getInputStream for a ZipEntry from ZipInputStream(不使用 ZipFile 类)
2022-09-01 07:42:53
如何在不使用该类的情况下从 a 获取 a for a?InputStream
ZipEntry
ZipInputStream
ZipFile
如何在不使用该类的情况下从 a 获取 a for a?InputStream
ZipEntry
ZipInputStream
ZipFile
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();
}
呃,已经是一个你不需要另一个。获取下一个位置将流定位在条目的开头。请参阅 Javadoc。ZipInputStream
InputStream.
ZipEntry