如何在android中访问资产？

Question

编辑：修复它

ZipInputStream zin = new ZipInputStream(getAssets().open("totalkeys.zip")); 

---

我从一个例子中得到了一个解压缩器（ decompress ），它把一个字符串作为一个压缩文件的路径。但是由于我的资产中有该文件，所以我需要从那里读取它......好吧，到目前为止。

不幸的是它抛出我“错误/解压缩（24122）：java.lang.ClassCastException：android.content.res.AssetManager$AssetInputStream”任何想法如何解决它？)

public class dec extends Activity {
/** Called when the activity is first created. */
@Override
    public void onCreate(Bundle savedInstanceState) {

        super.onCreate(savedInstanceState);

        Toast.makeText(this, "hello, starting to unZipp!", 15500).show();

        String location = Environment.getExternalStorageDirectory() + "/unzipped/"; 
        /////////////////////////////////////////////////////////////////////

        try  { 

            AssetManager mgr = getBaseContext().getAssets();

            FileInputStream fin = (FileInputStream)mgr.open("totalkeys.zip");
            // throws ERROR/Decompress(24122): java.lang.ClassCastException: android.content.res.AssetManager$AssetInputStream

            //FileInputStream fin = new FileInputStream(_zipFile); /// old one, that wanted a String.
            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(location + 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) { 

          String location = Environment.getExternalStorageDirectory() + "/unzipped/"; 
          File f = new File(location + dir); 

          if(!f.isDirectory()) { 
            f.mkdirs(); 
          } 

        ////////////////////////////////////////////////////

        finish();

}

}

谢谢！

Answer 1

使用您的上下文：

InputStream is = myContext.getAssets().open("totalkeys.zip");

这将返回一个可以读取到缓冲区的输入流。

// Open the input stream
InputStream is = mContext.getAssets().open(FILE_NAME);

byte[] buffer = new byte[1024];
int length;
while ((length = is.read(buffer))>0){
          // write buffer some where, e.g. to an output stream
          // as 'myOutputStream.write(buffer, 0, length);'
}
// Close the stream
try{
    is.close();
} catch(IOException e){
    Log.e(this.getLocalClassName().toString(), e.getMessage());
    //this.getLocalClassName().toString() could be replaced with any (string) tag
}

如果您正在从事一项活动，则可以使用this.getAssets()因为Activityextends Context。Context如果您不在活动中工作，也可以将 的实例传递给自定义构造函数，并在以后需要时将其分配给成员。