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jcifs copyTo from one network to other network path not working

I'm using the Java CIFS Client Library but facing the problem and problem is copyTo function is not working.
I have one folder which contains files. I want to read these files to other network path.
String path1 = "//MACHINE-NAME/SHARE-FOLDER"
NtlmPasswordAuthentication auth = new
NtlmPasswordAuthentication(DOMAIN;USERNAME:PASSWORD)
SmbFile readFolder = new SmbFile("smb://MACHINE-NAME/SHARE-FOLDER/",auth)
This is working fine.
Then i have another network path and define like this and ShareFolder2 is have the read/write access to 'everyone' user.
String path2 = "//MACHINE-NAME/SHARE-FOLDER2"
NtlmPasswordAuthentication auth = new
NtlmPasswordAuthentication(DOMAIN;USERNAME:PASSWORD)
SmbFile destinationFolder = new SmbFile("smb://MACHINE-NAME/SHARE-FOLDER2/",auth)
ArrayList<SmbFile> readFiles = readFolder?.listFiles()
for(file in readFiles ){
file.copyTo(destinationFolder)
}

If you wanted to copy a file from one shared location to another shared location. You can this like this
ArrayList<SmbFile> readFiles = readFolder?.listFiles()
for(file in readFiles ){
String name = file.properties.getKey("name")
destinationFolder = new SmbFile(foldersInfo?.destinationFolder+"/"+name,auth)
destinationFolder.createNewFile()
file.copyTo(destinationFolder)
}
The file which you want to copy that file must be in the destination folder.
First we will create a file with same name in the destination folder then copy to that folder

Copy one file from a folder to another folder in java

I am trying to copy a file form a folder to another folder
i have tried what was suggested in other posts but i have not been successful
Copying files from one directory to another in Java
this has not worked for me
the file is C:/Users/win7/Desktop/G1_S215075820014_T111_N20738-A_D2015-01-26_P_H0.xml
the destination folder is C:/Users/win7/Desktop/destiny
this is the copy code
String origen = "C:/Users/win7/Desktop/G1_S215075820014"
+"_T111_N20738-A_D2015-01-26_P_H0.xml";
String destino = "C:/Users/win7/Desktop/destiny";
private void copiarArchivoACarpeta(String origen, String destino) throws IOException {
Path FROM = Paths.get(origen);
Path TO = Paths.get(destino);
CopyOption[] options =
new CopyOption[] {StandardCopyOption.REPLACE_EXISTING,
StandardCopyOption.COPY_ATTRIBUTES };
java.nio.file.Files.copy(FROM, TO, options);
}

Try:
java.nio.file.Files.copy(FROM, TO.resolve(FROM.getFileName()),
StandardCopyOption.REPLACE_EXISTING, StandardCopyOption.COPY_ATTRIBUTES);
Because the second parameter must be a Path to a file that not yet exists.
Just like the docu sais:

renaming file name inside a zip file

trying to rename internal file within a zip file without having to extract and then re-zip programatically.
example. test.zip contains test.txt, i want to change it so that test.zip will contain newtest.txt(test.txt renamed to newtest.txt, contents remain the same)
came across this link that works but unfortunately it expects test.txt to exist on the system. In the example the srcfile should exist on the server.
Blockquote Rename file in zip with zip4j
Then icame across zipnote on Linux that does the trick but unfortunately the version i have doesnt work for files >4GB.
Any suggestions on how to accomplish this? prefereably in java.

This should be possible using Java 7 Zip FileSystem provider, something like:
// syntax defined in java.net.JarURLConnection
URI uri = URI.create("jar:file:/directoryPath/file.zip");
try (FileSystem zipfs = FileSystems.newFileSystem(uri, Collections.<String, Object>emptyMap())) {
Path sourceURI = zipfs.getPath("/pathToDirectoryInsideZip/file.txt");
Path destinationURI = zipfs.getPath("/pathToDirectoryInsideZip/renamed.txt");
Files.move(sourceURI, destinationURI);
}

Using zip4j, I am modifying and re-writing the file headers inside of the central directory section to avoid rewriting the entire zip file:
ArrayList<FileHeader> FHs = (ArrayList<FileHeader>) zipFile.getFileHeaders();
FHs.get(0).setFileName("namename.mp4");
FHs.get(0).setFileNameLength("namename.mp4".getBytes("UTF-8").length);
zipFile.updateHeaders ();
//where updateHeaders is :
public void updateHeaders() throws ZipException, IOException {
checkZipModel();
if (this.zipModel == null) {
throw new ZipException("internal error: zip model is null");
}
if (Zip4jUtil.checkFileExists(file)) {
if (zipModel.isSplitArchive()) {
throw new ZipException("Zip file already exists. Zip file format does not allow updating split/spanned files");
}
}
long offset = zipModel.getEndCentralDirRecord().getOffsetOfStartOfCentralDir();
HeaderWriter headerWriter = new HeaderWriter();
SplitOutputStream splitOutputStream = new SplitOutputStream(new File(zipModel.getZipFile()), -1);
splitOutputStream.seek(offset);
headerWriter.finalizeZipFile(zipModel, splitOutputStream);
splitOutputStream.close();
}
The name field in the local file header section remains unchanged, so there will be a mismatch exception in this library.
It's tricky but maybe problematic, I don't know..

