This question already has answers here:
Write chinese characters from one file to another
(2 answers)
Closed 7 years ago.
copying Chinese file in java using this code . but the destination file contains question mark (?) instead of Chinese character . is there any way in java to achieve this functionality..
File source = new File("H:\\work-temp\\file");
File dest = new File("H:\\work-temp\\file2");
try {
FileUtils.copyDirectory(source, dest);
} catch (IOException e) {
e.printStackTrace();
}
First of all, while copying a file, the destination should be a folder not file... so please change File dest = new File("H:\\work-temp\\file2"); to File dest = new File("H:\\work-temp");
Next you should use FileUtils.copyFile(source, dest); instead FileUtils.copyDirectory(source, dest);
Use one of the JDK 7 Files.copy methods. They create a binary copy of your file.
You are miss the extension of files for ex: file.txt and file2.txt:
File source = new File("H:\\work-temp\\file.txt");
File dest = new File("H:\\work-temp\\file2.txt");
For coping use this:
FileChannel inputChannel = null;
FileChannel outputChannel = null;
try{
inputChannel = new FileInputStream(source).getChannel();
outputChannel = new FileOutputStream(dest).getChannel();
outputChannel.transferFrom(inputChannel, 0, inputChannel.size());
} finally {
inputChannel.close();
outputChannel.close();
}
For more info go to this link
Related
How do you move a file from one location to another? When I run my program any file created in that location automatically moves to the specified location. How do I know which file is moved?
myFile.renameTo(new File("/the/new/place/newName.file"));
File#renameTo does that (it can not only rename, but also move between directories, at least on the same file system).
Renames the file denoted by this abstract pathname.
Many aspects of the behavior of this method are inherently platform-dependent: The rename operation might not be able to move a file from one filesystem to another, it might not be atomic, and it might not succeed if a file with the destination abstract pathname already exists. The return value should always be checked to make sure that the rename operation was successful.
If you need a more comprehensive solution (such as wanting to move the file between disks), look at Apache Commons FileUtils#moveFile
With Java 7 or newer you can use Files.move(from, to, CopyOption... options).
E.g.
Files.move(Paths.get("/foo.txt"), Paths.get("bar.txt"), StandardCopyOption.REPLACE_EXISTING);
See the Files documentation for more details
Java 6
public boolean moveFile(String sourcePath, String targetPath) {
File fileToMove = new File(sourcePath);
return fileToMove.renameTo(new File(targetPath));
}
Java 7 (Using NIO)
public boolean moveFile(String sourcePath, String targetPath) {
boolean fileMoved = true;
try {
Files.move(Paths.get(sourcePath), Paths.get(targetPath), StandardCopyOption.REPLACE_EXISTING);
} catch (Exception e) {
fileMoved = false;
e.printStackTrace();
}
return fileMoved;
}
File.renameTo from Java IO can be used to move a file in Java. Also see this SO question.
To move a file you could also use Jakarta Commons IOs FileUtils.moveFile
On error it throws an IOException, so when no exception is thrown you know that that the file was moved.
Just add the source and destination folder paths.
It will move all the files and folder from source folder to
destination folder.
File destinationFolder = new File("");
File sourceFolder = new File("");
if (!destinationFolder.exists())
{
destinationFolder.mkdirs();
}
// Check weather source exists and it is folder.
if (sourceFolder.exists() && sourceFolder.isDirectory())
{
// Get list of the files and iterate over them
File[] listOfFiles = sourceFolder.listFiles();
if (listOfFiles != null)
{
for (File child : listOfFiles )
{
// Move files to destination folder
child.renameTo(new File(destinationFolder + "\\" + child.getName()));
}
// Add if you want to delete the source folder
sourceFolder.delete();
}
}
else
{
System.out.println(sourceFolder + " Folder does not exists");
}
Files.move(source, target, REPLACE_EXISTING);
You can use the Files object
Read more about Files
You could execute an external tool for that task (like copy in windows environments) but, to keep the code portable, the general approach is to:
read the source file into memory
write the content to a file at the new location
delete the source file
File#renameTo will work as long as source and target location are on the same volume. Personally I'd avoid using it to move files to different folders.
Try this :-
boolean success = file.renameTo(new File(Destdir, file.getName()));
Wrote this method to do this very thing on my own project only with the replace file if existing logic in it.
