本文介绍了来自ZipInputStream的ZipEntry的getInputStream(不使用ZipFile类)的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
如何从 ZipInputStream $获取
ZipEntry
的 InputStream
c $ c>不使用 ZipFile
类?
How can I get an InputStream
for a ZipEntry
from a ZipInputStream
without using the ZipFile
class?
推荐答案
它有效这样
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();
}
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