Where to put a file to read from a class under a package in java?

I have a properties file contains the file name only say file=fileName.dat. I've put the properties file under the class path and could read the file name(file.dat) properly from it in the mainClass. After reading the file name I passed the file name(just name not the path) to another class under a package say pack.myClass to read that file. But the problem is pack.myClass could not get the file path properly. I've put the file fileName.dat both inside and outside the packagepack but couldn't make it work.
Can anybody suggest me that where to put the file fileName.dat so I can read it properly and the whole application would be portable too.
Thanks!
The code I'm using to read the config file and getting the file name:
Properties prop = new Properties();
InputStream in = mainClass.class.getResourceAsStream("config.properties");
prop.load(in);
in.close();
myClass mc = new myClass();
mc.readTheFile(prop.getProperty("file"));
/*until this code is working good*/
Then in myClass which is under package named pack I am doing:
public void readTheFile(String filename) throws IOException {
FileReader fileReader = new FileReader(filename); /*this couldn't get the file whether i'm putting the file inside or outside the package folder */
/*after reading the file I've to do the BufferReader for further operation*/
BufferedReader bufferedReader = new BufferedReader(fileReader);

I assume that you are trying to read properties file using getResource method of class. If you put properties file on root of the classpath you should prefix file name with '/' to indicate root of classpath, for example getResource("/file.dat"). If properties file is under the same folder with the class you on which you invoke getResource method, than you should not use '/' prefix.

When you use a relative file name such as fileName.dat, you're asking for a file with this name in the current directory. The current directory has nothing to do with packages. It's the directory from which the JVM is started.
So if you're in the directory c:\foo\bar when you launch your application (using java -cp ... pack.MyClass), it will look for the file c:\foo\bar\fileName.dat.

Try..
myClass mc = new myClass();
InputStream in = mc.getClass().getResourceAsStream("/pack/config.properties");
..or simply
InputStream in = mc.getClass().getResourceAsStream("config.properties");
..for the last line if the main is in myClass The class loader available in the main() will often be the bootstrap class-loader, as opposed to the class-loader intended for application resources.

Class.getResource will look in your package directory for a file of the specified name.
JavaDocs here
Or getResourceAsStream is sometimes more convenient as you probably want to read the contents of the resource.

Most of the time it would be best to look for the "fileName.dat" somewhere in the "user.home" folder, which is a system property. First create a File path from the "user.home" and then try to find the file there. This is a bit of a guess as you don't provide the exact user of the application, but this would be the most common place.
You are currently reading from the current folder which is determined by
String currentDir = new File(".").getAbsolutePath();
or
System.getProperty("user.dir")

To read a file, even from within a jar archive:
readTheFile(String package, String filename) throws MalformedURLException, IOException
{
String filepath = package+"/"+filename;
// like "pack/fileName.dat" or "fileName.dat"
String s = (new SourceBase()).getSourceBase() + filepath;
URL url = new URL(s);
InputStream ins = url.openStream();
BufferedReader rdr = new BufferedReader(new InputStreamReader(ins, "utf8"));
do {
s = rdr.readLine();
if(s!= null) System.out.println(s);
}
while(s!=null);
rdr.close();
}
with
class SourceBase
{
public String getSourceBase()
{
String cn = this.getClass().getName().replace('.', '/') + ".class";
// like "packagex/SourceBase.class"
String s = this.getClass().getResource('/' + cn).toExternalForm();
// like "file:/javadir/Projects/projectX/build/classes/packagex/SourceBase.class"
// or "jar:file:/opt/java/PROJECTS/testProject/dist/
// testProject.jar!/px/SourceBase.class"
return s.substring(0, s.lastIndexOf(cn));
// like "file:/javadir/Projects/projectX/build/classes/"
// or "jar:file:/opt/java/PROJECTS/testProject/dist/testProject.jar!/"
}
}

Deleting random access file in java

I've created a random access file as follows:
RandomAccessFile aFile = null;
aFile = new RandomAccessFile(NetSimView.filename, "rwd");
I want to delete the file "afile". can anyone suggest me how to do it?

You can do that:
File f = new File(NetSimView.filename);
f.delete();
Edit, regarding your comment:
The parameter NetSimView.filename seems to be a File and not a String that contains the path to the file. So simply do:
NetSimView.filename.delete();