如何在不使用InputStream
课程的情况下从ZipEntry
获得ZipInputStream
的{{1}}?
答案 0 :(得分:18)
它以这种方式工作
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();
}
答案 1 :(得分:16)
错误,ZipInputStream
已经是InputStream.
你不需要另一个。在条目开头获取流的下一个ZipEntry
位置。见Javadoc。
答案 2 :(得分:1)
要返回稍后可以使用的输入流列表,我使用了以下
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());
}