I have cheerapp.wav or cheerapp.mp3 or some other format.
InputStream in = context.getResources().openRawResource(R.raw.cheerapp);
BufferedInputStream bis = new BufferedInputStream(in, 8000);
// Create a DataInputStream to read the audio data from the saved file
DataInputStream dis = new DataInputStream(bis);
byte[] music = null;
music = new byte[??];
int i = 0; // Read the file into the "music" array
while (dis.available() > 0) {
// dis.read(music[i]); // This assignment does not reverse the order
music[i]=dis.readByte();
i++;
}
dis.close();
For the music byte array which takes the data from the DataInputStream. I don't know what the length of that to allocate.
This is raw file from resource not a file therefore I wouldn't know the size of that thing.
You do have byte array length as you can see:
InputStream inStream = context.getResources().openRawResource(R.raw.cheerapp);
byte[] music = new byte[inStream.available()];
And then you can read whole Stream into byte array easily.
Of course I would recommend that you do check when it comes to the size and use ByteArrayOutputStream with smaller byte[] buffer if needed:
public static byte[] convertStreamToByteArray(InputStream is) throws IOException {
ByteArrayOutputStream baos = new ByteArrayOutputStream();
byte[] buff = new byte[10240];
int i = Integer.MAX_VALUE;
while ((i = is.read(buff, 0, buff.length)) > 0) {
baos.write(buff, 0, i);
}
return baos.toByteArray(); // be sure to close InputStream in calling function
}
If you'll be doing lots of IO operations I recommend that you make use of org.apache.commons.io.IOUtils. That way you won't need to worry too much about quality of your IO implementation and once you import JAR into your project you would just do:
byte[] payload = IOUtils.toByteArray(context.getResources().openRawResource(R.raw.cheerapp));
Hope it will help.
Create an sdcard path:
String outputFile =
Environment.getExternalStorageDirectory().getAbsolutePath() + "/recording.3gp";
Convert as a file and have to call the byte array method:
byte[] soundBytes;
try {
InputStream inputStream =
getContentResolver().openInputStream(Uri.fromFile(new File(outputFile)));
soundBytes = new byte[inputStream.available()];
soundBytes = toByteArray(inputStream);
Toast.makeText(this, "Recordin Finished"+ " " + soundBytes, Toast.LENGTH_LONG).show();
} catch(Exception e) {
e.printStackTrace();
}
method:
public byte[] toByteArray(InputStream in) throws IOException {
ByteArrayOutputStream out = new ByteArrayOutputStream();
int read = 0;
byte[] buffer = new byte[1024];
while (read != -1) {
read = in.read(buffer);
if (read != -1)
out.write(buffer,0,read);
}
out.close();
return out.toByteArray();
}
In Kotlin use
InputStream.readBytes()
Related
I'm trying to store an audio file that is picked by the user from his own music player into sqlite database and I want to know is there a way to convert audio files to byte array.
String path = ""; // Audio File path
InputStream inputStream = new FileInputStream(path);
byte[] arr = readByte(inputStream);
Log.d("byte: ", "" + Arrays.toString(arr));
or
public static byte[] getBytesFromInputStream(InputStream is) throws IOException {
ByteArrayOutputStream os = new ByteArrayOutputStream();
byte[] buffer = new byte[0xFFFF];
for (int len = is.read(buffer); len != -1; len = is.read(buffer)) {
os.write(buffer, 0, len);
}
return os.toByteArray();
}
try {
String path = ""; // Audio File path
InputStream inputStream = new FileInputStream(path);
byte[] myByteArray = getBytesFromInputStream(inputStream);
// ...
} catch(IOException e) {
// Handle error...
}
I'm using an InjectCSS script to use an extra css file on a webview.