// we use the older file i/o operations for this rather than the newer jdk7+ Files.move() operation
private boolean moveFileToDirectory(File sourceFile, String targetPath) {
File tDir = new File(targetPath);
if (tDir.exists()) {
String newFilePath = targetPath+File.separator+sourceFile.getName();
File movedFile = new File(newFilePath);
if (movedFile.exists())
movedFile.delete();
return sourceFile.renameTo(new File(newFilePath));
} else {
LOG.warn("unable to move file "+sourceFile.getName()+" to directory "+targetPath+" -> target directory does not exist");
return false;
}
}
Please try this.
private boolean filemovetoanotherfolder(String sourcefolder, String destinationfolder, String filename) {
boolean ismove = false;
InputStream inStream = null;
OutputStream outStream = null;
try {
File afile = new File(sourcefolder + filename);
File bfile = new File(destinationfolder + filename);
inStream = new FileInputStream(afile);
outStream = new FileOutputStream(bfile);
byte[] buffer = new byte[1024 * 4];
int length;
// copy the file content in bytes
while ((length = inStream.read(buffer)) > 0) {
outStream.write(buffer, 0, length);
}
// delete the original file
afile.delete();
ismove = true;
System.out.println("File is copied successful!");
} catch (IOException e) {
e.printStackTrace();
}finally{
inStream.close();
outStream.close();
}
return ismove;
}
I have a .jar that has two .dll files that it is dependent on. I would like to know if there is any way for me to copy these files from within the .jar to a users temp folder at runtime. here is the current code that I have (edited to just one .dll load to reduce question size):
public String tempDir = System.getProperty("java.io.tmpdir");
public String workingDir = dllInstall.class.getProtectionDomain().getCodeSource().getLocation().getPath();
public boolean installDLL() throws UnsupportedEncodingException {
try {
String decodedPath = URLDecoder.decode(workingDir, "UTF-8");
InputStream fileInStream = null;
OutputStream fileOutStream = null;
File fileIn = new File(decodedPath + "\\loadAtRuntime.dll");
File fileOut = new File(tempDir + "loadAtRuntime.dll");
fileInStream = new FileInputStream(fileIn);
fileOutStream = new FileOutputStream(fileOut);
byte[] bufferJNI = new byte[8192000013370000];
int lengthFileIn;
while ((lengthFileIn = fileInStream.read(bufferJNI)) > 0) {
fileOutStream.write(bufferJNI, 0, lengthFileIn);
}
//close all steams
} catch (IOException e) {
e.printStackTrace();
return false;
} catch (UnsupportedEncodingException e) {
System.out.println(e);
return false;
}
My main problem is getting the .dll files out of the jar at runtime. Any way to retrieve the path from within the .jar would be helpful.
Thanks in advance.
Since your dlls are bundeled inside your jar file you could just try to acasses them as resources using ClassLoader#getResourceAsStream and write them as binary files any where you want on the hard drive.
Here is some sample code:
InputStream ddlStream = <SomeClassInsideTheSameJar>.class
.getClassLoader().getResourceAsStream("some/pack/age/somelib.dll");
try (FileOutputStream fos = new FileOutputStream("somelib.dll");){
byte[] buf = new byte[2048];
int r;
while(-1 != (r = ddlStream.read(buf))) {
fos.write(buf, 0, r);
}
}
The code above will extract the dll located in the package some.pack.age to the current working directory.
Use a class loader that is able to locate resources in this JAR file. Either you can use the class loader of a class as Peter Lawrey suggested, or you can also create a URLClassLoader with the URL to that JAR.
Once you have that class loader you can retrieve a byte input stream with ClassLoader.getResourceAsStream. On the other hand you just create a FileOutputStream for the file you want to create.
The last step then is to copy all bytes from the input stream to the output stream, as you already did in your code example.
Use myClass.getClassLoader().getResourceAsStream("loadAtRuntime.dll"); and you will be able to find and copy DLLs in the JAR. You should pick a class which will also be in the same JAR.
This question already has answers here:
Copying files from one directory to another in Java
(34 answers)
Closed 9 years ago.
I want to copy a zip file from one folder to another folder in java.
i have a migrate.zip file in sourcefolder .i need to copy that migrat.zip file to destination
folder.
can any one help me on this.
Thanks&Regards,
sivakrishna.m
apache-commons-io library is helpful in you problem
org.apache.commons.io.FileUtils.copyFile(File, File)
FileUtils.copyFile(new File("/sourcefolder/migrate.zip"),
new File("/destination/migrate.zip"))
Please check the below question and answer. This may help you.
