I have a jar file that contains a folder structure ie...
jar/com/foo/bar/images
Inside the Images directory I have several images. I have tried to read them out with getResourceAsStream() and also attempted to get at them via ZipEntries.
All of my attempts have been fruitless, if someone could please give me an example on how this would be done.
P.S don't worry about closing resources and error handling etc. I will be able to put that part in.
Thanks for the help.
here is the zipfile entry code I tried
private File[] getFilesFromPath(String location) {
File[] files = null;
try {
CodeSource src = this.getClass().getProtectionDomain().getCodeSource();
if (src != null) {
URL jar = src.getLocation();
ZipInputStream zip = new ZipInputStream(jar.openStream());
while(true) {
ZipEntry e = zip.getNextEntry();
if (e == null)
break;
String name = e.getName();
if (name.endsWith(location)) {
files = new File(name).listFiles((new FileFilter() {
#Override
public boolean accept(File f) {
return f.isFile();
}
}));
for (File file : files){
System.out.println(file.getName());
}
}
}
}
}catch (IOException e) {
e.printStackTrace();
}
return files;
}
The following code should work in your case as I already used following code to read hbm files from an existing jar to create dynamic session factory. It is simply reading the jar file and yhen looking image folder for given location(com/foo/bar/images) then it assume that all the file in this folder are image file. It convert file to Input stream.
Change the JARCLASSNAME to the one of your class name from jar file.
CodeSource src = JARCLASSNAME.class.getProtectionDomain().getCodeSource();
if (src != null) {
URL jar = src.getLocation();
ZipInputStream zip = new ZipInputStream(jar.openStream());
while(true) {
ZipEntry e = zip.getNextEntry();
if (e == null)
break;
String name = e.getName();
// Check the folder name if folder name match then only read the file
if (name.startsWith("com/foo/bar/images")) {
InputStream inputStream =JARCLASSNAME.class.getClassLoader().getResourceAsStream(name);
}
}
}
How about constructin a URL image?
URL imageURL = JARFILENAME.class.getResource.("pic1.gif");
then make the image like this:
Image yourImage = Toolkit.getDefaultToolKit().getImage(imageURL);
URL: http://www.daniweb.com/software-development/java/threads/21906/accessing-images-in-jar-file
Related
I have some Json files in the folder "resource/json/templates". I want to read these Json files. So far, the snippet of code below is allowing me to do so when running the program in IDE but it fails when I run it in the jar.
JSONParser parser = new JSONParser();
ClassLoader loader = getClass().getClassLoader();
URL url = loader.getResource(templateDirectory);
String path = url.getPath();
File[] files = new File(path).listFiles();
PipelineTemplateRepo pipelineTemplateRepo = new PipelineTemplateRepoImpl();
File templateFile;
JSONObject templateJson;
PipelineTemplateVo templateFromFile;
PipelineTemplateVo templateFromDB;
String templateName;
for (int i = 0; i < files.length; i++) {
if (files[i].isFile()) {
templateFile = files[i];
templateJson = (JSONObject) parser.parse(new FileReader(templateFile));
//Other logic
}
}
}
catch (Exception e) {
e.printStackTrace();
}
Any help would be greatly appreciated.
Thanks a lot.
Assuming that in the class path, in the jar the directory starts with /json (/resource is a root directory), it could be as such:
URL url = getClass().getResource("/json");
Path path = Paths.get(url.toURI());
Files.walk(path, 5).forEach(p -> System.out.printf("- %s%n", p.toString()));
This uses a jar:file://... URL, and opens a virtual file system on it.
Inspect that the jar indeed uses that path.
Reading can be done as desired.
BufferedReader in = Files.newBufferedReader(p, StandardCharsets.UTF_8);
Firstly, remember that Jars are Zip files so you can't get an individual File out of it without unzipping it. Zip files don't exactly have directories, so it's not as simple as getting the children of a directory.
This was a bit of a difficult one but I too was curious, and after researching I have come up with the following.
Firstly, you could try putting the resources into a flat Zip file (resource/json/templates.zip) nested in the Jar, then loading all the resources from that zip file since you know all the zip entries will be the resources you want. This should work even in the IDE.
String path = "resource/json/templates.zip";
ZipInputStream zis = new ZipInputStream(getClass().getResourceAsStream(path));
for (ZipEntry ze = zis.getNextEntry(); ze != null; ze = zis.getNextEntry()) {
// 'zis' is the input stream and will yield an 'EOF' before the next entry
templateJson = (JSONObject) parser.parse(zis);
}
Alternatively, you could get the running Jar, iterate through its entries, and collect the ones that are children of resource/json/templates/ then get the streams from those entries. NOTE: This will only work when running the Jar, add a check to run something else while running in the IDE.
public void runOrSomething() throws IOException, URISyntaxException {
// ... other logic ...
final String path = "resource/json/templates/";
Predicate<JarEntry> pred = (j) -> !j.isDirectory() && j.getName().startsWith(path);
try (JarFile jar = new Test().getThisJar()) {
List<JarEntry> resources = getEntriesUnderPath(jar, pred);
for (JarEntry entry : resources) {
System.out.println(entry.getName());
try (InputStream is = jar.getInputStream(entry)) {
// JarEntry streams are closed when their JarFile is closed,
// so you must use them before closing 'jar'
templateJson = (JSONObject) parser.parse(is);
// ... other logic ...
