Develop Reference

how to decode/ get encoding of file (Power BI desktop file)

I am having power BI desktop report(pbix) internal file (DataMashup), which i am trying to decode.
My Aim is to create Power-BI desktop report, Data Model using any programming language. I am using Java for initial.
files are encoded with some encoding technique.
I tried to get encoding of file and it is returning windows 1254. but decoding is not happening.
File f = new File("example.txt");
String[] charsetsToBeTested = {"UTF-8", "windows-1254", "ISO-8859-7"};
CharsetDetector cd = new CharsetDetector();
Charset charset = cd.detectCharset(f, charsetsToBeTested);
if (charset != null) {
try {
InputStreamReader reader = new InputStreamReader(new FileInputStream(f), charset);
int c = 0;
while ((c = reader.read()) != -1) {
System.out.print((char)c);
}
reader.close();
} catch (FileNotFoundException fnfe) {
fnfe.printStackTrace();
}catch(IOException ioe){
ioe.printStackTrace();
}
}else{
System.out.println("Unrecognized charset.");
}
Unzipping of file is also not working
public void unZipIt(String zipFile, String outputFolder)
{
byte buffer[] = new byte[1024];
try
{
File folder = new File(outputFolder);
if(!folder.exists())
{
folder.mkdir();
}
ZipInputStream zis = new ZipInputStream(new FileInputStream(zipFile));
System.out.println(zis);
System.out.println(zis.getNextEntry());
for(ZipEntry ze = zis.getNextEntry(); ze != null; ze = zis.getNextEntry())
{
String fileName = ze.getName();
System.out.println(ze);
File newFile = new File((new StringBuilder(String.valueOf(outputFolder))).append(File.separator).append(fileName).toString());
System.out.println((new StringBuilder("file unzip : ")).append(newFile.getAbsoluteFile()).toString());
(new File(newFile.getParent())).mkdirs();
FileOutputStream fos = new FileOutputStream(newFile);
int len;
while((len = zis.read(buffer)) > 0)
{
fos.write(buffer, 0, len);
}
fos.close();
}
zis.closeEntry();
zis.close();
System.out.println("Done");
}
catch(IOException ex)
{
ex.printStackTrace();
}
}

The file contains a binary header and then XML with UTF-8 specified.
The header data seems to hold the file name (Config/Package.xml), so assuming a zip format is understandable. With a zip format also there would be binary data at the end of file.
Maybe the file was downloaded using FTP, and a text conversion ("\n" to "\r\n") was done. Then the zip would be corrupted. Renaming the file to .zip might help testing the file with zip tools.
Try first the .tar format. This would be logical as the XML file is not compressed. Add .tar to the file ending.
Otherwise, if the content is always UTF-8 XML:
Path f = Paths.get("example.txt");
String start ="<?xml";
String end = ">";
byte[] bytes = Files.readAllBytes(f);
String s = new String(bytes, StandardCharsets.ISO_8859_1); // Single byte encoding.
int startI = s.indexOf(start);
int endI = s.lastIndexOf(end) + end.length();
//bytes = Arrays.copyOfRange(bytes, startI, endI);
String xml = new String(bytes, startI, endI - startI, StandardCharsets.UTF_8);

You can use the System.IO.Packaging library to extract the Power BI data mashup. It uses the OPC package standard, see here.

How to create a ZIP InputStream in Android without creating a ZIP file first?

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();

Zip Jasper reports [duplicate]

I am trying to create a zip file of multiple image files. I have succeeded in creating the zip file of all the images but somehow all the images have been hanged to 950 bytes. I don't know whats going wrong here and now I can't open the images were compressed into that zip file.
Here is my code. Can anyone let me know what's going here?
String path="c:\\windows\\twain32";
File f=new File(path);
f.mkdir();
File x=new File("e:\\test");
x.mkdir();
byte []b;
String zipFile="e:\\test\\test.zip";
FileOutputStream fout=new FileOutputStream(zipFile);
ZipOutputStream zout=new ZipOutputStream(new BufferedOutputStream(fout));
File []s=f.listFiles();
for(int i=0;i<s.length;i++)
{
b=new byte[(int)s[i].length()];
FileInputStream fin=new FileInputStream(s[i]);
zout.putNextEntry(new ZipEntry(s[i].getName()));
int length;
while((length=fin.read())>0)
{
zout.write(b,0,length);
}
zout.closeEntry();
fin.close();
}
zout.close();

This is my zip function I always use for any file structures:
public static File zip(List<File> files, String filename) {
File zipfile = new File(filename);
// Create a buffer for reading the files
byte[] buf = new byte[1024];
try {
// create the ZIP file
ZipOutputStream out = new ZipOutputStream(new FileOutputStream(zipfile));
// compress the files
for(int i=0; i<files.size(); i++) {
FileInputStream in = new FileInputStream(files.get(i).getCanonicalName());
// add ZIP entry to output stream
out.putNextEntry(new ZipEntry(files.get(i).getName()));
// transfer bytes from the file to the ZIP file
int len;
while((len = in.read(buf)) > 0) {
out.write(buf, 0, len);
}
// complete the entry
out.closeEntry();
in.close();
}
// complete the ZIP file
out.close();
return zipfile;
} catch (IOException ex) {
System.err.println(ex.getMessage());
}
return null;
}

Change this:
while((length=fin.read())>0)
to this:
while((length=fin.read(b, 0, 1024))>0)
And set buffer size to 1024 bytes:
b=new byte[1024];

How to convert a byte array to ZIP file and download it using Java?

