This is my code, which I want to create method, that accept file and move it in my pc of specified folder. I just make this copy existing file of text to another text file, but I want to move in specified folder, not copy. How to solve this problem?
public static void main(String[] args) {
InputStream inStream = null;
OutputStream outStream = null;
try {
File afile = new File("C:\\Users\\anar.memmedov\\Desktop\\test.txt");
File bfile = new File("C:\\Users\\anar.memmedov\\Desktop\\ok\\test3.txt");
inStream = new FileInputStream(afile);
outStream = new FileOutputStream(bfile);
byte[] buffer = new byte[1024];
int length;
//copy the file content in bytes
while ((length = inStream.read(buffer)) > 0) {
outStream.write(buffer, 0, length);
}
inStream.close();
outStream.close();
//delete the original file
afile.delete();
System.out.println("File is copied successful!");
} catch (IOException e) {
e.printStackTrace();
}
}
You can simply use Files.move: https://docs.oracle.com/javase/7/docs/api/java/nio/file/Files.html#move(java.nio.file.Path,%20java.nio.file.Path,%20java.nio.file.CopyOption...)
Move or rename a file to a target file.
By default, this method attempts to move the file to the target file, failing if the target file exists except if the source and target are the same file, in which case this method has no effect. If the file is a symbolic link then the symbolic link itself, not the target of the link, is moved. This method may be invoked to move an empty directory. In some implementations a directory has entries for special files or links that are created when the directory is created. In such implementations a directory is considered empty when only the special entries exist. When invoked to move a directory that is not empty then the directory is moved if it does not require moving the entries in the directory. For example, renaming a directory on the same FileStore will usually not require moving the entries in the directory. When moving a directory requires that its entries be moved then this method fails (by throwing an IOException). To move a file tree may involve copying rather than moving directories and this can be done using the copy method in conjunction with the Files.walkFileTree utility method.
Path sourcePath = Paths.get("sourceFile.txt");
Path targetPath = Paths.get("targetFolder\\" + sourcePath.getFileName());
Files.move(sourcePath, targetPath);
I'm using a library which wants a File() as an argument.
The file I want to pass it is one I want to package with my app, as part of the .jar
Is there any way to convert the JarEntry that I get from within my .jar to a File object I can pass?
If not and I have to copy the resource to disk temporarily, where's the best place to put the temporary file?
Thanks.
You cannot get a path to a file within a JARFile, only a stream, so you should extract it to the temporary directory and then pass that extracted file.
Here's a function I wrote to do this when I provided a db with a jar previously.
/**
* This method is responsible for extracting resource files from within the .jar to the temporary directory.
* #param filePath The filepath relative to the 'Resources/' directory within the .jar from which to extract the file.
* #return A file object to the extracted file
**/
public File extract(String filePath)
{
try
{
File f = File.createTempFile(filePath, null);
FileOutputStream resourceOS = new FileOutputStream(f);
byte[] byteArray = new byte[1024];
int i;
InputStream classIS = getClass().getClassLoader().getResourceAsStream("Resources/"+filePath);
//While the input stream has bytes
while ((i = classIS.read(byteArray)) > 0)
{
//Write the bytes to the output stream
resourceOS.write(byteArray, 0, i);
}
//Close streams to prevent errors
classIS.close();
resourceOS.close();
return f;
}
catch (Exception e)
{
System.out.println("An error has occurred while extracting the database. This may mean the program is unable to have any database interaction, please contact the developer.\nError Description:\n"+e.getMessage());
return null;
}
}
A File represents a real entry in the filesystem; a JarEntry doesn't exist on the file system. The mapping won't be there unless you extract the JAR entry to an actual file.
You can create a temp file using File.createTempFile. More details are available at this SO answer.
I am trying to add a txt file into a folder which is inside a zip file.
First, I was extracting all the contents of zip file then adding the txt file and then zipping back.
Then I read about the nio method which I can modify the zip without extracting it. Using this method I can add the txt file to the main folder of zip but I can't go deeper.
testing.zip file has res folder in it.
Here is my code:
Path txtFilePath = Paths.get("\\test\\prefs.txt");
Path zipFilePath = Paths.get("\\test\\testing.zip");
FileSystem fs;
try {
fs = FileSystems.newFileSystem(zipFilePath, null);
Path fileInsideZipPath = fs.getPath("res/prefs.txt"); //when I remover "res/" code works.
