我在 android 资产中放了一个 zip 文件。如何提取android内部存储中的文件？我知道如何获取文件，但我不知道如何提取它。这是我的代码..
Util zip ;
zip = new Util();
zip.copyFileFromAsset(this, "myfile.zip", getExternalStorage()+
"/android/data/edu.binus.profile/");
感谢您的帮助：D
问问题
20280 次
5 回答
15
这段代码将帮助您....只需将 zipfile 位置和您希望将提取的文件保存到此类的位置传递给此类，同时制作对象...并调用 unzip 方法...
public class Decompress {
private String zip;
private String loc;
public Decompress(String zipFile, String location) {
zip = zipFile;
loc = location;
dirChecker("");
}
public void unzip() {
try {
FileInputStream fin = new FileInputStream(zip);
ZipInputStream zin = new ZipInputStream(fin);
ZipEntry ze = null;
while ((ze = zin.getNextEntry()) != null) {
Log.v("Decompress", "Unzipping " + ze.getName());
if(ze.isDirectory()) {
dirChecker(ze.getName());
} else {
FileOutputStream fout = new FileOutputStream(loc + ze.getName());
for (int c = zin.read(); c != -1; c = zin.read()) {
fout.write(c);
}
zin.closeEntry();
fout.close();
}
}
zin.close();
} catch(Exception e) {
Log.e("Decompress", "unzip", e);
}
}
private void dirChecker(String dir) {
File f = new File(_location + dir);
if(!f.isDirectory()) {
f.mkdirs();
}
}
}
于 2013-03-26T05:09:43.680 回答
13
基于 Sreedev R 解决方案，我添加了从资产中读取文件并使用缓冲区的选项：
package com.pixoneye.api.utils;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.util.zip.ZipEntry;
import java.util.zip.ZipInputStream;
import android.content.Context;
import android.util.Log;
public class Decompress {
private static final int BUFFER_SIZE = 1024 * 10;
private static final String TAG = "Decompress";
public static void unzipFromAssets(Context context, String zipFile, String destination) {
try {
if (destination == null || destination.length() == 0)
destination = context.getFilesDir().getAbsolutePath();
InputStream stream = context.getAssets().open(zipFile);
unzip(stream, destination);
} catch (IOException e) {
e.printStackTrace();
}
}
public static void unzip(String zipFile, String location) {
try {
FileInputStream fin = new FileInputStream(zipFile);
unzip(fin, location);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
public static void unzip(InputStream stream, String destination) {
dirChecker(destination, "");
byte[] buffer = new byte[BUFFER_SIZE];
try {
ZipInputStream zin = new ZipInputStream(stream);
ZipEntry ze = null;
while ((ze = zin.getNextEntry()) != null) {
Log.v(TAG, "Unzipping " + ze.getName());
if (ze.isDirectory()) {
dirChecker(destination, ze.getName());
} else {
File f = new File(destination, ze.getName());
if (!f.exists()) {
boolean success = f.createNewFile();
if (!success) {
Log.w(TAG, "Failed to create file " + f.getName());
continue;
}
FileOutputStream fout = new FileOutputStream(f);
int count;
while ((count = zin.read(buffer)) != -1) {
fout.write(buffer, 0, count);
}
zin.closeEntry();
fout.close();
}
}
}
zin.close();
} catch (Exception e) {
Log.e(TAG, "unzip", e);
}
}
private static void dirChecker(String destination, String dir) {
File f = new File(destination, dir);
if (!f.isDirectory()) {
boolean success = f.mkdirs();
if (!success) {
Log.w(TAG, "Failed to create folder " + f.getName());
}
}
}
}
于 2014-12-09T09:36:30.387 回答
1
也许您应该尝试将 FileOutputStream 与 zip 文件中的输入流结合使用。使用包文件，这应该可以工作。
从这个问题引用@wordy ：
PackageManager pm = context.getPackageManager();
String apkFile = pm.getApplicationInfo(context.getPackageName(), 0).sourceDir;
ZipFile zipFile = new ZipFile(apkFile);
ZipEntry entry = zipFile.getEntry("assets/FILENAME");
myInput = zipFile.getInputStream(entry);
myOutput = new FileOutputStream(file);
byte[] buffer = new byte[1024*4];
int length;
int total = 0;
int counter = 1;
while ((length = myInput.read(buffer)) > 0) {
total += length;
counter++;
if (counter % 32 == 0) {
publishProgress(total);
}
myOutput.write(buffer, 0, length);
}
看起来 ProGuard 可能存在问题，但希望代码示例对您有用。
于 2013-03-26T02:54:22.757 回答
0
您还可以使用提供加密等附加功能的 zip4j 外部库。此外，它还具有将文件提取到提供路径的特定位置的功能。
于 2013-03-26T04:25:09.440 回答