getInputStream for a ZipEntry from ZipInputStream (without using the ZipFile class)

前提是你 提交于 2019-12-04 15:25:55

问题


How can I get an InputStream for a ZipEntry from a ZipInputStream without using the ZipFile class?


回答1:


it works this way

static InputStream getInputStream(File zip, String entry) throws IOException {
    ZipInputStream zin = new ZipInputStream(new FileInputStream(zip));
    for (ZipEntry e; (e = zin.getNextEntry()) != null;) {
        if (e.getName().equals(entry)) {
            return zin;
        }
    }
    throw new EOFException("Cannot find " + entry);
}

public static void main(String[] args) throws Exception {
    InputStream in = getInputStream(new File("f:/1.zip"), "launch4j/LICENSE.txt");
    Scanner sc = new Scanner(in);
    while(sc.hasNextLine()) {
        System.out.println(sc.nextLine());
    }
    in.close();
}



回答2:


Err, the ZipInputStream already is an InputStream. You don't need another one. Getting the next ZipEntry positions the stream at the beginning of the entry. See the Javadoc.




回答3:


To return a List of Input Streams that can be used later I used the following

public static List<InputStream> listResourcesInJar(URL jar) throws IOException{
    ZipInputStream zipInputStream = new ZipInputStream(jar.openStream());
    ZipEntry zipEntry = null;

    List<InputStream> inputStreams = new ArrayList<>();

    while ((zipEntry = zipInputStream.getNextEntry()) != null) {
        String entryName = zipEntry.getName();
        if (entryName.endsWith(".xsd")) {
            inputStreams.add(convertToInputStream(zipInputStream));
        }
    }
    return inputStreams;
}

private static InputStream convertToInputStream(final ZipInputStream inputStreamIn) throws IOException {
    ByteArrayOutputStream out = new ByteArrayOutputStream();
    IOUtils.copy(inputStreamIn, out);
    return new ByteArrayInputStream(out.toByteArray());
}


来源:https://stackoverflow.com/questions/14603319/getinputstream-for-a-zipentry-from-zipinputstream-without-using-the-zipfile-cla

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