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java socket file transfer output path

I'm working on a java project that transfers files between server and client and I've managed to send a file to a desired output location, but the only problem is I have to include the full file name in the output path to save it successfully. My program runs in this way:
First it gets the path of the file to be transferred as input to the console, and then it gets the output path, again as an input to the console.
here are the codes of corresponding file name import and exports(I think the problem is somewhere here and posting this part will be sufficed)
Server Side
....
String in_filePath = null;
System.out.print("enter the file name: ");
in_filePath = sc.nextLine();
File myFile = new File( in_filePath );
System.out.println("The file chosen is being sent...");
byte[] mybytearray = new byte[(int) myFile.length()];
FileInputStream fis = null;
try {
fis = new FileInputStream(myFile);
sc.close();
} catch (FileNotFoundException ex) {
ex.printStackTrace();
}
Client Side
.....
int bufferSize = clientSocket.getReceiveBufferSize();
is = clientSocket.getInputStream();
DataInputStream clientdata = new DataInputStream(is);
String fileName = clientdata.readUTF();
System.out.println("file to be transferred is: " + fileName );
System.out.print("file output path: ");
String out_filePath;
out_filePath = sc.nextLine();
File file = new File( out_filePath );
fos = new FileOutputStream( file );
bos = new BufferedOutputStream(fos);
bytesRead = is.read(aByte, 0, aByte.length);
do {
baos.write(aByte);
bytesRead = is.read(aByte);
} while (bytesRead != -1);
bos.write(baos.toByteArray());
System.out.println(fileName + " transferred successfully");
At first I haven't included the output path in my program; as expected the output path was the root project folder and it was working great as it was reading the filename and sending the file with the same name without problems. But when I implemented the output path query, the output paths I choose like "C:\" or "C:\blabla\" gives me the exception as I stated above. Moreover giving the output path as "C:\image.jpg" or "blablabla\image.jpg" works perfectly well(assuming the name of the file to be copied as image.jpg) can it be a problem with reading the file name? any help would be appreciated
edit: now I'm receiving a socket write error if I had given "c:\" (or any kind of paths like that) as output path, yet it still works well if the output path is given like "c:\image.jpg"

Assume you're executing your program on directory "C:\", and your file "image.jpg" is located on "C:\images\image.jpg".
If you provide your input as "image.jpg", it won't work. You'd have to provide "images\image.jpg" instead.
You can easily check if you've hit an actual file by adding the following line to your Server Side code:
System.out.println(myFile.isFile());

Zipping Files using util.zip No directory

I have the following situation:
I am able to zip my files with the following method:
public boolean generateZip(){
byte[] application = new byte[100000];
ByteArrayOutputStream baos = new ByteArrayOutputStream();
// These are the files to include in the ZIP file
String[] filenames = new String[]{"/subdirectory/index.html", "/subdirectory/webindex.html"};
// Create a buffer for reading the files
try {
// Create the ZIP file
ZipOutputStream out = new ZipOutputStream(baos);
// Compress the files
for (int i=0; i<filenames.length; i++) {
byte[] filedata = VirtualFile.fromRelativePath(filenames[i]).content();
ByteArrayInputStream in = new ByteArrayInputStream(filedata);
// Add ZIP entry to output stream.
out.putNextEntry(new ZipEntry(filenames[i]));
// Transfer bytes from the file to the ZIP file
int len;
while ((len = in.read(application)) > 0) {
out.write(application, 0, len);
}
// Complete the entry
out.closeEntry();
in.close();
}
// Complete the ZIP file
out.close();
} catch (IOException e) {
System.out.println("There was an error generating ZIP.");
e.printStackTrace();
}
downloadzip(baos.toByteArray());
}
This works perfectly and I can download the xy.zip which contains the following directory and file structure:
subdirectory/
----index.html
----webindex.html
My aim is to completely leave out the subdirectory and the zip should only contain the two files. Is there any way to achieve this?
(I am using Java on Google App Engine).
Thanks in advance

If you are sure the files contained in the filenames array are unique if you leave out the directory, change your line for constructing ZipEntrys:
String zipEntryName = new File(filenames[i]).getName();
out.putNextEntry(new ZipEntry(zipEntryName));
This uses java.io.File#getName()

You can use Apache Commons io to list all your files, then read them to an InputStream
Replace the line below
String[] filenames = new String[]{"/subdirectory/index.html", "/subdirectory/webindex.html"}
with the following
Collection<File> files = FileUtils.listFiles(new File("/subdirectory"), new String[]{"html"}, true);
for (File file : files)
{
FileInputStream fileStream = new FileInputStream(file);
byte[] filedata = IOUtils.toByteArray(fileStream);
//From here you can proceed with your zipping.
}
Let me know if you have issues.

You could use the isDirectory() method on VirtualFile

FileNotFoundException when trying to unzip an archive with java.util.zip.ZipFile

I have a silly problem i haven't been able to figure out. Can anyone help me?
My Code is as:
String zipname = "C:/1100.zip";
String output = "C:/1100";
BufferedInputStream bis = null;
BufferedOutputStream bos = null;
try {
ZipFile zipFile = new ZipFile(zipname);
Enumeration<?> enumeration = zipFile.entries();
while (enumeration.hasMoreElements()) {
ZipEntry zipEntry = (ZipEntry) enumeration.nextElement();
System.out.println("Unzipping: " + zipEntry.getName());
bis = new BufferedInputStream(zipFile.getInputStream(zipEntry));
int size;
byte[] buffer = new byte[2048];
It doesn't create a folder but debugging shows all the contents being generated.
In Order to create a folder i used the code
if(!output.exists()){ output.mkdir();} // here i get an error saying filenotfoundexception
bos = new BufferedOutputStream(new FileOutputStream(new File(outPut)));
while ((size = bis.read(buffer)) != -1) {
bos.write(buffer, 0, size);
}
}
} catch (Exception ex) {
ex.printStackTrace();
} finally {
bos.flush();
bos.close();
bis.close();
}
My zip file contains images: a.jpg b.jpg... and in the same hierarchy, I have abc.xml.
I need to extract the content as is in the zip file.
Any helps here.

