Develop Reference

Set filename of an only streamed zip file

i am currently creating a zip file and filling it with various json files and images. All this should only run in memory and not on the hard disk. Therefore I have the following construct so far:
ByteArrayOutputStream baos = new ByteArrayOutputStream();
ZipOutputStream zip = null;
String FILE_NAME = "file.zip";
try {
zip = new ZipOutputStream(baos);
//now the critical part where the name of the file should be set
ZipEntry entry = new ZipEntry(FILE_NAME);
zip.putNextEntry(entry);
byte[] data = FILE_NAME.getBytes();
zip.write(data, 0, data.length);
zip.closeEntry();
//end of critical part and filling the rest of the zip
//...
//
}finally{
IOUtils.closeQuietly(zip);
byte[] byteFile = baos.toByteArray();
IOUtils.closeQuietly(baos);}
The problem is that the zip-file is called file.zip, but also contains a file.zip itself.
How can I name my Zip file from the ZipOutputStream without packing into this one file with the same name? Unfortunately I only found this solution here.

public byte[] zipBytesFile(List<byte[]> files) throws IOException {
ByteArrayOutputStream baos = new ByteArrayOutputStream();
ZipOutputStream zos = new ZipOutputStream(baos);
int i = 0;
for (byte[] file : files) {
ZipEntry entry = new ZipEntry(++i + ".pdf");
entry.setSize(file.length);
zos.putNextEntry(entry);
zos.write(file);
}
zos.closeEntry();
zos.close();
return baos.toByteArray();
}

only first few hundred samples are read from input stream into byte array(the rest are zeros)

i'm trying to convert audio(mp3/wav etc.) to byte array. i did it using inputStream to byte array conversion.
the problem is that after few hundred samples i recieve only zeroes.
at first i thought the problem was with the file so i tried debugging with another file and had the same problem.
I thought the problem was with the code so i tried using IOUtils and got the exact same resualts.
can anyone tell me what i'm doing wrong?
the code i used:
File file = new File(path);
final InputStream inputStream = new FileInputStream(file);
byte[] byteSamples = inputStreamToByteArray(inputStream);
public byte[] inputStreamToByteArray(InputStream inStream) throws IOException {
ByteArrayOutputStream baos = new ByteArrayOutputStream();
byte[] buffer = new byte[8192];
int bytesRead;
while ((bytesRead = inStream.read(buffer)) > 0) {
baos.write(buffer, 0, bytesRead);
}
return baos.toByteArray();
}
the alternate version using IOUtils:
byte[] byteSamples = IOUtils.toByteArray(inputStream);
update : i tried doing it using BufferedInputStream, still the exact same results.
byte[] byteSamples = new byte[(int)file.length()];
try {
BufferedInputStream buf = new BufferedInputStream(new FileInputStream(file));
buf.read(byteSamples, 0, byteSamples.length);
buf.close();
} catch (FileNotFoundException e) {
e.printStackTrace();}

You need to close the streams when done.

Saving ZipOutputStream to zip file with Java

I have to create ZIP archive in memory
But now, I need it to be saved in a real .zip file in a disk. How to do it?
Pseudocode:
public byte[] crtZipByteArray(ByteArrayInputStream data,ZipEntry entry) throws IOException{
ByteArrayOutputStream zipout = new ByteArrayOutputStream();
ZipOutputStream zos = new ZipOutputStream(zipout);
byte[] buffer = new byte[1024];
int len;
zos.putNextEntry(entry);
while ((len = data.read(buffer)) > 0) {
zos.write(buffer, 0, len);
}
zos.closeEntry();
zos.close();
data.close();
return zipout.toByteArray();
}

