeExtracted);
FileInputStream fileInputStream = new FileInputStream(zipPackage);
BufferedInputStream bufferedInputStream = new BufferedInputStream(fileInputStream );
ZipInputStream zin = new ZipInputStream(bufferedInputStream);
ZipEntry ze = null;
while ((ze = zin.getNextEntry()) != null) {
if (ze.getName().equals(fileToBeExtracted)) {
byte[] buffer = new byte[9000];
int len;
while ((len = zin.read(buffer)) != -1) {
out.write(buffer, 0, len);
}
out.close();
break;
}
}
zin.close();
}
Also refer this link: How to extract a single file from a remote archive file?
另请参阅此链接：如何从远程存档文件中提取单个文件？
回答by Garry
You can try like this:
你可以这样试试：
ZipFile zf = new ZipFile(file);
try {
InputStream in = zf.getInputStream(zf.getEntry("file.txt"));
// ... read from 'in' as normal
} finally {
zf.close();
}
I havent tried it but in Java 7 ZipFileSystem you can try like this to extract file.TXT file from the zip file.
我还没有尝试过，但是在 Java 7 ZipFileSystem 中，您可以尝试这样从 zip 文件中提取 file.TXT 文件。
Path zipfile = Paths.get("/samples/ziptest.zip");
FileSystem fs = FileSystems.newFileSystem(zipfile, env, null);
final Path root = fs.getPath("/file.TXT");
