I have a Java program that needs to read a file from a resource within the JAR and it only takes it through byte[]. My problem is converting the resource file from a folder within the project (i.e. tools/test.txt) into byte[]. I have tried the following (gave an "undefined for type" error):
final byte[] temp = new File("tools/test.txt").getBytes();
Another method I tried resulted in not being able to find the file:
FileOutputStream fos = new FileOutputStream("tools/test.txt");
byte[] myByteArray = null;
fos.write(myByteArray);
fos.close();
System.out.println("Results = " + myByteArray);
And lastly using Inputstream and BufferedReader. This actually gave the content of the file when running the program from Eclipse, but came out as null when running it as a jar (I am assuming that it is also not reading the file).
InputStream is = null;
BufferedReader br = null;
String line;
ArrayList list = new ArrayList();
try {
is = Main.class.getResourceAsStream("tools/test.txt");
br = new BufferedReader(new InputStreamReader(is));
while (null != (line = br.readLine())) {
list.add(line);
System.out.println("Output:" + line);
}
while (null == (line = br.readLine())) {
System.out.println("Error loading file:" + line);
}
}
catch (Exception ef) {
ef.printStackTrace();
System.out.println("Output:" + ef);
}
So my question is, if I have a folder named "tools" and have a file called "test.txt", what code would I use to turn it into byte[] and still work when compiled into a Jar file?
ByteArrayOutputStream baos = new ByteArrayOutputStream();
InputStream in = Main.class.getResourceAsStream("/tools/test.txt");
byte[] buffer = new byte[4096];
for (;;) {
int nread = in.read(buffer);
if (nread <= 0) {
break;
}
baos.write(buffer, 0, nread);
}
byte[] data = baos.toByteArray();
String text = new String(data, "Windows-1252");
Byte[] asByteObjects = new Byte[data.length];
for (int i # 0; i < data.length: ++i) {
asByteObjects[i] = data[i];
}
Without the heading slash the path would be relative to the package of the class. A ByteArrayOutputStream serves to collect for a byte[].
If the bytes represent text is some encoding, one can turn it into a String. Here with Windows Latin-1.
have you tried Scanner.nextByte()? make a new scanner with the file you want to parse as the input and use a for loop to create your array.
Related
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.
I'm trying read a file(doesn't matter the extension) and write after this, but when I do it, the output file is different from the input.
my code is the next:
OutputStream outputStream = null;
FileReader fr = new FileReader("rute\\inputfile.PNG");
BufferedReader br = new BufferedReader(fr);
String line;
while ((line= br.readLine()) != null) {
content += line;
}
byte[] toBytes= content.getBytes();
InputStream inputStream = new ByteArrayInputStream(toBytes);
try {
outputStream = new FileOutputStream(new File("rute\\output.PNG"));
int read = 0;
byte[] bytes = new byte[1024];
while ((read = inputStream.read(bytes)) != -1) {
outputStream.write(bytes, 0, read);
}
outputStream.close();
} catch (Exception e) {
e.printStackTrace();
}
inputStream.close();
If you ask me why convert into bytes and write from this form, is because I need do something with the data, and I need this conversion.
If you tell me that i cant load an image on a String, yes I can do something like that:
File fil = ~~~~;
FileInputStream fis = null;
fis = new FileInputStream(fil);
byte[] bytess = IOUtils.toByteArray(fis);
But I dont want do it by this way because if I want load big files, the heap size is not enough an this could be solved by the "line per line" read.
Thanks for your answers
I will recommend read this question before. Since you are reading binary data into a String you are changing the encoding of that data. So the output will be different.
Best approach is read binary files as byte arrays. But I will depend which type of transformation/edition/changes you need to do with them.
UPDATE
And, of course, you are editing your content before writing
while ((line= br.readLine()) != null) {
content += line + "\n";
}
so the your output file will be different always.
UPDATE 2
Since the question/problem is how to read a big binary file, google is usually your friend.
Or you can check this other question
I´m trying to implement some "over the air" update mechanism for OSGi bundles. For that, I need to be able to create a jar file from a String (basically the content of the jar file read by JarInputStream). The following example code should illustrate my needs:
//read bundle to be copied!
