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Unzip content from zipped file in memory [duplicate]

I must get file content from ZIP archive (only one file, I know its name) using SFTP. The only thing I'm having is ZIP's InputStream. Most examples show how get content using this statement:
ZipFile zipFile = new ZipFile("location");
But as I said, I don't have ZIP file on my local machine and I don't want to download it. Is an InputStream enough to read?
UPD: This is how I do:
import java.util.zip.ZipInputStream;
import com.jcraft.jsch.Channel;
import com.jcraft.jsch.ChannelSftp;
import com.jcraft.jsch.JSch;
import com.jcraft.jsch.Session;
public class SFTP {
public static void main(String[] args) {
String SFTPHOST = "host";
int SFTPPORT = 3232;
String SFTPUSER = "user";
String SFTPPASS = "mypass";
String SFTPWORKINGDIR = "/dir/work";
Session session = null;
Channel channel = null;
ChannelSftp channelSftp = null;
try {
JSch jsch = new JSch();
session = jsch.getSession(SFTPUSER, SFTPHOST, SFTPPORT);
session.setPassword(SFTPPASS);
java.util.Properties config = new java.util.Properties();
config.put("StrictHostKeyChecking", "no");
session.setConfig(config);
session.connect();
channel = session.openChannel("sftp");
channel.connect();
channelSftp = (ChannelSftp) channel;
channelSftp.cd(SFTPWORKINGDIR);
ZipInputStream stream = new ZipInputStream(channelSftp.get("file.zip"));
ZipEntry entry = zipStream.getNextEntry();
System.out.println(entry.getName); //Yes, I got its name, now I need to get content
} catch (Exception ex) {
ex.printStackTrace();
} finally {
session.disconnect();
channelSftp.disconnect();
channel.disconnect();
}
}
}

Below is a simple example on how to extract a ZIP File, you will need to check if the file is a directory. But this is the simplest.
The step you are missing is reading the input stream and writing the contents to a buffer which is written to an output stream.
// Expands the zip file passed as argument 1, into the
// directory provided in argument 2
public static void main(String args[]) throws Exception
{
if(args.length != 2)
{
System.err.println("zipreader zipfile outputdir");
return;
}
// create a buffer to improve copy performance later.
byte[] buffer = new byte[2048];
// open the zip file stream
InputStream theFile = new FileInputStream(args[0]);
ZipInputStream stream = new ZipInputStream(theFile);
String outdir = args[1];
try
{
// now iterate through each item in the stream. The get next
// entry call will return a ZipEntry for each file in the
// stream
ZipEntry entry;
while((entry = stream.getNextEntry())!=null)
{
String s = String.format("Entry: %s len %d added %TD",
entry.getName(), entry.getSize(),
new Date(entry.getTime()));
System.out.println(s);
// Once we get the entry from the stream, the stream is
// positioned read to read the raw data, and we keep
// reading until read returns 0 or less.
String outpath = outdir + "/" + entry.getName();
FileOutputStream output = null;
try
{
output = new FileOutputStream(outpath);
int len = 0;
while ((len = stream.read(buffer)) > 0)
{
output.write(buffer, 0, len);
}
}
finally
{
// we must always close the output file
if(output!=null) output.close();
}
}
}
finally
{
// we must always close the zip file.
stream.close();
}
}
Code excerpt came from the following site:
http://www.thecoderscorner.com/team-blog/java-and-jvm/12-reading-a-zip-file-from-java-using-zipinputstream#.U4RAxYamixR

Well, I've done this:
zipStream = new ZipInputStream(channelSftp.get("Port_Increment_201405261400_2251.zip"));
zipStream.getNextEntry();
sc = new Scanner(zipStream);
while (sc.hasNextLine()) {
System.out.println(sc.nextLine());
}
It helps me to read ZIP's content without writing to another file.

