我在资产文件夹里有几个文件。我需要把它们都复制到一个文件夹,比如/sdcard/folder。我想从一个线程中做这件事。我该怎么做?
当前回答
基于Yoram Cohen的回答,这里有一个支持非静态目标目录的版本。
使用copyFileOrDir(getDataDir(), "")来写入应用程序内部存储文件夹/data/data/pkg_name/
Supports subfolders. Supports custom and non-static target directory Avoids copying "images" etc fake asset folders like private void copyFileOrDir(String TARGET_BASE_PATH, String path) { AssetManager assetManager = this.getAssets(); String assets[] = null; try { Log.i("tag", "copyFileOrDir() "+path); assets = assetManager.list(path); if (assets.length == 0) { copyFile(TARGET_BASE_PATH, path); } else { String fullPath = TARGET_BASE_PATH + "/" + path; Log.i("tag", "path="+fullPath); File dir = new File(fullPath); if (!dir.exists() && !path.startsWith("images") && !path.startsWith("sounds") && !path.startsWith("webkit")) if (!dir.mkdirs()) Log.i("tag", "could not create dir "+fullPath); for (int i = 0; i < assets.length; ++i) { String p; if (path.equals("")) p = ""; else p = path + "/"; if (!path.startsWith("images") && !path.startsWith("sounds") && !path.startsWith("webkit")) copyFileOrDir(TARGET_BASE_PATH, p + assets[i]); } } } catch (IOException ex) { Log.e("tag", "I/O Exception", ex); } } private void copyFile(String TARGET_BASE_PATH, String filename) { AssetManager assetManager = this.getAssets(); InputStream in = null; OutputStream out = null; String newFileName = null; try { Log.i("tag", "copyFile() "+filename); in = assetManager.open(filename); if (filename.endsWith(".jpg")) // extension was added to avoid compression on APK file newFileName = TARGET_BASE_PATH + "/" + filename.substring(0, filename.length()-4); else newFileName = TARGET_BASE_PATH + "/" + filename; out = new FileOutputStream(newFileName); byte[] buffer = new byte[1024]; int read; while ((read = in.read(buffer)) != -1) { out.write(buffer, 0, read); } in.close(); in = null; out.flush(); out.close(); out = null; } catch (Exception e) { Log.e("tag", "Exception in copyFile() of "+newFileName); Log.e("tag", "Exception in copyFile() "+e.toString()); } }
其他回答
轻微修改以上回答复制文件夹递归和适应自定义目的地。
public void copyFileOrDir(String path, String destinationDir) {
AssetManager assetManager = this.getAssets();
String assets[] = null;
try {
assets = assetManager.list(path);
if (assets.length == 0) {
copyFile(path,destinationDir);
} else {
String fullPath = destinationDir + "/" + path;
File dir = new File(fullPath);
if (!dir.exists())
dir.mkdir();
for (int i = 0; i < assets.length; ++i) {
copyFileOrDir(path + "/" + assets[i], destinationDir + path + "/" + assets[i]);
}
}
} catch (IOException ex) {
Log.e("tag", "I/O Exception", ex);
}
}
private void copyFile(String filename, String destinationDir) {
AssetManager assetManager = this.getAssets();
String newFileName = destinationDir + "/" + filename;
InputStream in = null;
OutputStream out = null;
try {
in = assetManager.open(filename);
out = new FileOutputStream(newFileName);
byte[] buffer = new byte[1024];
int read;
while ((read = in.read(buffer)) != -1) {
out.write(buffer, 0, read);
}
in.close();
in = null;
out.flush();
out.close();
out = null;
} catch (Exception e) {
Log.e("tag", e.getMessage());
}
new File(newFileName).setExecutable(true, false);
}
你可以用Kotlin在几个步骤中做到这一点,在这里我只复制几个文件,而不是所有从资产到我的应用程序文件目录。
private fun copyRelatedAssets() {
val assets = arrayOf("myhome.html", "support.css", "myscript.js", "style.css")
assets.forEach {
val inputStream = requireContext().assets.open(it)
val nameSplit = it.split(".")
val name = nameSplit[0]
val extension = nameSplit[1]
val path = inputStream.getFilePath(requireContext().filesDir, name, extension)
Log.v(TAG, path)
}
}
这是扩展函数,
fun InputStream.getFilePath(dir: File, name: String, extension: String): String {
val file = File(dir, "$name.$extension")
val outputStream = FileOutputStream(file)
this.copyTo(outputStream, 4096)
return file.absolutePath
}
洛格猫
/data/user/0/com.***.***/files/myhome.html
/data/user/0/com.***.***/files/support.css
/data/user/0/com.***.***/files/myscript.js
/data/user/0/com.***.***/files/style.css
使用这个问题答案中的一些概念,我编写了一个名为assetcopyer的类来简化复制/assets/。它在github上可用,可以通过jitpack.io访问:
new AssetCopier(MainActivity.this)
.withFileScanning()
.copy("tocopy", destDir);
详情见https://github.com/flipagram/android-assetcopier。
基于Rohith Nandakumar的解决方案,我自己做了一些事情,从资产的子文件夹中复制文件。“资产/ MyFolder”)。此外,在尝试再次复制之前,我正在检查文件是否已经存在于sdcard中。
private void copyAssets() {
AssetManager assetManager = getAssets();
String[] files = null;
try {
files = assetManager.list("MyFolder");
} catch (IOException e) {
Log.e("tag", "Failed to get asset file list.", e);
}
if (files != null) for (String filename : files) {
InputStream in = null;
OutputStream out = null;
try {
in = assetManager.open("MyFolder/"+filename);
File outFile = new File(getExternalFilesDir(null), filename);
if (!(outFile.exists())) {// File does not exist...
out = new FileOutputStream(outFile);
copyFile(in, out);
}
} catch(IOException e) {
Log.e("tag", "Failed to copy asset file: " + filename, e);
}
finally {
if (in != null) {
try {
in.close();
} catch (IOException e) {
// NOOP
}
}
if (out != null) {
try {
out.close();
} catch (IOException e) {
// NOOP
}
}
}
}
}
private void copyFile(InputStream in, OutputStream out) throws IOException {
byte[] buffer = new byte[1024];
int read;
while((read = in.read(buffer)) != -1){
out.write(buffer, 0, read);
}
}
这在Kotlin中是一种简洁的方式。
fun AssetManager.copyRecursively(assetPath: String, targetFile: File) {
val list = list(assetPath)
if (list.isEmpty()) { // assetPath is file
open(assetPath).use { input ->
FileOutputStream(targetFile.absolutePath).use { output ->
input.copyTo(output)
output.flush()
}
}
} else { // assetPath is folder
targetFile.delete()
targetFile.mkdir()
list.forEach {
copyRecursively("$assetPath/$it", File(targetFile, it))
}
}
}
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