File originalFile = new File(
"/Users/stefan/Documents/Projects/OSGi/SimpleBundle_1.0.0.201404.jar");
JarInputStream fis = new JarInputStream(new FileInputStream(originalFile));
StringBuilder stringBuilder = new StringBuilder();
int ch;
while ((ch = fis.read()) != -1) {
stringBuilder.append((char) ch);
}
fis.close();
//Create content string
String content = stringBuilder.toString();
if (logger.isInfoEnabled()) {
logger.info(content);
}
//Init new jar input stream
JarInputStream jarInputStream = new JarInputStream(
new ByteArrayInputStream(content.getBytes()));
if (logger.isInfoEnabled()) {
logger.info("Save content to disc!");
}
File newFile = new File(
"/Users/stefan/Documents/Projects/OSGi/equinox/SimpleBundle_1.0.0.201404.jar");
//Init new jar output stream
JarOutputStream fos = new JarOutputStream(
new FileOutputStream(newFile));
if (!newFile.exists()) {
newFile.createNewFile();
}
int BUFFER_SIZE = 10240;
byte buffer[] = new byte[BUFFER_SIZE];
while (true) {
int nRead = jarInputStream.read(buffer, 0,
buffer.length);
if (nRead <= 0)
break;
fos.write(buffer, 0, nRead);
}
//Write content to new jar file.
fos.flush();
fos.close();
jarInputStream.close();
Unfortunately, the created jar file is empty and throws an "Invalid input file" error if I try to open it with JD-GUI. Is it possible to create a jar file from the String "content"?
Best regards and thank you very much
Stefan
Your jar is empty because you do not read anything from the JarInputStream. If you want to read JarInputStream, you should iterate its entries. If you want to change the Manifest, the first entry should be skipped, use the getManifest() of the jarInputStream and the constructor of the JarOutputStream, where Manifest can be specified. Based on your code (no manifest change but plain jar copy):
ZipEntry zipEntry = jarInputStream.getNextEntry();
while (zipEntry != null) {
fos.putNextEntry(zipEntry);
// Simple stream copy comes here
int BUFFER_SIZE = 10240;
byte buffer[] = new byte[BUFFER_SIZE];
int l = jarInputStream.read(buffer);
while(l >= 0) {
fos.write(buffer, 0, l);
l = jarInputStream.read(buffer);
}
zipEntry = jarInputStream.getNextEntry();
}
You only need this if you want to change the content (Manifest or entries) of the JAR file during the copy. Otherwise, simple InputStream and FileOutputStream will do the work (as Tim said).
From a DB2 table I've got blob which I'm converting to a byte array so I can work with it. I need to take the byte array and create a PDF out of it.
This is what I have:
static void byteArrayToFile(byte[] bArray) {
try {
// Create file
FileWriter fstream = new FileWriter("out.pdf");
BufferedWriter out = new BufferedWriter(fstream);
for (Byte b: bArray) {
out.write(b);
}
out.close();
} catch (Exception e) {
System.err.println("Error: " + e.getMessage());
}
}
But the PDF it creates is not right, it has a bunch of black lines running from top to bottom on it.
I was actually able to create the correct PDF by writing a web application using essentially the same process. The primary difference between the web application and the code about was this line:
response.setContentType("application/pdf");
So I know the byte array is a PDF and it can be done, but my code in byteArrayToFile won't create a clean PDF.
Any ideas on how I can make it work?
Sending your output through a FileWriter is corrupting it because the data is bytes, and FileWriters are for writing characters. All you need is:
OutputStream out = new FileOutputStream("out.pdf");
out.write(bArray);
out.close();
One can utilize the autoclosable interface that was introduced in java 7.
try (OutputStream out = new FileOutputStream("out.pdf")) {
out.write(bArray);
}
Read from file or string to bytearray.