The ZipInputStream is an InputStream by itself and delivers the contents of each entry after each call to getNextEntry(). Special care must be taken, not to close the stream from which the contents is read, since it is the same as the ZIP stream:
public void readZipStream(InputStream in) throws IOException {
ZipInputStream zipIn = new ZipInputStream(in);
ZipEntry entry;
while ((entry = zipIn.getNextEntry()) != null) {
System.out.println(entry.getName());
readContents(zipIn);
zipIn.closeEntry();
}
}
private void readContents(InputStream contentsIn) throws IOException {
byte contents[] = new byte[4096];
int direct;
while ((direct = contentsIn.read(contents, 0, contents.length)) >= 0) {
System.out.println("Read " + direct + "bytes content.");
}
}
When delegating reading contents to other logic, it can be necessary to wrap the ZipInputStream with a FilterInputStream to close only the entry instead of the whole stream as in:
public void readZipStream(InputStream in) throws IOException {
ZipInputStream zipIn = new ZipInputStream(in);
ZipEntry entry;
while ((entry = zipIn.getNextEntry()) != null) {
System.out.println(entry.getName());
readContents(new FilterInputStream(zipIn) {
#Override
public void close() throws IOException {
zipIn.closeEntry();
}
});
}
}

OP was close. Just need to read the bytes. The call to getNextEntry positions the stream at the beginning of the entry data (docs). If that's the entry we want (or the only entry), then the InputStream is in the right spot. All we need to do is read that entry's decompressed bytes.
byte[] bytes = new byte[(int) entry.getSize()];
int i = 0;
while (i < bytes.length) {
// .read doesn't always fill the buffer we give it.
// Keep calling it until we get all the bytes for this entry.
i += zipStream.read(bytes, i, bytes.length - i);
}
So if these bytes really are text, then we can decode those bytes to a String. I'm just assuming utf8 encoding.
new String(bytes, "utf8")
Side note: I personally use apache commons-io IOUtils to cut down on this kind of lower level stuff. The docs for ZipInputStream.read seem to imply that read will stop at the end of the current zip entry. If that is true, then reading the current textual entry is one line with IOUtils.
String text = IOUtils.toString(zipStream)

Unzip archive (zip) with preserving file structure into given directory.
Note; this code use deps on "org.apache.commons.io.IOUtils"), but you can replace it by yours custom 'read-stream' code
public static void unzipDirectory(File archiveFile, File destinationDir) throws IOException
{
Path destPath = destinationDir.toPath();
try (ZipInputStream zis = new ZipInputStream(new FileInputStream(archiveFile)))
{
ZipEntry zipEntry;
while ((zipEntry = zis.getNextEntry()) != null)
{
Path resolvedPath = destPath.resolve(zipEntry.getName()).normalize();
if (!resolvedPath.startsWith(destPath))
{
throw new IOException("The requested zip-entry '" + zipEntry.getName() + "' does not belong to the requested destination");
}
if (zipEntry.isDirectory())
{
Files.createDirectories(resolvedPath);
} else
{
if(!Files.isDirectory(resolvedPath.getParent()))
{
Files.createDirectories(resolvedPath.getParent());
}
try (FileOutputStream outStream = new FileOutputStream(resolvedPath.toFile()))
{
IOUtils.copy(zis, outStream);
}
}
}
}
}

Here a more generic solution to process a zip inputstream with a BiConsumer. It's nearly the same solution that was used by haui
private void readZip(InputStream is, BiConsumer<ZipEntry,InputStream> consumer) throws IOException {
try (ZipInputStream zipFile = new ZipInputStream(is);) {
ZipEntry entry;
while((entry = zipFile.getNextEntry()) != null){
consumer.accept(entry, new FilterInputStream(zipFile) {
#Override
public void close() throws IOException {
zipFile.closeEntry();
}
});
}
}
}
You can use it by just calling
readZip(<some inputstream>, (entry, is) -> {
/* don't forget to close this stream after processing. */
is.read() // ... <- to read each entry
});

If content of your ZIP consist of 1 file (for example, zipped content of HTTP response), you can read text content using Kotlin as follows:
#Throws(IOException::class)
fun InputStream.readZippedContent() = ZipInputStream(this).use { stream ->
stream.nextEntry?.let { stream.bufferedReader().readText() } ?: String()
}
This extension function unzips first ZIP entry of Zip file and read content as plain text.
Usage:
val inputStream: InputStream = ... // your zipped InputStream
val textContent = inputStream.readZippedContent()

