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Spring rest upload file is corrupted

I have a java application which uses spring rest to upload a jar file.But the uploaded file is corrupted and I am not able to access the jar file from the server.Please help.
fileloc = fileloc.replace("$", "/");
String filename = uploadedFileRef.getOriginalFilename();
String path = fileloc + filename;
byte[] buffer = new byte[1000];
File outputFile = new File(path);
FileInputStream reader = null;
FileOutputStream writer = null;
int totalBytes = 0;
try {
outputFile.createNewFile();
reader = (FileInputStream) uploadedFileRef.getInputStream();
writer = new FileOutputStream(outputFile);
int bytesRead = 0;
while ((bytesRead = reader.read(buffer)) != -1) {
writer.write(buffer);
totalBytes += bytesRead;
}
} catch (IOException e) {
e.printStackTrace();
} finally {
try {
reader.close();
writer.close();
} catch (IOException e) {
e.printStackTrace();
}
}

You call
writer.write(buffer);
so you write always the buffer size. Imagine the last read reads 10 bytes only but you will write 1000 bytes anyway.
Use
write(buffer, 0, bytesRead );
Or just check the question

Output file is getting large size than input (original) file

I am working on reading a file and write same file, but the problem is the downloaded file is 2kb larger than input original file.
Some piece of code
#Override
public void run() {
try {
BufferedInputStream bis;
ArrayList<byte[]> al =new ArrayList<byte[]>();
File file = new File(Environment.getExternalStorageDirectory(), "test.mp3");
byte[] bytes = new byte[2048];
bis = new BufferedInputStream(new FileInputStream(file));
OutputStream os = socket.getOutputStream();
int read ;
int fileSize = (int) file.length();
int readlen=1024;
while (fileSize>0) {
if(fileSize<1024){
readlen=fileSize;
System.out.println("Hello.........");
}
bytes=new byte[readlen];
read = bis.read(bytes, 0, readlen);
fileSize-=read;
al.add(bytes);
}
ObjectOutputStream out1 = new ObjectOutputStream(new FileOutputStream(Environment.getExternalStorageDirectory()+"/newfile.mp3"));
for(int ii=1;ii<al.size();ii++){
out1.write(al.get(ii));
// out1.flush();
}
out1.close();
File file1 = new File(Environment.getExternalStorageDirectory(), "newfile.mp3");

Don't use an ObjectOutputStream. Just use the FileOutputStream, or a BufferedOutputStream wrapped around it.
The correct way to copy streams in Java is as follows:
byte[] buffer = new byte[8192]; // or more, or even less, anything > 0
int count;
while ((count = in.read(buffer)) > 0)
{
out.write(buffer, 0, count);
}
out.close();
Note that you don't need a buffer the size of the input, and you don't need to read the entire input before writing any of the output.
Wish I had $1 for every time I've posted this.

I think you should use ByteArrayOutputStream not an ObjectOutputStream.
I belive this is not a raw code, but the parts of the code, placed in different procedures, otherwise it is meaningless.
For example, in case you want to stream some data from a file, process this data, and then write the data to another file.
BufferedInputStream bis = null;
ByteArrayOutputStream al = new ByteArrayOutputStream();
FileOutputStream out1 = null;
byte[] bytes;
try {
File file = new File("testfrom.mp3");
bis = new BufferedInputStream(new FileInputStream(file));
int fileSize = (int) file.length();
int readLen = 1024;
bytes = new byte[readLen];
while (fileSize > 0) {
if (fileSize < readLen) {
readLen = fileSize;
}
bis.read(bytes, 0, readLen);
al.write(bytes, 0, readLen);
fileSize -= readLen;
}
bis.close();
} catch (IOException e){
e.printStackTrace();
}
//proceed the data from al here
//...
//finish to proceed
try {
out1 = new FileOutputStream("testto.mp3");
al.writeTo(out1);
out1.close();
} catch (IOException e){
e.printStackTrace();
}
Don't forget to use try-catch directives where it needed
http://codeinventions.blogspot.ru/2014/08/creating-file-from-bytearrayoutputstrea.html

Optimizing Ungunzipping process in Java?

