How can I get an InputStream
for a ZipEntry
from a ZipInputStream
without using the ZipFile
class?
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试):
问题:
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());
}