Java write exe file

is it possible to write/create an exe file in Java?
I can successfully read it but writing the exact same data that has been read to a new file seems to create some trouble because Windows tell's me it's not supported for my pc anymore.
This is the code I'm using to read the file where path is a String given with the actual path (it's in the .jar itself that's why I'm using ResourceAsStream()):
try {
InputStream inputStream = FileIO.class.getResourceAsStream(path);
BufferedReader reader = new BufferedReader(new InputStreamReader(inputStream));
ArrayList<String> _final = new ArrayList<String>();
String line;
while ((line = reader.readLine()) != null) {
_final.add(line);
}
inputStream.close();
return _final.toArray(new String[_final.size()]);
}catch(Exception e) {
return null;
}
This is the code I'm using to write the file:
public static void writeFileArray(String path, String[] data) {
String filename = path;
try{
FileWriter fileWriter = new FileWriter(filename);
BufferedWriter bufferedWriter = new BufferedWriter(fileWriter);
for(String d : data) {
bufferedWriter.write(d + "\n");
}
bufferedWriter.close();
}
catch(IOException ex){
System.out.println("FileIO failed to write file, IO exception");
}
}
So it doesn't give me any error's or something and the file size of the original .exe and the 'transferred' .exe stays the same, but it doesn't work anymore. Am I just doing it wrong? Did I forget something? Can u even do this with Java?
Btw I'm not that experienced with reading/writing files..
Thanks for considering my request.

I'm going to guess that you're using a Reader when you should be using a raw input stream. Use BufferedInputStream instead of BufferedReader.
BufferedInputStream in = new BufferedInputStream( inputStream );
The problem is that Reader interprets the binary as your local character set instead of the data you want.
Edit: if you need a bigger hint start with this. I just noticed you're using a BufferedWriter too, that won't work either.
try {
InputStream inputStream = FileIO.class.getResourceAsStream(path);
BufferedInputStream in = new BufferedInputStream( inputStream );
ByteArrayOutputStream bos = new ByteArrayOutputStream();
byte[] bytes = new byte[ 1024 ];
for( int length; ( length = ins.read( bytes ) ) != -1; )
bos.write( bytes, 0, length );
}
inputStream.close();
return bos;

When you are using Java 7 or newer, you should copy a resource to a file using
public static void copyResourceToFile(String resourcePath, String filePath) {
try(InputStream inputStream = FileIO.class.getResourceAsStream(resourcePath)) {
Files.copy(inputStream, Paths.get(filePath));
}
catch(IOException ex){
System.out.println("Copying failed. "+ex.getMessage());
}
}
This construct ensures correct closing of the resources even in the exceptional case and the JRE method ensures correct and efficient copying of the data.
It accepts additional options, e.g. to specify that the target file should be overwritten in case it already exists, you would use
public static void copyResourceToFile(String resourcePath, String filePath) {
try(InputStream inputStream = FileIO.class.getResourceAsStream(resourcePath)) {
Files.copy(inputStream, Paths.get(filePath), StandardCopyOption.REPLACE_EXISTING);
}
catch(IOException ex){
System.out.println("Copying failed. "+ex.getMessage());
}
}

You are using InputStreams for strings, .exe files are bytes!
Try using a ByteArrayInputStream and ByteArrayOutputStream.
Edit: completing with markspace's answer:
new BufferedInputStream(new ByteArrayInputStream( ... ) )

How to read file from ZIP using InputStream?

I must get file content from ZIP archive (only one file, I know its name) using SFTP. The only thing I'm having is ZIP's InputStream. Most examples show how get content using this statement:
ZipFile zipFile = new ZipFile("location");
But as I said, I don't have ZIP file on my local machine and I don't want to download it. Is an InputStream enough to read?
UPD: This is how I do:
import java.util.zip.ZipInputStream;
import com.jcraft.jsch.Channel;
import com.jcraft.jsch.ChannelSftp;
import com.jcraft.jsch.JSch;
import com.jcraft.jsch.Session;
public class SFTP {
public static void main(String[] args) {
String SFTPHOST = "host";
int SFTPPORT = 3232;
String SFTPUSER = "user";
String SFTPPASS = "mypass";
String SFTPWORKINGDIR = "/dir/work";
Session session = null;
Channel channel = null;
ChannelSftp channelSftp = null;
try {
JSch jsch = new JSch();
session = jsch.getSession(SFTPUSER, SFTPHOST, SFTPPORT);
session.setPassword(SFTPPASS);
java.util.Properties config = new java.util.Properties();
config.put("StrictHostKeyChecking", "no");
session.setConfig(config);
session.connect();
channel = session.openChannel("sftp");
channel.connect();
channelSftp = (ChannelSftp) channel;
channelSftp.cd(SFTPWORKINGDIR);
ZipInputStream stream = new ZipInputStream(channelSftp.get("file.zip"));
ZipEntry entry = zipStream.getNextEntry();
System.out.println(entry.getName); //Yes, I got its name, now I need to get content
} catch (Exception ex) {
ex.printStackTrace();
} finally {
session.disconnect();
channelSftp.disconnect();
channel.disconnect();
}
}
}