Files.copy(txtFilePath, fileInsideZipPath);
fs.close();
} catch (IOException e) {
e.printStackTrace();
}
I get the following exception:
java.nio.file.NoSuchFileException: res/
(edit to give the actual answer)
Do:
fs.getPath("res").resolve("prefs.txt")
instead of:
fs.getPath("res/prefs.txt")
The .resolve() method will do the correct thing with regards to file separators etc.
The fs.getPath("res/prefs.txt") should certainly work and you don't need to split it to fs.getPath("res").resolve("prefs.txt") as the approved answer says.
The exception java.nio.file.NoSuchFileException: res/ is slightly confusing because it mentions file but in fact directory is missing.
I had a similar problem and all I had to do was:
if (fileInsideZipPath.getParent() != null)
Files.createDirectories(fileInsideZipPath.getParent());
See full example:
#Test
public void testAddFileToArchive() throws Exception {
Path fileToAdd1 = rootTestFolder.resolve("notes1.txt");
addFileToArchive(archiveFile, "notes1.txt", fileToAdd1);
Path fileToAdd2 = rootTestFolder.resolve("notes2.txt");
addFileToArchive(archiveFile, "foo/bar/notes2.txt", fileToAdd2);
. . .
}
public void addFileToArchive(Path archiveFile, String pathInArchive, Path srcFile) throws Exception {
FileSystem fs = FileSystems.newFileSystem(archiveFile, null);
Path fileInsideZipPath = fs.getPath(pathInArchive);
if (fileInsideZipPath.getParent() != null) Files.createDirectories(fileInsideZipPath.getParent());
Files.copy(srcFile, fileInsideZipPath, StandardCopyOption.REPLACE_EXISTING);
fs.close();
}
If I remove Files.createDirectories() bit, and ensure clear start with clear test directory, I get:
java.nio.file.NoSuchFileException: foo/bar/
at com.sun.nio.zipfs.ZipFileSystem.checkParents(ZipFileSystem.java:863)
at com.sun.nio.zipfs.ZipFileSystem.newOutputStream(ZipFileSystem.java:528)
at com.sun.nio.zipfs.ZipPath.newOutputStream(ZipPath.java:792)
at com.sun.nio.zipfs.ZipFileSystemProvider.newOutputStream(ZipFileSystemProvider.java:285)
at java.nio.file.Files.newOutputStream(Files.java:216)
at java.nio.file.Files.copy(Files.java:3016)
at java.nio.file.CopyMoveHelper.copyToForeignTarget(CopyMoveHelper.java:126)
at java.nio.file.Files.copy(Files.java:1277)
at my.home.test.zipfs.TestBasicOperations.addFileToArchive(TestBasicOperations.java:111)
at my.home.test.zipfs.TestBasicOperations.testAddFileToArchive(TestBasicOperations.java:51)
I am trying to unzip a zip file which is stored in the raw folder. Code is as follows
try
{
File myDir = new File(getFilesDir().getAbsolutePath());
File newFile = new File(myDir + "/imageFolder");
if(!newFile.exists())
{
newFile.mkdir();
}
ZipInputStream zipIs = new ZipInputStream(con
.getResources().openRawResource(R.raw.images));
ZipEntry ze = null;
while ((ze = zipIs.getNextEntry()) != null)
{
Log.v("Name", ze.getName());
Log.v("Size", "" + ze.getSize());
if(ze.getSize() >0)
{
FileOutputStream fout = new FileOutputStream(newFile
+ "/" + ze.getName());
byte[] buffer = new byte[1024];
int length = 0;
while ((length = zipIs.read(buffer)) > 0)
{
fout.write(buffer, 0, length);
}
zipIs.closeEntry();
fout.close();
}
}
zipIs.close();
} catch (Exception e)
{
e.printStackTrace();
}
But I keep getting this error
01-18 11:24:28.301: W/System.err(2285): java.io.FileNotFoundException:
/data/data/com.example.ziptests/files/imageFolder/TestImages/background.png
(Not a directory)
I have absolutely no idea why it is causing this, it finds the files, but when it comes to writing them out, it brings up that error. Originally I found a problem that was caused by having the zip file zipped up on the mac, so I zipped up the file on my windows machine instead, that got rid of one problem (when you zip on a mac, it adds these extra folders and files such s store.ds which causes an error when trying to unzip), but this not a directory error keeps coming up.
Any ideas why this is happening?
Please try below link code for unzip zip file.
Code for Extract Zip File
Unzip Zip File
The Problem is I am Uploading zip File which is not made using winrar software, so it is not proper extracted and give me error.
It will solve your problem.
You can't write files to the raw folder. It is a read only dir intended to contain resource files included in your apk.