Best Way to copy a Zip File via Java
try this set of lines.
String sourceFilePath =" Source path";
File f = new File(sourceFilePath);
File f1 = new File(destinationFilePath);
File fCopy = new File(destinationFilePath);
if (f1.exists()) {
// Don't do anything..
f1.delete();
}
FileUtils.copyFile(f, fCopy)
Use java.util.ZipInputStream class to read migrate.zip file from source folder and use java.util.ZipOutputStream class to write migrate.zip to the destination folder....
public class CopyZip
{
public static void main(String[] args)
{
FileInputStream fin = new FileInputStream(new File("source_folder\migrate.zip"));
ZipInputStream zin = new ZipInputStream(fin);
byte[] in_bytes = new bytes[1000];
zin.read(in_bytes,0,1000);
FileOutputStream fout = new FileOutputStream(new File("dest_folder\migrate.zip"));
ZipOutputSrream zout = new ZipOutputStream(fout);
zout.write(in_bytes,0,in_bytes.length);
}
}
i need some help with creating file
Im trying in the last hours to work with RandomAccessFile and try to achieve the next logic:
getting a file object
creating a temporary file with similar name (how do i make sure the temp file will be created in same place as the given original one?)
write to this file
replace the original file on the disk with the temporary one (should be in original filename).
I look for a simple code who does that preferring with RandomAccessFile
I just don't how to solve these few steps right..
edited:
Okay so ive attachted this part of code
my problem is that i can't understand what should be the right steps..
the file isn't being created and i don't know how to do that "switch"
File tempFile = null;
String[] fileArray = null;
RandomAccessFile rafTemp = null;
try {
fileArray = FileTools.splitFileNameAndExtension(this.file);
tempFile = File.createTempFile(fileArray[0], "." + fileArray[1],
this.file); // also tried in the 3rd parameter this.file.getParentFile() still not working.
rafTemp = new RandomAccessFile(tempFile, "rw");
rafTemp.writeBytes("temp file content");
tempFile.renameTo(this.file);
} catch (IOException ex) {
ex.printStackTrace();
} finally {
rafTemp.close();
}
try {
// Create temp file.
File temp = File.createTempFile("TempFileName", ".tmp", new File("/"));
// Delete temp file when program exits.
temp.deleteOnExit();
// Write to temp file
BufferedWriter out = new BufferedWriter(new FileWriter(temp));
out.write("Some temp file content");
out.close();
// Original file
File orig = new File("/orig.txt");
// Copy the contents from temp to original file
FileChannel src = new FileInputStream(temp).getChannel();
FileChannel dest = new FileOutputStream(orig).getChannel();
dest.transferFrom(src, 0, src.size());
} catch (IOException e) { // Handle exceptions here}
you can direct overwrite file. or do following
create file in same directory with diff name
delete old file
rename new file
I am trying to create a back up file for an html file on a web server.
I want the backup to be in the same location as the existing file (it's a quick fix). I want to create the file using File file = new File(PathName);
public void backUpOldPage(String oldContent) throws IOException{
// this.uri is a class variable with the path of the file to be backed up
String fileName = new File(this.uri).getName();
String pathName = new File(this.uri).getPath();
System.out.println(pathName);
String bckPath = pathName+"\\"+bckName;
FileOutputStream fout;
try
{
// Open an output stream
fout = new FileOutputStream (bckFile);
fout.close();
}
// Catches any error conditions
catch (IOException e)
{
System.err.println ("Unable to write to file");
System.exit(-1);
}
}
But if instead I was to set bckPath like this, it will work.
String bckPath = "C://dev/server/tomcat6/webapps/sample-site/index_sdjf---sd.html";
I am working on Windows, not sure if that makes a difference.
The result of String bckPath = pathName+"\"+bckName;
is bckPath = C:\dev\server\tomcat6\webapps\sample-site\filename.html - this doesn't result in a new file.
Use File.pathSeparator, that way you dont need to worry what OS you are using.
Try to use File.getCanonicalPath() instead of plain getPath(). This helps if the orginal path is not fully specified.
Regarding slashes, / or \ or File.pathSeparator is not causing the problem, because they are all the same on Windows and Java. (And you do not define bckFile in your code, only bckPath. Also use getCanonicalPath() on the new created bckPath.)