Below is a simple example on how to extract a ZIP File, you will need to check if the file is a directory. But this is the simplest.
The step you are missing is reading the input stream and writing the contents to a buffer which is written to an output stream.
// Expands the zip file passed as argument 1, into the
// directory provided in argument 2
public static void main(String args[]) throws Exception
{
if(args.length != 2)
{
System.err.println("zipreader zipfile outputdir");
return;
}
// create a buffer to improve copy performance later.
byte[] buffer = new byte[2048];
// open the zip file stream
InputStream theFile = new FileInputStream(args[0]);
ZipInputStream stream = new ZipInputStream(theFile);
String outdir = args[1];
try
{
// now iterate through each item in the stream. The get next
// entry call will return a ZipEntry for each file in the
// stream
ZipEntry entry;
while((entry = stream.getNextEntry())!=null)
{
String s = String.format("Entry: %s len %d added %TD",
entry.getName(), entry.getSize(),
new Date(entry.getTime()));
System.out.println(s);
// Once we get the entry from the stream, the stream is
// positioned read to read the raw data, and we keep
// reading until read returns 0 or less.
String outpath = outdir + "/" + entry.getName();
FileOutputStream output = null;
try
{
output = new FileOutputStream(outpath);
int len = 0;
while ((len = stream.read(buffer)) > 0)
{
output.write(buffer, 0, len);
}
}
finally
{
// we must always close the output file
if(output!=null) output.close();
}
}
}
finally
{
// we must always close the zip file.
stream.close();
}
}
Code excerpt came from the following site:
http://www.thecoderscorner.com/team-blog/java-and-jvm/12-reading-a-zip-file-from-java-using-zipinputstream#.U4RAxYamixR

Well, I've done this:
zipStream = new ZipInputStream(channelSftp.get("Port_Increment_201405261400_2251.zip"));
zipStream.getNextEntry();
sc = new Scanner(zipStream);
while (sc.hasNextLine()) {
System.out.println(sc.nextLine());
}
It helps me to read ZIP's content without writing to another file.

The ZipInputStream is an InputStream by itself and delivers the contents of each entry after each call to getNextEntry(). Special care must be taken, not to close the stream from which the contents is read, since it is the same as the ZIP stream:
public void readZipStream(InputStream in) throws IOException {
ZipInputStream zipIn = new ZipInputStream(in);
ZipEntry entry;
while ((entry = zipIn.getNextEntry()) != null) {
System.out.println(entry.getName());
readContents(zipIn);
zipIn.closeEntry();
}
}
private void readContents(InputStream contentsIn) throws IOException {
byte contents[] = new byte[4096];
int direct;
while ((direct = contentsIn.read(contents, 0, contents.length)) >= 0) {
System.out.println("Read " + direct + "bytes content.");
}
}
When delegating reading contents to other logic, it can be necessary to wrap the ZipInputStream with a FilterInputStream to close only the entry instead of the whole stream as in:
public void readZipStream(InputStream in) throws IOException {
ZipInputStream zipIn = new ZipInputStream(in);
ZipEntry entry;
while ((entry = zipIn.getNextEntry()) != null) {
System.out.println(entry.getName());
readContents(new FilterInputStream(zipIn) {
#Override
public void close() throws IOException {
zipIn.closeEntry();
}
});
}
}

OP was close. Just need to read the bytes. The call to getNextEntry positions the stream at the beginning of the entry data (docs). If that's the entry we want (or the only entry), then the InputStream is in the right spot. All we need to do is read that entry's decompressed bytes.
byte[] bytes = new byte[(int) entry.getSize()];
int i = 0;
while (i < bytes.length) {
// .read doesn't always fill the buffer we give it.
// Keep calling it until we get all the bytes for this entry.
i += zipStream.read(bytes, i, bytes.length - i);
}
So if these bytes really are text, then we can decode those bytes to a String. I'm just assuming utf8 encoding.
new String(bytes, "utf8")
Side note: I personally use apache commons-io IOUtils to cut down on this kind of lower level stuff. The docs for ZipInputStream.read seem to imply that read will stop at the end of the current zip entry. If that is true, then reading the current textual entry is one line with IOUtils.
String text = IOUtils.toString(zipStream)