byte[] filedata = null;
String content = new String(bytearray);
content = content.replace("\r", "").replace("\uf8ff", "").replace("'", "").replace("\"", "").replace("`", "");
String[] arrOfStr = content.split("\n");
PDDocument document = new PDDocument();
PDPage page = new PDPage();
document.addPage(page);
try (PDPageContentStream cs = new PDPageContentStream(document, page)) {
// setting font family and font size
cs.beginText();
cs.setFont(PDType1Font.HELVETICA, 14);
cs.setNonStrokingColor(Color.BLACK);
cs.newLineAtOffset(20, 750);
for (String str: arrOfStr) {
cs.newLineAtOffset(0, -15);
cs.showText(str);
}
cs.newLine();
cs.endText();
}
document.save(znaFile);
document.close();
public static String getPDF() throws IOException {
File file = new File("give complete path of file which must be read");
FileInputStream stream = new FileInputStream(file);
byte[] buffer = new byte[8192];
ByteArrayOutputStream baos = new ByteArrayOutputStream();
int bytesRead;enter code here
while ((bytesRead = stream.read(buffer)) != -1) {
baos.write(buffer, 0, bytesRead);
}
System.out.println("it came back"+baos);
byte[] buffer1= baos.toByteArray();
String fileName = "give your filename with location";
//stream.close();
FileOutputStream outputStream =
new FileOutputStream(fileName);
outputStream.write(buffer1);
return fileName;
}
I call a service which returns a gzipped file. I have the data as an InputStream (courtesy of javax.activation.DataHandler.getInputStream();) from the response.
What I would like to do is, without writing anything to disk, get an InputStream of the decompressed data in the file that is in the archive. The compressed file in this case is an xml document that I am trying to unmarshal using javax.xml.bind.Unmarshaller which takes an InputStream.
I'm currently trying to write the InputStream to an OutputStream (decompressing the data) and then I'll need to write it back to an InputStream. It's not working yet so I thought I would see if there was a better (I would hope so) approach.
I can write the initial InputStream to disk and get a gz file, and then read that file, get the compressed file out of it and go from there but I'd rather keep it all in memory is possible.
Update 1: Here is my current (not working - get a "Not in GZIP format" exception):
ByteArrayInputStream xmlInput = null;
try {
InputStream in = dh.getInputStream(); //dh is a javax.activation.DataHandler
BufferedInputStream bis = new BufferedInputStream(in);
ByteArrayOutputStream bo = new ByteArrayOutputStream();
int bytes_read = 0;
byte[] dataBuf = new byte[4096];
while ((bytes_read = bis.read(dataBuf)) != -1) {
bo.write(dataBuf, 0, bytes_read);
}
ByteArrayInputStream bin = new ByteArrayInputStream(bo.toByteArray());
GZIPInputStream gzipInput = new GZIPInputStream(bin);
ByteArrayOutputStream out = new ByteArrayOutputStream();
dataBuf = new byte[4096];;
bytes_read = 0;
while ((bytes_read = gzipInput.read(dataBuf)) > 0) {
out.write(dataBuf, 0, bytes_read);
}
xmlInput = new ByteArrayInputStream(out.toByteArray());
If instead of writing to a ByteArrayOutputStream I write to a FileOutputStream the first time around I get a compressed file (which I can manually open to get the xml file within) and the service (eBay) says it should be a gzip file so I'm not sure why I get a "Not in GZIP format" error.
Update 2: I tried something a little different - same error ("Not in GZIP format"). Wow, I just tried to end that parenthesis with a semi-colon. Anyways, here is my second attempt, which still does not work:
ByteArrayInputStream xmlInput = null;
try {
GZIPInputStream gzipInput = new GZIPInputStream(dh.getInputStream());
ByteArrayOutputStream bo = new ByteArrayOutputStream();
int bytes_read = 0;
byte[] dataBuf = new byte[4096];
while ((bytes_read = gzipInput.read(dataBuf)) != -1) {
bo.write(dataBuf, 0, bytes_read);
}
xmlInput = new ByteArrayInputStream(bo.toByteArray());
Decorate the input stream with a GZIPInputStream.
InputStream decompressed = new GZIPInputStream(compressed);
The following code should work. Keep in mind you'll have to handle exceptions properly.
OutputStream out = null;
InputStream in = null;
try {
out = /* some output stream */;
in = new java.util.GZIPInputStream(/*some stream*/);
byte[] buffer = new byte[4096];
int c = 0;
while (( c = in.read(buffer, 0, 4096)) > 0) {
out.write(buffer, 0, c);
}
} finally {
if (in != null) {
in.close();
}
if (out != null) {
out.close();
}
}
Take a look at GZIPInputStream. Here's an example; the class handles this very transparently, it's almost no work to use.