But the script takes the CSS file from the assets folder, I want the css file externally hosted.
private void injectCSS() {
try {
InputStream inputStream = getAssets().open("style.css");
byte[] buffer = new byte[inputStream.available()];
inputStream.read(buffer);
inputStream.close();
String encoded = Base64.encodeToString(buffer, Base64.NO_WRAP);
wv.loadUrl("javascript:(function() {" +
"var parent = document.getElementsByTagName('head').item(0);" +
"var style = document.createElement('style');" +
"style.type = 'text/css';" +
// Tell the browser to BASE64-decode the string into your script !!!
"style.innerHTML = window.atob('" + encoded + "');" +
"parent.appendChild(style)" +
"})();");
} catch (Exception e) {
e.printStackTrace();
}
}
I tried to change the inputstream to url but that didnt work.
InputStream inputSteam = new URL("http://www.website.com/folder/style.css").openStream();
If you need to use BASE64-encode, you need this
InputStream inputSteam = new URL("http://www.website.com/folder/style.css").openStream();
String encoded = new String(readBytes(inputStream), Base64.NO_WRAP);
// ...
public byte[] readBytes(InputStream inputStream) throws IOException {
// this dynamically extends to take the bytes you read
ByteArrayOutputStream byteBuffer = new ByteArrayOutputStream();
// this is storage overwritten on each iteration with bytes
int bufferSize = 1024;
byte[] buffer = new byte[bufferSize];
// we need to know how may bytes were read to write them to the byteBuffer
int len = 0;
while ((len = inputStream.read(buffer)) != -1) {
byteBuffer.write(buffer, 0, len);
}
// and then we can return your byte array.
return byteBuffer.toByteArray();
}
I am trying to compress a large string object. This is what i tried, but i am unable to understand how to get compressed data, and how to define different type of compression tools.
This is what i got from Android docs.
byte[] input = jsonArray.getBytes("UTF-8");
byte[] output = new byte[100];
Deflater compresser = new Deflater();
compresser.setInput(input);
compresser.finish();
int compressedDataLength = compresser.deflate(output);
compresser.end();
compresser.deflate(output) gives me a int number, 100
but i am unable to understand which method will give me the compressed output that i can send to service.
The algorithm that I compress my data with is Huffman. You can find it by a simple search. But in your case maybe it helps you:
public static byte[] compress(String data) throws IOException {
ByteArrayOutputStream bos = new ByteArrayOutputStream(data.length());
GZIPOutputStream gzip = new GZIPOutputStream(bos);
gzip.write(data.getBytes());
gzip.close();
byte[] compressed = bos.toByteArray();
bos.close();
return compressed;
}
And to decompress it you can use:
public static String decompress(byte[] compressed) throws IOException {
ByteArrayInputStream bis = new ByteArrayInputStream(compressed);
GZIPInputStream gis = new GZIPInputStream(bis);
BufferedReader br = new BufferedReader(new InputStreamReader(gis, "UTF-8"));
StringBuilder sb = new StringBuilder();
String line;
while((line = br.readLine()) != null) {
sb.append(line);
}
br.close();
gis.close();
bis.close();
return sb.toString();
}
The documentation for Deflator shows that the output gets put into the buffer output
try {
// Encode a String into bytes
String inputString = "blahblahblah";
byte[] input = inputString.getBytes("UTF-8");
// Compress the bytes
byte[] output = new byte[100];
Deflater compresser = new Deflater();
compresser.setInput(input);
compresser.finish();
int compressedDataLength = compresser.deflate(output);
compresser.end();
// Decompress the bytes
Inflater decompresser = new Inflater();
decompresser.setInput(output, 0, compressedDataLength);
byte[] result = new byte[100];
int resultLength = decompresser.inflate(result);
decompresser.end();
// Decode the bytes into a String
String outputString = new String(result, 0, resultLength, "UTF-8");
} catch(java.io.UnsupportedEncodingException ex) {
// handle
} catch (java.util.zip.DataFormatException ex) {
// handle
}
all code you need to ENCODE, COMPRESS , DECOMPRESS , DECODE
I'm new at android development and I'm creating simple bluetooth app that can receive xml file and save xml file values to database. But how can I receive xml file from bytes array? Is it possible? After searchinf I found this question and based ont that question I try to save byte array to file. But how I need to test it? I can't find my file in my phone.
