I just want to create a File object like this
File myImageFile = new File ("image1") ;
but it is giving me exception of FileNotFoundException
How can i reference a file inside my raw Folder
EDIT:
Actually i wanted to do something like this
MultipartEntity multipartEntity= new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE);
multipartEntity.addPart("uploaded", new FileBody(new File("myimage")));
Generally you access the files through getResources().openRawResource(R.id._your_id). If you absolutely need a File reference to it, one option is to copy it over to internal storage:
File file = new File(this.getFilesDir() + File.separator + "DefaultProperties.xml");
try {
InputStream inputStream = resources.openRawResource(R.id._your_id);
FileOutputStream fileOutputStream = new FileOutputStream(file);
byte buf[]=new byte[1024];
int len;
while((len=inputStream.read(buf))>0) {
fileOutputStream.write(buf,0,len);
}
fileOutputStream.close();
inputStream.close();
} catch (IOException e1) {}
Now you have a File that you can access anywhere you need it.
here are 2 functions. one to read from RAW and one from the Assets
/**
* Method to read in a text file placed in the res/raw directory of the
* application. The method reads in all lines of the file sequentially.
*/
public static void readRaw(Context ctx,int res_id) {
InputStream is = ctx.getResources().openRawResource(res_id);
InputStreamReader isr = new InputStreamReader(is);
BufferedReader br = new BufferedReader(isr, 8192); // 2nd arg is buffer
// size
// More efficient (less readable) implementation of above is the
// composite expression
/*
* BufferedReader br = new BufferedReader(new InputStreamReader(
* this.getResources().openRawResource(R.raw.textfile)), 8192);
*/
try {
String test;
while (true) {
test = br.readLine();
// readLine() returns null if no more lines in the file
if (test == null)
break;
}
isr.close();
is.close();
br.close();
} catch (IOException e) {
e.printStackTrace();
}
}
and from Assets folder
/**
* Read a file from assets
*
* #return the string from assets
*/
public static String getQuestions(Context ctx,String file_name) {
AssetManager assetManager = ctx.getAssets();
ByteArrayOutputStream outputStream = null;
InputStream inputStream = null;
try {
inputStream = assetManager.open(file_name);
outputStream = new ByteArrayOutputStream();
byte buf[] = new byte[1024];
int len;
try {
while ((len = inputStream.read(buf)) != -1) {
outputStream.write(buf, 0, len);
}
outputStream.close();
inputStream.close();
} catch (IOException e) {
}
} catch (IOException e) {
}
return outputStream.toString();
}
You can open it as InputStream, I don't know if possible as a file:
int rid = resources.getIdentifier(packageName + ":raw/" + fileName, null, null);
//get the file as a stream
InputStrea is = resources.openRawResource(rid);
You can use InputStreamBody instead of FileBody so you can use it like this:
InputStream inputStream = resources.openRawResource(R.raw.yourresource);
MultipartEntity multipartEntity= new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE);
multipartEntity.addPart("uploaded", new InputStreamBody(inputStream));
Related
I am trying to download a PDF file with HttpClient, it is downloading the PDF file but pages are blank. I can see the bytes on console from response if I print them. But when I try to write it to file it is producing a blank file.
FileUtils.writeByteArrayToFile(new File(outputFilePath), bytes);
However the file is showing correct size of 103KB and 297KB as expected but its just blank!!
I tried with Output stream as well like:
FileOutputStream fileOutputStream = new FileOutputStream(outFile);
fileOutputStream.write(bytes);
Also tried to write with UTF-8 coding like:
Writer out = new BufferedWriter( new OutputStreamWriter(
new FileOutputStream(outFile), "UTF-8"));
String str = new String(bytes, StandardCharsets.UTF_8);
try {
out.write(str);
} finally {
out.close();
}
Nothing is working for me. Any suggestion is highly appreciated..
Update: I am using DefaultHttpClient.
HttpGet httpget = new HttpGet(targetURI);
HttpResponse response = null;
String htmlContents = null;
try {
httpget = new HttpGet(url);
response = httpclient.execute(httpget);
InputStreamReader dataStream=new InputStreamReader(response.getEntity().getContent());
byte[] bytes = IOUtils.toByteArray(dataStream);
...
You do
InputStreamReader dataStream=new InputStreamReader(response.getEntity().getContent());
byte[] bytes = IOUtils.toByteArray(dataStream);
As has already been mentioned in comments, using a Reader class can damage binary data, e.g. PDF files. Thus, you should not wrap your content in an InputStreamReader.