UPDATE
That file would be better in the assets directory. You can access it through the AssetManager. If not, leave it in the res/raw dir, but access it through Resources.openRawResource. Either way they are Read-Only.
I am trying to get a path to a Resource but I have had no luck.
This works (both in IDE and with the JAR) but this way I can't get a path to a file, only the file contents:
ClassLoader classLoader = getClass().getClassLoader();
PrintInputStream(classLoader.getResourceAsStream("config/netclient.p"));
If I do this:
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("config/netclient.p").getFile());
The result is:
java.io.FileNotFoundException: file:/path/to/jarfile/bot.jar!/config/netclient.p (No such file or directory)
Is there a way to get a path to a resource file?
This is deliberate. The contents of the "file" may not be available as a file. Remember you are dealing with classes and resources that may be part of a JAR file or other kind of resource. The classloader does not have to provide a file handle to the resource, for example the jar file may not have been expanded into individual files in the file system.
Anything you can do by getting a java.io.File could be done by copying the stream out into a temporary file and doing the same, if a java.io.File is absolutely necessary.
When loading a resource make sure you notice the difference between:
getClass().getClassLoader().getResource("com/myorg/foo.jpg") //relative path
and
getClass().getResource("/com/myorg/foo.jpg")); //note the slash at the beginning
I guess, this confusion is causing most of problems when loading a resource.
Also, when you're loading an image it's easier to use getResourceAsStream():
BufferedImage image = ImageIO.read(getClass().getResourceAsStream("/com/myorg/foo.jpg"));
When you really have to load a (non-image) file from a JAR archive, you might try this:
File file = null;
String resource = "/com/myorg/foo.xml";
URL res = getClass().getResource(resource);
if (res.getProtocol().equals("jar")) {
try {
InputStream input = getClass().getResourceAsStream(resource);
file = File.createTempFile("tempfile", ".tmp");
OutputStream out = new FileOutputStream(file);
int read;
byte[] bytes = new byte[1024];
while ((read = input.read(bytes)) != -1) {
out.write(bytes, 0, read);
}
out.close();
file.deleteOnExit();
} catch (IOException ex) {
Exceptions.printStackTrace(ex);
}
} else {
//this will probably work in your IDE, but not from a JAR
file = new File(res.getFile());
}
if (file != null && !file.exists()) {
throw new RuntimeException("Error: File " + file + " not found!");
}
The one line answer is -
String path = this.getClass().getClassLoader().getResource(<resourceFileName>).toExternalForm()
Basically getResource method gives the URL. From this URL you can extract the path by calling toExternalForm().
I spent a while messing around with this problem, because no solution I found actually worked, strangely enough! The working directory is frequently not the directory of the JAR, especially if a JAR (or any program, for that matter) is run from the Start Menu under Windows. So here is what I did, and it works for .class files run from outside a JAR just as well as it works for a JAR. (I only tested it under Windows 7.)
try {
//Attempt to get the path of the actual JAR file, because the working directory is frequently not where the file is.
//Example: file:/D:/all/Java/TitanWaterworks/TitanWaterworks-en.jar!/TitanWaterworks.class
//Another example: /D:/all/Java/TitanWaterworks/TitanWaterworks.class
PROGRAM_DIRECTORY = getClass().getClassLoader().getResource("TitanWaterworks.class").getPath(); // Gets the path of the class or jar.
//Find the last ! and cut it off at that location. If this isn't being run from a jar, there is no !, so it'll cause an exception, which is fine.
try {
PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(0, PROGRAM_DIRECTORY.lastIndexOf('!'));
} catch (Exception e) { }
//Find the last / and cut it off at that location.
PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(0, PROGRAM_DIRECTORY.lastIndexOf('/') + 1);
//If it starts with /, cut it off.
if (PROGRAM_DIRECTORY.startsWith("/")) PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(1, PROGRAM_DIRECTORY.length());
//If it starts with file:/, cut that off, too.
if (PROGRAM_DIRECTORY.startsWith("file:/")) PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(6, PROGRAM_DIRECTORY.length());
} catch (Exception e) {
PROGRAM_DIRECTORY = ""; //Current working directory instead.
}
This is same code from user Tombart with stream flush and close to avoid incomplete temporary file content copy from jar resource and to have unique temp file names.