There are a few problems with your code: Where is outPut declared? output is not a file but a string, so exists() and mkdir() do not exist. Start by declaring output like:
File output = new File("C:/1100");
Furthermore, outPut (with big P) is not declared. It be something like output + File.seprator + zipEntry.getName().
bos = new BufferedOutputStream(new FileOutputStream(output + File.seprator + zipEntry.getName()));
Note that you don't need to pass a File to FileOutputStream, as constructors show in the documentation.
At this point, your code should work if your Zip file does not contain directory. However, when opening the output stream, if zipEntry.getName() has a directory component (for instance somedir/filename.txt), opening the stream will result in a FileNotFoundException, as the parent directory of the file you try to create does not exist. If you want to be able to handle such zip files, you will find your answer in: How to unzip files recursively in Java?

Create a file object from a resource path to an image in a jar file

I need to create a File object out of a file path to an image that is contained in a jar file after creating a jar file. If tried using:
URL url = getClass().getResource("/resources/images/image.jpg");
File imageFile = new File(url.toURI());
but it doesn't work. Does anyone know of another way to do it?

To create a file on Android from a resource or raw file I do this:
try{
InputStream inputStream = getResources().openRawResource(R.raw.some_file);
File tempFile = File.createTempFile("pre", "suf");
copyFile(inputStream, new FileOutputStream(tempFile));
// Now some_file is tempFile .. do what you like
} catch (IOException e) {
throw new RuntimeException("Can't create temp file ", e);
}
private void copyFile(InputStream in, OutputStream out) throws IOException {
byte[] buffer = new byte[1024];
int read;
while((read = in.read(buffer)) != -1){
out.write(buffer, 0, read);
}
}
Don't forget to close your streams etc

This should work.
String imgName = "/resources/images/image.jpg";
InputStream in = getClass().getResourceAsStream(imgName);
ImageIcon img = new ImageIcon(ImageIO.read(in));

Usually, you can't directly get a java.io.File object, since there is no physical file for an entry within a compressed archive. Either you live with a stream (which is best most in the cases, since every good API can work with streams) or you can create a temporary file:
URL imageResource = getClass().getResource("image.gif");
File imageFile = File.createTempFile(
FilenameUtils.getBaseName(imageResource.getFile()),
FilenameUtils.getExtension(imageResource.getFile()));
IOUtils.copy(imageResource.openStream(),
FileUtils.openOutputStream(imageFile));

You cannot create a File object to a reference inside an archive. If you absolutely need a File object, you will need to extract the file to a temporary location first. On the other hand, most good API's will also take an input stream instead, which you can get for a file in an archive.

How to use java.util.zip to archive/deflate string in java for use in Google Earth?

Use Case
I need to package up our kml which is in a String into a kmz response for a network link in Google Earth. I would like to also wrap up icons and such while I'm at it.
Problem
Using the implementation below I receive errors from both WinZip and Google Earth that the archive is corrupted or that the file cannot be opened respectively. The part that deviates from other examples I'd built this from are the lines where the string is added:
ZipEntry kmlZipEntry = new ZipEntry("doc.kml");
out.putNextEntry(kmlZipEntry);
out.write(kml.getBytes("UTF-8"));
Please point me in the right direction to correctly write the string so that it is in doc.xml in the resulting kmz file. I know how to write the string to a temporary file, but I would very much like to keep the operation in memory for understandability and efficiency.
private static final int BUFFER = 2048;
private static void kmz(OutputStream os, String kml)
{
try{
BufferedInputStream origin = null;
ZipOutputStream out = new ZipOutputStream(os);
out.setMethod(ZipOutputStream.DEFLATED);
byte data[] = new byte[BUFFER];
File f = new File("./icons"); //folder containing icons and such
String files[] = f.list();
if(files != null)
{
for (String file: files) {
LOGGER.info("Adding to KMZ: "+ file);
FileInputStream fi = new FileInputStream(file);
origin = new BufferedInputStream(fi, BUFFER);
ZipEntry entry = new ZipEntry(file);
out.putNextEntry(entry);
int count;
while((count = origin.read(data, 0, BUFFER)) != -1) {
out.write(data, 0, count);
}
origin.close();
}
}
ZipEntry kmlZipEntry = new ZipEntry("doc.kml");
out.putNextEntry(kmlZipEntry);
out.write(kml.getBytes("UTF-8"));
}
catch(Exception e)
{
LOGGER.error("Problem creating kmz file", e);
}
}
Bonus points for showing me how to put the supplementary files from the icons folder into a similar folder within the archive as opposed to at the same layer as the doc.kml.
Update Even when saving the string to a temp file the errors occur. Ugh.
Use Case Note The use case is for use in a web app, but the code to get the list of files won't work there. For details see how-to-access-local-files-on-server-in-jboss-application

You forgot to call close() on ZipOutputStream. Best place to call it is the finally block of the try block where it's been created.
Update: To create a folder, just prepend its name in the entry name.
ZipEntry entry = new ZipEntry("icons/" + file);