As your content can be used to construct an InputStreamReader, though, I assume response.getEntity().getContent() returns an InputStream. Such an InputStream usually can be directly used as IOUtils.toByteArray argument.
So:
InputStream dataStream=response.getEntity().getContent();
byte[] bytes = IOUtils.toByteArray(dataStream);
should already work for you!
Here is a method I use to download a PDF file from a specific URL. The method requires two string arguments, an url string (example: "https://www.ibm.com/support/knowledgecenter/SSWRCJ_4.1.0/com.ibm.safos.doc_4.1/Planning_and_Installation.pdf") and a destination folder path to download the PDF file (or whatever) into. If the destination path does not exist within the local file system then it is automatically created:
public boolean downloadFile(String urlString, String destinationFolderPath) {
boolean result = false; // will turn to true if download is successful
if (!destinationFolderPath.endsWith("/") && !destinationFolderPath.endsWith("\\")) {
destinationFolderPath+= "/";
}
// If the destination path does not exist then create it.
File foldersToMake = new File(destinationFolderPath);
if (!foldersToMake.exists()) {
foldersToMake.mkdirs();
}
try {
// Open Connection
URL url = new URL(urlString);
// Get just the file Name from URL
String fileName = new File(url.getPath()).getName();
// Try with Resources....
try (InputStream in = url.openStream(); FileOutputStream outStream =
new FileOutputStream(new File(destinationFolderPath + fileName))) {
// Read from resource and write to file...
int length = -1;
byte[] buffer = new byte[1024]; // buffer for portion of data from connection
while ((length = in.read(buffer)) > -1) {
outStream.write(buffer, 0, length);
}
}
// File Successfully Downloaded");
result = true;
}
catch (MalformedURLException ex) { ex.printStackTrace(); }
catch (IOException ex) { ex.printStackTrace(); }
return result;
}
I use NanoHTTPD as web server in my Android APP, I hope to compress some files and create a InputStream in server side, and I download the InputStream in client side using Code A.
I have read Code B at How to zip and unzip the files?, but how to create a ZIP InputStream in Android without creating a ZIP file first?
BTW, I don't think Code C is good way, because it make ZIP file first, then convert ZIP file to FileInputStream , I hope to create a ZIP InputStream directly!
Code A
private Response ActionDownloadSingleFile(InputStream fis) {
Response response = null;
response = newChunkedResponse(Response.Status.OK, "application/octet-stream",fis);
response.addHeader("Content-Disposition", "attachment; filename="+"my.zip");
return response;
}
Code B
public static void zip(String[] files, String zipFile) throws IOException {
BufferedInputStream origin = null;
ZipOutputStream out = new ZipOutputStream(new BufferedOutputStream(new FileOutputStream(zipFile)));
try {
byte data[] = new byte[BUFFER_SIZE];
for (int i = 0; i < files.length; i++) {
FileInputStream fi = new FileInputStream(files[i]);
origin = new BufferedInputStream(fi, BUFFER_SIZE);
try {
ZipEntry entry = new ZipEntry(files[i].substring(files[i].lastIndexOf("/") + 1));
out.putNextEntry(entry);
int count;
while ((count = origin.read(data, 0, BUFFER_SIZE)) != -1) {
out.write(data, 0, count);
}
}
finally {
origin.close();
}
}
}
finally {
out.close();
}
}
Code C
File file= new File("my.zip");
FileInputStream fis = null;
try
{
fis = new FileInputStream(file);
} catch (FileNotFoundException ex)
{
}
ZipInputStream as per the documentation ZipInputStream
ZipInputStream is an input stream filter for reading files in the ZIP file format. Includes support for both compressed and uncompressed entries.
Earlier I answered to this question in a way that it is not possible using ZipInputStream. I am Sorry.
But after investing some time I found that it is possible as per the below code
It is very much obvious that since you are sending files in zip format
over the network.