case Constants.MESSAGE_READ:
byte[] readBuffer = (byte[]) msg.obj;
try {
String path = activity.getFilesDir() + "/myFile.xml";
Log.d("MuTestClass", path);
FileOutputStream stream = new FileOutputStream(path);
stream.write(readBuffer);
stream.close();
} catch (Exception e1) {
e1.printStackTrace();
}
break;
You can use:
class Utils{
public static InputStream openFile(String filename) throws IOException{
AssetManager assManager = getApplicationContext().getAssets();
InputStream is = null;
is = assManager.open(filename);
return new BufferedInputStream(is);
}
public static byte[] readBytes(InputStream inputStream) throws IOException {
ByteArrayOutputStream byteBuffer = new ByteArrayOutputStream();
int bufferSize = 1024;
byte[] buffer = new byte[bufferSize];
int len = 0;
while ((len = inputStream.read(buffer)) != -1) {
byteBuffer.write(buffer, 0, len);
}
return byteBuffer.toByteArray();
}
}
like this:
try {
Utils.readBytes(Utils.openFile("something.xml"));
} catch (IOException e) {
e.printStackTrace();
}
I am trying to convert a file from the sdcard to Base64 but it seems the file is too big and i get an OutOfMemoryError.
Here is my code :
InputStream inputStream = null;//You can get an inputStream using any IO API
inputStream = new FileInputStream(file.getAbsolutePath());
byte[] bytes;
byte[] buffer = new byte[8192];
int bytesRead;
ByteArrayOutputStream output = new ByteArrayOutputStream();
try {
while ((bytesRead = inputStream.read(buffer)) != -1) {
output.write(buffer, 0, bytesRead);
}
} catch (IOException e) {
e.printStackTrace();
}
bytes = output.toByteArray();
attachedFile = Base64.encodeToString(bytes, Base64.DEFAULT);
Is there a way to go around the OutOfMemoryError while filing the String attachedFile ?
Base64 encoding takes 3 input bytes and converts them to 4 bytes. So if you have 100 Mb file that will end up to be 133 Mb in Base64. When you convert it to Java string (UTF-16) it size will be doubled. Not to mention that during conversion process at some point you will hold multiple copies in memory. No matter how you turn this it is hardly going to work.
This is slightly more optimized code that uses Base64OutputStream and will need less memory than your code, but I would not hold my breath. My advice would be to improve that code further by skipping conversion to string, and using temporary file stream as output instead of ByteArrayOutputStream.
InputStream inputStream = null;//You can get an inputStream using any IO API
inputStream = new FileInputStream(file.getAbsolutePath());
byte[] buffer = new byte[8192];
int bytesRead;
ByteArrayOutputStream output = new ByteArrayOutputStream();
Base64OutputStream output64 = new Base64OutputStream(output, Base64.DEFAULT);
try {
while ((bytesRead = inputStream.read(buffer)) != -1) {
output64.write(buffer, 0, bytesRead);
}
} catch (IOException e) {
e.printStackTrace();
}
output64.close();
attachedFile = output.toString();
// Converting File to Base64.encode String type using Method
public String getStringFile(File f) {
InputStream inputStream = null;
String encodedFile = "", lastVal;
try {
inputStream = new FileInputStream(f.getAbsolutePath());
byte[] buffer = new byte[10240]; //specify the size to allow
int bytesRead;
ByteArrayOutputStream output = new ByteArrayOutputStream();
Base64OutputStream output64 = new Base64OutputStream(output, Base64.DEFAULT);
while ((bytesRead = inputStream.read(buffer)) != -1) {
output64.write(buffer, 0, bytesRead);
}
output64.close();
encodedFile = output.toString();
} catch (FileNotFoundException e1) {
e1.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
lastVal = encodedFile;
return lastVal;
}