Replace ByteArrayOutputStream zipout with FileOutputStream zipout.
If you sill need to return byte array as method result use apache commons TeeOutputStream to duplicate output for two streams.
public byte[] crtZipByteArray(ByteArrayInputStream data,ZipEntry entry) throws IOException{
ByteArrayOutputStream byteArrayOut = new ByteArrayOutputStream();
OutputStream fileOut = new FileOutputStream("filename.zip");
OutputStream teeOut = new TeeOutputStream(byteArrayOut, fileOut);
ZipOutputStream zos = new ZipOutputStream(teeOut);
.....
}

how to convert image to byte array in java?(With out using buffered image)

Hi could anyone please explain me how to convert the image data to byte array in java I am trying like this.I do not need to use buffered image here.
File file = new File("D:/img.jpg");
FileInputStream imageInFile = new FileInputStream(file);
byte imageData[] = new byte[(int) file.length()];
imageInFile.read(imageData);

You can convert your image data using FileInputStream also.
File file = new File("D:\\img.jpg");
FileInputStream fis = new FileInputStream(file);
//Now try to create FileInputStream which obtains input bytes from a file.
//FileInputStream is meant for reading streams of raw bytes,in this case its image data.
//For reading streams of characters, consider using FileReader.
ByteArrayOutputStream bos = new ByteArrayOutputStream();
byte[] buf = new byte[1024];
try {
for (int readNum; (readNum = fis.read(buf)) != -1;) {
//Now Write to this byte array output stream
bos.write(buf, 0, readNum);
System.out.println("read " + readNum + " bytes,");
}
} catch (IOException ex) {
Logger.getLogger(ConvertImage.class.getName()).log(Level.SEVERE, null, ex);
}
byte[] bytes = bos.toByteArray();

or you could use:
Image image = Toolkit.getDefaultToolkit().getImage("D:/img.jpg");
byte[] imageBytes = getImageBytes(image);
private byte[] getImageBytes(Image image) throws IOException {
try (ByteArrayOutputStream baos = new ByteArrayOutputStream()) {
ImageIO.write(image, "bmp", baos);
baos.flush();
return baos.toByteArray();
}
}

Not able to process big file inside a zip file using ZipInputStream

I am having a below mentioned java class which extracts a zip, and one by one convert its content to string and print to console.
Problem is, when the file present inside the zip is big ~80KB. Entire content is not getting displayed (only 3/4 of the data is getting converted to string and displayed in console).
Secondly below mentioned code is introducing null/space in between and also if the file size is small ~1KB
what is wrong in below mentioned code.
public static void main(String[] args) throws Exception {
byte[] buf = new byte[1024];
final int BUFFER = 1024;
String fName = "c:\\DOC00001.zip";
ZipInputStream zinstream = new ZipInputStream(
new FileInputStream(fName));
ZipEntry zentry = zinstream.getNextEntry();
while (zentry != null) {
byte data[] = new byte[BUFFER];
ByteArrayOutputStream out = new ByteArrayOutputStream();
while ((zinstream.read(data, 0, BUFFER)) != -1) {
out.write(data);
}
InputStream is = new ByteArrayInputStream(out.toByteArray());
StringWriter writer = new StringWriter();
IOUtils.copy(is, writer, "UTF-8");
String response = writer.toString();
System.out.println(response);
zentry = zinstream.getNextEntry();
}
zinstream.close();
}

The read method is not guaranteed to read a full buffer; the number of bytes that have been read is returned. The correct way to extract data from a zip file, or any InputStream in general, would be:
byte[] data = new byte[BUFFER];
ByteArrayOutputStream out = new ByteArrayOutputStream();
int bytesRead;
while ((bytesRead = zinstream.read(data, 0, BUFFER)) != -1) {
out.write(data, 0, bytesRead);
}
Or, since you are already using IOUtils,
ByteArrayOutputStream out = new ByteArrayOutputStream();
IOUtils.copy(zinstream, out);
Or, given that you write to a ByteArrayOutputStream only to later write to a String, you can skip the ByteArrayOutputStream entirely:
while (zentry != null) {
StringWriter writer = new StringWriter();
IOUtils.copy(zinstream, writer, "UTF-8");
String response = writer.toString();
System.out.println(response);
zentry = zinstream.getNextEntry();
}