//Create proper background thread pool. Not best but just for solution
new Thread(new Runnable() {
#Override
public void run() {
// Moves the current Thread into the background
android.os.Process.setThreadPriority(android.os.Process.THREAD_PRIORITY_BACKGROUND);
HttpURLConnection httpURLConnection = null;
byte[] buffer = new byte[2048];
try {
//Your http connection
httpURLConnection = (HttpURLConnection) new URL("https://s3-ap-southeast-1.amazonaws.com/uploads-ap.hipchat.com/107225/1251522/SFSCjI8ZRB7FjV9/zvsd.zip").openConnection();
//Change below path to Environment.getExternalStorageDirectory() or something of your
// own by creating storage utils
File outputFilePath = new File ("/mnt/sdcard/Android/data/somedirectory/");
ZipInputStream zipInputStream = new ZipInputStream(new BufferedInputStream(httpURLConnection.getInputStream()));
ZipEntry zipEntry = zipInputStream.getNextEntry();
int readLength;
while(zipEntry != null){
File newFile = new File(outputFilePath, zipEntry.getName());
if (!zipEntry.isDirectory()) {
FileOutputStream fos = new FileOutputStream(newFile);
while ((readLength = zipInputStream.read(buffer)) > 0) {
fos.write(buffer, 0, readLength);
}
fos.close();
} else {
newFile.mkdirs();
}
Log.i("zip file path = ", newFile.getPath());
zipInputStream.closeEntry();
zipEntry = zipInputStream.getNextEntry();
}
// Close Stream and disconnect HTTP connection. Move to finally
zipInputStream.closeEntry();
zipInputStream.close();
} catch (IOException e) {
e.printStackTrace();
}finally {
// Close Stream and disconnect HTTP connection.
if (httpURLConnection != null) {
httpURLConnection.disconnect();
}
}
}
}).start();
This is my test program. I need it to apply somewhere.This may be small, sorry for that. But I'm a starter still. So kindly help me.
try{
File file1 = new File("c:\\Users\\prasad\\Desktop\\bugatti.jpg");
File file2 = new File("c:\\Users\\prasad\\Desktop\\hello.jpg");
file2.createNewFile();
BufferedReader reader = new BufferedReader(new InputStreamReader(new FileInputStream(file1)));
String data = null;
StringBuilder imageBuild = new StringBuilder();
while((data = reader.readLine())!=null){
imageBuild.append(data);
}
reader.close();
BufferedWriter writer = new BufferedWriter(new PrintWriter(new FileOutputStream(file2)));
writer.write(imageBuild.toString());
writer.close();
}catch(IOException e){
e.printStackTrace();
}
This is file1
and This is file2
You can do either of these two:
private static void copyFile(File source, File dest) throws IOException {
Files.copy(source.toPath(), dest.toPath());
}
or maybe this if you want to use streams:
private static void copyFile(File source, File dest)
throws IOException {
InputStream input = null;
OutputStream output = null;
try {
input = new FileInputStream(source);
output = new FileOutputStream(dest);
byte[] buf = new byte[1024];
int bytesRead;
while ((bytesRead = input.read(buf)) > 0) {
output.write(buf, 0, bytesRead);
}
} finally {
input.close();
output.close();
}
}
Images do not contain lines or even characters. You therefore should not be using readLine() or even Readers or Writers. You should rewrite the copy loop using input and output streams directly.
Right now I am working on a task to convert binary data in to a zip file
I am calling a url and getting a response from server like
A#B�ArE⏾�7�ϫ���f�걺N�����Yg���o_M^�D�T�U X_���e?� hi\ � �ڂ(� �0 rm��'�ed���� �:6h�k�ڗ� ���fnp���7��)��:��N�U�viR�,) II����M��Np�M��7��
n��
!A!) )AAFAq)Q)�y
y� ��.�����?���
��֞��ͅ��Ɲ_�O�����nc��f��w��ʰ�6��3 2�ƢZZ��N0� O{� mC� ��$��,>����������
���CW/)?�?٥��ߗ�d�=�R�J*E{2L���ח�W���ӑ_PRR�_#�_H��:������Ə�Ջ�J�^v�0wo��+�o���
�-Ä#�R6��P�(���0�WPj�k�
C�E
now I want to save this data to zip file i have searched a lot and find some links but not meet the goal.
here i have done
OutputStreamWriter osw = new OutputStreamWriter(openFileOutput(
"products.zip", Context.MODE_PRIVATE));
osw.write(data);
osw.close();
please guid me if you have any idea about this.
OutputStreamWriter osw
NO!
A Writer is made to write text, not binary.
In the first place, it looks like you read text as well, which you shouldn't.
Use an InputStream to read the original content, and an OutputStream to write into the file:
final OutputStream out = /* open your file using a FileOutputStream here */;
final byte[] buf = new byte[8096]; // size as appropriate
// "in" is the InputStream from the socket
int count;
try {
while ((count = in.read(buf)) != -1)
out.write(buf, 0, count);
out.flush();
} finally {
out.close();
}
Readers are not meant to read octet streams.
Reads text from a character-input stream, buffering characters so as to provide for the efficient reading of characters, arrays, and lines.
You're looking for a BufferedInputStream.