Unzip archive (zip) with preserving file structure into given directory.
Note; this code use deps on "org.apache.commons.io.IOUtils"), but you can replace it by yours custom 'read-stream' code
public static void unzipDirectory(File archiveFile, File destinationDir) throws IOException
{
Path destPath = destinationDir.toPath();
try (ZipInputStream zis = new ZipInputStream(new FileInputStream(archiveFile)))
{
ZipEntry zipEntry;
while ((zipEntry = zis.getNextEntry()) != null)
{
Path resolvedPath = destPath.resolve(zipEntry.getName()).normalize();
if (!resolvedPath.startsWith(destPath))
{
throw new IOException("The requested zip-entry '" + zipEntry.getName() + "' does not belong to the requested destination");
}
if (zipEntry.isDirectory())
{
Files.createDirectories(resolvedPath);
} else
{
if(!Files.isDirectory(resolvedPath.getParent()))
{
Files.createDirectories(resolvedPath.getParent());
}
try (FileOutputStream outStream = new FileOutputStream(resolvedPath.toFile()))
{
IOUtils.copy(zis, outStream);
}
}
}
}
}

Here a more generic solution to process a zip inputstream with a BiConsumer. It's nearly the same solution that was used by haui
private void readZip(InputStream is, BiConsumer<ZipEntry,InputStream> consumer) throws IOException {
try (ZipInputStream zipFile = new ZipInputStream(is);) {
ZipEntry entry;
while((entry = zipFile.getNextEntry()) != null){
consumer.accept(entry, new FilterInputStream(zipFile) {
#Override
public void close() throws IOException {
zipFile.closeEntry();
}
});
}
}
}
You can use it by just calling
readZip(<some inputstream>, (entry, is) -> {
/* don't forget to close this stream after processing. */
is.read() // ... <- to read each entry
});

If content of your ZIP consist of 1 file (for example, zipped content of HTTP response), you can read text content using Kotlin as follows:
#Throws(IOException::class)
fun InputStream.readZippedContent() = ZipInputStream(this).use { stream ->
stream.nextEntry?.let { stream.bufferedReader().readText() } ?: String()
}
This extension function unzips first ZIP entry of Zip file and read content as plain text.
Usage:
val inputStream: InputStream = ... // your zipped InputStream
val textContent = inputStream.readZippedContent()

Reading bytes from a file?

I need to read some data until file is opened at different times, but I'm not sure if pointer to data that have not been read yet is automatic increased?
My method:
//method for copy binary data from file to binaryDataBuffer
void readcpy(String fileName, int pos, int len) {
try {
File lxDirectory = new File(Environment.getExternalStorageDirectory().getPath() + "/DATA/EXAMPLE/");
File lxFile = new File(lxDirectory, (fileName);
FileInputStream mFileInputStream = new FileInputStream(lxFile);
mFileInputStream.read(binaryDataBuffer, pos, len);
}
catch (Exception e) {
Log.d("Exception", e.getMessage());
}
}
So, if I call this method first time and read and save 5 bytes for example, will be on next call of the method read out bytes from 5th byte? I don't close file after reading.

When you create an InputStream (because a FileInputStream is an InputStream), the stream is created anew each time, and starts at the beginning of the stream (therefore the file).
If you want to read from where you left off the last time, you need to retain the offset and seek -- or retain the initial input stream you have opened.
While you can seek into a stream (using .skip()), it is in any event NOT recommended to reopen each time, it is costly; also, when you are done with a stream, you should close it:
// with Java 7: in automatically closed
try (InputStream in = ...;) {
// do stuff
} catch (WhateverException e) {
// handle exception
}
// with Java 6
InputStream in = ...;
try {
// do stuff
} catch (WhateverException e) {
// handle exception
} finally {
in.close();
}

Try this code:
public String getStringFromFile (String filePath) throws Exception {
File fl = new File(filePath);
FileInputStream fin = new FileInputStream(fl);
BufferedReader reader = new BufferedReader(new InputStreamReader(fin));
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line).append("\n");
}
String ret = sb.toString();
//Make sure you close all streams.
fin.close();
reader.close();
return ret;
}

I find RandomAccessFile, it has offset which I need in my case.