I have a problem with the time it requires for .gz files to get uncompressed using the following code:
import java.io.*;
import java.util.zip.*;
public class UnGunzipClass{
public static boolean ungunzip(String compressedFile, String decompressedFile){
try{
// in
FileInputStream fileIn = new FileInputStream(compressedFile);
GZIPInputStream gZipIn = new GZIPInputStream(fileIn);
BufferedInputStream in = new BufferedInputStream(gZipIn);
// out
FileOutputStream fileOut = new FileOutputStream(decompressedFile);
BufferedOutputStream out = new BufferedOutputStream(fileOut);
int n = 0;
int len = 1024*1024*1024;
byte[] buffer = new byte[len];
while((n = in.read(buffer,0,len)) > 0){
out.write(buffer,0,n);
}
gZipIn.close();
fileOut.close();
return true;
} catch (IOException e){
e.printStackTrace();
return false;
}
}
}
Note: files are up to 100MB, but it is still taking me tens of minutes to run, so I am trying to get something faster. Speed is good :)

You created your BufferedInputStream from the GZIPInputStream, for performance you would do that in the reverse order. Also, I suggest you shrink your buffer size (and use your buffered streams). Finally, I would use a try-with-resources Statement and that might look something like
public static boolean ungunzip(String compressedFile,
String decompressedFile) {
final int BUFFER_SIZE = 32768;
try (InputStream in = new GZIPInputStream(new BufferedInputStream(
new FileInputStream(compressedFile)), BUFFER_SIZE);
OutputStream out = new BufferedOutputStream(
new FileOutputStream(decompressedFile), BUFFER_SIZE)) {
int n = 0;
byte[] buffer = new byte[BUFFER_SIZE];
while ((n = in.read(buffer, 0, BUFFER_SIZE)) > 0) {
out.write(buffer, 0, n);
}
return true;
} catch (IOException e) {
e.printStackTrace();
}
return false;
}

How I can divide a file to blocks in java?

I try make code by java able to read 100 kb from file then divide the file into blocks each block has 256 bit or 32 byte also I want conver each block to binary format or integer format the following is it
I need any suggest
public static void main(String[] args) {
ReadFileExample newclass = new ReadFileExample();
System.out.println("-----------Wellcome in ECC ENCRYPTION NEW--------");
File clearmsg = new File("F:/java_projects/clearmsg.txt");
File ciphermsg = new File("F:/java_projects/ciphermsg.txt");
byte[] block = new byte[32];
try {
FileInputStream fis = new FileInputStream(clearmsg);
FileOutputStream fos = new FileOutputStream(ciphermsg);
CipherOutputStream cos = new CipherOutputStream(fos);
System.out.println("Total file size to read (in bytes) : "
+ fis.available());
int i;
while ((i = fis.read(block)) != -1) {
System.out.println(block);
fos.write(block, 0, i);
}
fos.close();

A better way to convert a directory of files into bytes

I have been messing with this for some time and it's getting better and better, but it's still a little slow for me. Can anyone help speed this up / make the design better, please?
Also, the files must only be numbers and the file must end with the file extension ".dat"
I never added the checks because I didn't feel is was necessary.
public void preloadModels() {
try {
File directory = new File(signlink.findcachedir() + "raw", File.separator);
File[] modelFiles = directory.listFiles();
for (int modelIndex = modelFiles.length - 1;; modelIndex--) {
String modelFileName = modelFiles[modelIndex].getName();
byte[] buffer = getBytesFromInputStream(new FileInputStream(new File(directory, modelFileName)));
Model.method460(buffer, Integer.parseInt(modelFileName.replace(".dat", "")));
}
} catch (Throwable e) {
return;
}
}
public static final byte[] getBytesFromInputStream(InputStream inputStream) throws IOException {
byte[] buffer = new byte[32 * 1024];
int bufferSize = 0;
for (;;) {
int read = inputStream.read(buffer, bufferSize, buffer.length - bufferSize);
if (read == -1) {
return Arrays.copyOf(buffer, bufferSize);
}
bufferSize += read;
if (bufferSize == buffer.length) {
buffer = Arrays.copyOf(buffer, bufferSize * 2);
}
}
}

I would do the following.
public void preloadModels() throws IOException {
File directory = new File(signlink.findcachedir() + "raw");
for (File file : directory.listFiles()) {
if (!file.getName().endsWith(".dat")) continue;
byte[] buffer = getBytesFromFile(file);
Model.method460(buffer, Integer.parseInt(file.getName().replace(".dat", "")));
}
}
public static byte[] getBytesFromFile(File file) throws IOException {
byte[] buffer = new byte[(int) file.length()];
try (DataInputStream dis = new DataInputStream(new FileInputStream(file))) {
dis.readFully(buffer);
return buffer;
}
}
If this is still too slow, most likely the limitation is the speed of hard drive.

How about using Apache Commons IOUtils class.
IOUtils.toByteArray(InputStream input)

I think the easiest way is to add all directories content to archive. Have a look at java.util.zip. It has some bugs with file names before 7th version. There is also Apache Commons implementation.