File file = null;
String resource = "/view/Trial_main.html" ;
URL res = getClass().getResource(resource);
if (res.toString().startsWith("jar:")) {
try {
InputStream input = getClass().getResourceAsStream(resource);
file = File.createTempFile(new Date().getTime()+"", ".html");
OutputStream out = new FileOutputStream(file);
int read;
byte[] bytes = new byte[1024];
while ((read = input.read(bytes)) != -1) {
out.write(bytes, 0, read);
}
out.flush();
out.close();
input.close();
file.deleteOnExit();
} catch (IOException ex) {
ex.printStackTrace();
}
} else {
//this will probably work in your IDE, but not from a JAR
file = new File(res.getFile());
}
if netclient.p is inside a JAR file, it won't have a path because that file is located inside other file. in that case, the best path you can have is really file:/path/to/jarfile/bot.jar!/config/netclient.p.
You need to understand the path within the jar file.
Simply refer to it relative. So if you have a file (myfile.txt), located in foo.jar under the \src\main\resources directory (maven style). You would refer to it like:
src/main/resources/myfile.txt
If you dump your jar using jar -tvf myjar.jar you will see the output and the relative path within the jar file, and use the FORWARD SLASHES.
In my case, I have used a URL object instead Path.
File
File file = new File("my_path");
URL url = file.toURI().toURL();
Resource in classpath using classloader
URL url = MyClass.class.getClassLoader().getResource("resource_name")
When I need to read the content, I can use the following code:
InputStream stream = url.openStream();
And you can access the content using an InputStream.
follow code!
/src/main/resources/file
streamToFile(getClass().getClassLoader().getResourceAsStream("file"))
public static File streamToFile(InputStream in) {
if (in == null) {
return null;
}
try {
File f = File.createTempFile(String.valueOf(in.hashCode()), ".tmp");
f.deleteOnExit();
FileOutputStream out = new FileOutputStream(f);
byte[] buffer = new byte[1024];
int bytesRead;
while ((bytesRead = in.read(buffer)) != -1) {
out.write(buffer, 0, bytesRead);
}
return f;
} catch (IOException e) {
LOGGER.error(e.getMessage(), e);
return null;
}
}
A File is an abstraction for a file in a filesystem, and the filesystems don't know anything about what are the contents of a JAR.
Try with an URI, I think there's a jar:// protocol that might be useful for your purpouses.
The following path worked for me: classpath:/path/to/resource/in/jar
private static final String FILE_LOCATION = "com/input/file/somefile.txt";
//Method Body
InputStream invalidCharacterInputStream = URLClassLoader.getSystemResourceAsStream(FILE_LOCATION);
Getting this from getSystemResourceAsStream is the best option. By getting the inputstream rather than the file or the URL, works in a JAR file and as stand alone.
It may be a little late but you may use my library KResourceLoader to get a resource from your jar:
File resource = getResource("file.txt")
Inside your ressources folder (java/main/resources) of your jar add your file (we assume that you have added an xml file named imports.xml), after that you inject ResourceLoader if you use spring like bellow
#Autowired
private ResourceLoader resourceLoader;
inside tour function write the bellow code in order to load file:
Resource resource = resourceLoader.getResource("classpath:imports.xml");
try{
File file;
file = resource.getFile();//will load the file
...
}catch(IOException e){e.printStackTrace();}
Maybe this method can be used for quick solution.
public class TestUtility
{
public static File getInternalResource(String relativePath)
{
File resourceFile = null;
URL location = TestUtility.class.getProtectionDomain().getCodeSource().getLocation();
String codeLocation = location.toString();
try{
if (codeLocation.endsWith(".jar"){
//Call from jar
Path path = Paths.get(location.toURI()).resolve("../classes/" + relativePath).normalize();
resourceFile = path.toFile();
}else{
//Call from IDE
resourceFile = new File(TestUtility.class.getClassLoader().getResource(relativePath).getPath());
}
}catch(URISyntaxException ex){
ex.printStackTrace();
}
return resourceFile;
}
}
When in a jar file, the resource is located absolutely in the package hierarchy (not file system hierarchy). So if you have class com.example.Sweet loading a resource named ./default.conf then the resource's name is specified as /com/example/default.conf.
But if it's in a jar then it's not a File ...
This Class function can get absolute file path by relative file path in .jar file.
public class Utility {
public static void main(String[] args) throws Exception {
Utility utility = new Utility();
String absolutePath = utility.getAbsolutePath("./absolute/path/to/file");
}
public String getAbsolutePath(String relativeFilePath) throws IOException {
URL url = this.getClass().getResource(relativeFilePath);
return url.getPath();
}
}
If target file is same directory with Utility.java, your input will be ./file.txt.
When you input only / on getAbsolutePath(), it returns /Users/user/PROJECT_NAME/target/classes/. This means you can select the file like this /com/example/file.txt.