Prepend lines to file in Java

Is there a way to prepend a line to the File in Java, without creating a temporary file, and writing the needed content to it?

No, there is no way to do that SAFELY in Java. (Or AFAIK, any other programming language.)
No filesystem implementation in any mainstream operating system supports this kind of thing, and you won't find this feature supported in any mainstream programming languages.
Real world file systems are implemented on devices that store data as fixed sized "blocks". It is not possible to implement a file system model where you can insert bytes into the middle of a file without significantly slowing down file I/O, wasting disk space or both.
The solutions that involve an in-place rewrite of the file are inherently unsafe. If your application is killed or the power dies in the middle of the prepend / rewrite process, you are likely to lose data. I would NOT recommend using that approach in practice.
Use a temporary file and renaming. It is safer.

There is a way, it involves rewriting the whole file though (but no temporary file). As others mentioned, no file system supports prepending content to a file. Here is some sample code that uses a RandomAccessFile to write and read content while keeping some content buffered in memory:
public static void main(final String args[]) throws Exception {
File f = File.createTempFile(Main.class.getName(), "tmp");
f.deleteOnExit();
System.out.println(f.getPath());
// put some dummy content into our file
BufferedWriter w = new BufferedWriter(new OutputStreamWriter(new FileOutputStream(f)));
for (int i = 0; i < 1000; i++) {
w.write(UUID.randomUUID().toString());
w.write('\n');
}
w.flush();
w.close();
// append "some uuids" to our file
int bufLength = 4096;
byte[] appendBuf = "some uuids\n".getBytes();
byte[] writeBuf = appendBuf;
byte[] readBuf = new byte[bufLength];
int writeBytes = writeBuf.length;
RandomAccessFile rw = new RandomAccessFile(f, "rw");
int read = 0;
int write = 0;
while (true) {
// seek to read position and read content into read buffer
rw.seek(read);
int bytesRead = rw.read(readBuf, 0, readBuf.length);
// seek to write position and write content from write buffer
rw.seek(write);
rw.write(writeBuf, 0, writeBytes);
// no bytes read - end of file reached
if (bytesRead < 0) {
// end of
break;
}
// update seek positions for write and read
read += bytesRead;
write += writeBytes;
writeBytes = bytesRead;
// reuse buffer, create new one to replace (short) append buf
byte[] nextWrite = writeBuf == appendBuf ? new byte[bufLength] : writeBuf;
writeBuf = readBuf;
readBuf = nextWrite;
};
rw.close();
// now show the content of our file
BufferedReader reader = new BufferedReader(new InputStreamReader(new FileInputStream(f)));
String line;
while ((line = reader.readLine()) != null) {
System.out.println(line);
}
}

You could store the file content in a String and prepend the desired line by using a StringBuilder-Object. You just have to put the desired line first and then append the file-content-String.
No extra temporary file needed.

No. There are no "intra-file shift" operations, only read and write of discrete sizes.

It would be possible to do so by reading a chunk of the file of equal length to what you want to prepend, writing the new content in place of it, reading the later chunk and replacing it with what you read before, and so on, rippling down the to the end of the file.
However, don't do that, because if anything stops (out-of-memory, power outage, rogue thread calling System.exit) in the middle of that process, data will be lost. Use the temporary file instead.

private static void addPreAppnedText(File fileName) {
FileOutputStream fileOutputStream =null;
BufferedReader br = null;
FileReader fr = null;
String newFileName = fileName.getAbsolutePath() + "#";
try {
fileOutputStream = new FileOutputStream(newFileName);
fileOutputStream.write("preappendTextDataHere".getBytes());
fr = new FileReader(fileName);
br = new BufferedReader(fr);
String sCurrentLine;
while ((sCurrentLine = br.readLine()) != null) {
fileOutputStream.write(("\n"+sCurrentLine).getBytes());
}
fileOutputStream.flush();
} catch (IOException e) {
e.printStackTrace();
} finally {
try {
fileOutputStream.close();
if (br != null)
br.close();
if (fr != null)
fr.close();
new File(newFileName).renameTo(new File(newFileName.replace("#", "")));
} catch (IOException ex) {
ex.printStackTrace();
}
}
}