I need to write a code to convert a byte array to ZIP file and make it download in Spring MVC.
Byte array is coming from a webservice which is a ZIP file originally. ZIP file has a folder and the folder contains 2 files. I have written the below code to convert to byte array to ZipInputStream. But I am not able to convert into ZIP file. Please help me in this.
Here is my code.
ZipInputStream zipStream = new ZipInputStream(new ByteArrayInputStream(bytes));
ZipEntry entry = null;
while ((entry = zipStream.getNextEntry()) != null) {
String entryName = entry.getName();
FileOutputStream out = new FileOutputStream(entryName);
byte[] byteBuff = new byte[4096];
int bytesRead = 0;
while ((bytesRead = zipStream.read(byteBuff)) != -1)
{
out.write(byteBuff, 0, bytesRead);
}
out.close();
zipStream.closeEntry();
}
zipStream.close();

I am presuming here that you want to write a byte array to a ZIP file. As the data sent is also a ZIP file and to be saved is also ZIP file, shouldn't be a problem.
Two steps are needed: save it on disk and return the file.
1) Save on disk part:
File file = new File(/path/to/directory/save.zip);
if (file.exists() && file.isDirectory()) {
try {
OutputStream outputStream = new FileOutputStream(new File(/path/to/directory/save.zip));
outputStream.write(bytes);
outputStream.close();
} catch (IOException ignored) {
}
} else {
// create directory and call same code
}
}
2) Now to get it back and download it, you need a controller :
#RequestMapping(value = "/download/attachment/", method = RequestMethod.GET)
public void getAttachmentFromDatabase(HttpServletResponse response) {
response.setContentType("application/octet-stream");
response.setHeader("Content-Disposition", "attachment; filename=\"" + file.getFileName() + "\"");
response.setContentLength(file.length);
FileCopyUtils.copy(file as byte-array, response.getOutputStream());
response.flushBuffer();
}
I have edited the code I have, so you will have to make some changes before it suits you 100%. Let me know if this is what you were looking for. If not, I will delete my answer. Enjoy.

Java Connecting URL and downloading a zip but when extracting the zip it's not properly downloaded

I am sending a request XML to the URL and receiving a zip file to the given path.
Sometimes I'm facing troubles when the bandwidth is low this zip file, most likely 120MB size is not getting downloaded properly. And getting an error when extracting the zip file. Extracting happens from the code as well. When I download in high bandwidth this file gets download without issue.
I'm looking for a solution without making the bandwidth high, from program level are there any ways to download this zip file, may be part by part or something like that? Or anyother solution that you all are having is highly appreciated.
Downloading :
url = new URL(_URL);
sc = (HttpURLConnection) url.openConnection();
sc.setDoInput(true);
sc.setDoOutput(true);
sc.setRequestMethod("POST");
sc.connect();
OutputStream mOstr = sc.getOutputStream();
mOstr.write(request.getBytes());
InputStream in = sc.getInputStream();
FileOutputStream out = new FileOutputStream(path);
int count;
byte[] buffer = new byte[86384];
while ((count = in.read(buffer,0,buffer.length)) > 0)
out.write(buffer, 0, count);
out.close();
Extracting :
try {
ZipFile zipFile = new ZipFile(path+zFile);
Enumeration<?> enu = zipFile.entries();
while (enu.hasMoreElements()) {
ZipEntry zipEntry = (ZipEntry) enu.nextElement();
String name = path+"/data_FILES/"+zipEntry.getName();
long size = zipEntry.getSize();
long compressedSize = zipEntry.getCompressedSize();
System.out.printf("name: %-20s | size: %6d | compressed size: %6d\n", name, size, compressedSize);
File file = new File(name);
if (name.endsWith("/")) {
file.mkdirs();
continue;
}
File parent = file.getParentFile();
if (parent != null) {
parent.mkdirs();
}
InputStream is = zipFile.getInputStream(zipEntry);
FileOutputStream fos = new FileOutputStream(file);
byte[] bytes = new byte[86384];
int length;
while ((length = is.read(bytes)) >= 0) {
fos.write(bytes, 0, length);
}
is.close();
fos.close();
}
zipFile.close();
} catch (Exception e) {
log("Error in extracting zip file ");
e.printStackTrace();
}