The getContent() method on the HttpEntity returns an InputStream. Wrap this around a BufferedInputStream and write it to a file or a ByteArrayOutputStream.
byte[] buffer = new byte[5 * 1024];
int numRead = -1;
while( (numRead = bufferedInputStream.read(buffer))!= -1)
{
byteArrayOutputStream.write(buffer, 0, numRead);
}
byteArrayOutputStream.flush();
byteArrayOutputStream.close();
byte[] result = byteArrayOutputStream.toByteArray();
To save on memory I'd advise you to write to a BufferedOutputStream instead of trying to get the bytes from the stream into a data structure. The android device is likely to run out of memory for large zip files.
You might be able to avoid the trouble of converting Binary by adding options to your request.
In NodeJS I specify the responseType as arraybuffer
const res = await axios.get('/routToThat/file', {
headers: {
Accept: 'application/zip',
},
responseType: 'arraybuffer',
});
So instead of receiving the server response as a Binary string:
A#B�ArE⏾�7�ϫ���f�걺N�����Yg���o_M^�D�T�U X_���e?� hi\...
I get a Buffer:
Buffer(22781691) [80, 75, 3, …]
NOTE: In my case the response I get is already a ZIP file.
More details on this NodeJS answer.
https://stackoverflow.com/a/62460311/3645464
All I need to do is construct a BufferedInputStream from the entity. Just replace BufferedReader with a BufferedInputStream. I would recommend using ISO-8859-1 .An underlying streaming encoder to read binary data is a waste of processing power.
private class methodName extends
AsyncTask<String, Integer, byte[]> {
#Override
protected byte[] doInBackground(String... params) {
String uri = params[0];
try {
MultipartEntityBuilder entity;
File f;
FileBody fb;
entity = MultipartEntityBuilder.create();
entity.setMode(HttpMultipartMode.BROWSER_COMPATIBLE);
f = new File(zipImageFile);
fb = new FileBody(f);
entity.addPart("orderFile", fb);
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(uri);
Log.e("Uploload Missing Image URL", "" + uri);
httppost.setEntity(entity.build());
HttpResponse response = httpclient.execute(httppost);
// byte[] fileBites=null;
BufferedInputStream bufferedInputStream;
ByteArrayOutputStream byteArrayOutputStream;
byte[] buffer = new byte[5 * 1024];
int numRead = -1;
while( (numRead = bufferedInputStream.read(buffer))!= -1)
{
byteArrayOutputStream.write(buffer, 0, numRead);
}
byteArrayOutputStream.flush();
byteArrayOutputStream.close();
byte[] result = byteArrayOutputStream.toByteArray();
// fileBites=stringBuffer.toString().getBytes();
// Log.e("FILE BITES", fileBites+"=>"+fileBites.length);
return ;
// return stringBuffer.toString();
} catch (Exception e) {
return e.toString().getBytes();
}
}
#Override
protected void onPostExecute(byte[] result) {
// TODO Auto-generated method stub
super.onPostExecute(result);
Log.e("Response From Server", "" + result);
writeToFile(result);
}
}
private void writeToFile(byte[] data) {
try {
FileOutputStream fop = null;
File file;
file = new File(AppConstants.DataPath+"/products.zip");
fop = new FileOutputStream(file);
// if file doesnt exists, then create it
if (!file.exists()) {
file.createNewFile();
}
try {
fop.write(data);
} catch (IOException e) {
Log.e("Exception", "File write failed: " + e.toString());
}
unzipImage(AppConstants.DataPath + "/products.zip",
AppConstants.DataPath);
}catch (Exception E)
{
}
}
i have to retrieve some data from a txt file and then show those data inside my app.
My problem is that if i have the special char 'ø' inside my txt, this is not shown and a '?' is shown instead.
i tried to check data like
if(string.charAt(i) == 'ø') do sth
or
string.replace('ø' , 'O')
but none of them is working and i think that Java could not recognize that char at all.
Do you have any idea?
thanks
edit
this is how i read data
String[] obj = getText(getActivity(), myTXT.txt").split("\n");
where getText is:
public String getText(Context c, String fileName){
ByteArrayOutputStream outputStream = null;
try {
AssetManager am = c.getAssets();
InputStream is = am.open(fileName);
outputStream = new ByteArrayOutputStream();
byte buf[] = new byte[1024];
int len;
while ((len = is.read(buf)) != -1){
outputStream.write(buf,0,len);
}
outputStream.close();
is.close();
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
return outputStream.toString();
}
These chars must be in UTF-8 encoding, check for your file while its getting saved whether its incoding. Create an InputStreamReader instance that uses the constructor specifying encoding.
InputStreamReader r= new InputStreamReader(new FileInputStream(myFile),"UTF-8");
// read your contents here.
r.close();