}
}
}
}
// gets ALL the children, not just direct
// path should usually end in backslash
public static List<JarEntry> getEntriesUnderPath(JarFile jar, Predicate<JarEntry> pred)
{
List<JarEntry> list = new LinkedList<>();
Enumeration<JarEntry> entries = jar.entries();
// has to iterate through all the Jar entries
while (entries.hasMoreElements()) {
JarEntry entry = entries.nextElement();
if (pred.test(entry))
list.add(entry);
}
return list;
}
public JarFile getThisJar() throws IOException, URISyntaxException {
URL url = getClass().getProtectionDomain().getCodeSource().getLocation();
return new JarFile(new File(url.toURI()));
}
I hope this helps.
I don't get how to load files from the produced Jar.
This is my code and it works fine inside the IDE, but not when I run the Jar:
URL url = ClassLoader.getSystemResource(".");
try
{
File dir = new File(url.toURI());
for (File f : dir.listFiles())
{
String fn = f.getName();
if (fn.endsWith(".png"))
{
ImageView iv = new ImageView(fn);
// ...
}
}
}
catch (Exception e)
{
e.printStackTrace();
}
The structure of the Jar is:
META-INF
de (and the rest of the packages with the class files)
file1
file2
... and so on
So the files are directly in the jar not in any subfolder.
Your code doesn't work because File objects cannot be used to access files inside a jar. What you can do is use ZipInputStreams to open & read your jar file alongside ZipEntry's to read the individual files in your jar.
This code will work in a jar, but most likely not in an IDE. In which case, you can detect the current state (IDE or Jar) and execute the desired loading code accordingly.
CodeSource src = ClientMain.class.getProtectionDomain().getCodeSource();
URL jar = src.getLocation();
ZipInputStream zip = new ZipInputStream(jar.openStream());
ZipEntry entry = null;
while ((entry = zip.getNextEntry()) != null) {
String entryName = entry.getName();
if (entryName.endsWith(".png")) {
BufferedImage image = ImageIO.read(zip);
// ...
}
}
Using the URL's already setup, we can determine if the program is in a jar or not with this simple code:
new File(jar.getFile()).toString().endsWith("jar")
This works because when in an IDE, (eclipse in my case)
new File(jar.getFile()).toString() returns
"D:\Java\Current%20Projects\Test\bin"
where as in a jar, I got
"D:\Windows%20Folders\Desktop\Test.jar"
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 am trying to move files from one directory to another delete that file from source directory after moving.
for (File file : files) {
if (file != null) {
boolean status = moveFile(file, filePath, name, docGroupId);
if (status) {
//some operations....
}
}
}
public static boolean moveFile(final File file, final String filePath, final String groupName, Integer docGroupId) {
// TODO Auto-generated method stub
String selectedDirectory = filePath + File.separator + groupName;
InputStream in = null;
OutputStream out = null;
try {
if (!file.isDirectory()) {
File dir = new File(selectedDirectory);
if (!dir.exists()) {
dir.mkdirs();
}
String newFilString = dir.getAbsolutePath() +
File.separator + file.getName();
File newFile = new File(newFilString);
in = new FileInputStream(file);
out = new FileOutputStream(newFile);
byte[] moveBuff = new byte[1024];
int butesRead;
while ((butesRead = in.read(moveBuff)) > 0) {
out.write(moveBuff, 0, butesRead);
}
}
in.close();
out.close();
if(file.delete())
return true;
} catch (Exception e) {
return false;
}
}
The program works on Linux-Ubuntu and all files are moved to another directory and deleted from source directory, but in Windows system all files are moved but failed to delete one or two files from source directory. Please note that while debugging the program is working fine.
Consider using Files.delete instead of File.delete. The javadoc says:
Note that the Files class defines the delete method to throw an IOException when a file cannot be deleted. This is useful for error reporting and to diagnose why a file cannot be deleted.
This should provide the information necessary to diagnose the problem.
So, if problem comes with delete, possible explanations:
you do file.delete() on every files and directories. How do you know the directory is empty ? If not, it will fail, then what happen to next instructions ?
file deletion is OS-dependant. On Windows, you can have many security issues, depending on which user, which rights, which location. You should check with a file-delete-alone program;
last: files can be locked by other programs (even explorer), it is also OS-dependant.
You don't need any of this if the source and target are in the same file system. Just use File.renameTo().
I have a resources folder/package in the root of my project, I "don't" want to load a certain File. If I wanted to load a certain File, I would use class.getResourceAsStream and I would be fine!! What I actually want to do is to load a "Folder" within the resources folder, loop on the Files inside that Folder and get a Stream to each file and read in the content... Assume that the File names are not determined before runtime... What should I do? Is there a way to get a list of the files inside a Folder in your jar File?
Notice that the Jar file with the resources is the same jar file from which the code is being run...
Finally, I found the solution:
final String path = "sample/folder";
final File jarFile = new File(getClass().getProtectionDomain().getCodeSource().getLocation().getPath());
if(jarFile.isFile()) { // Run with JAR file
final JarFile jar = new JarFile(jarFile);
final Enumeration<JarEntry> entries = jar.entries(); //gives ALL entries in jar
while(entries.hasMoreElements()) {
final String name = entries.nextElement().getName();
if (name.startsWith(path + "/")) { //filter according to the path
System.out.println(name);
}
}
jar.close();
} else { // Run with IDE
final URL url = Launcher.class.getResource("/" + path);
if (url != null) {
try {
final File apps = new File(url.toURI());
for (File app : apps.listFiles()) {
System.out.println(app);
}
} catch (URISyntaxException ex) {
// never happens
}
}
}
The second block just work when you run the application on IDE (not with jar file), You can remove it if you don't like that.
Try the following.
Make the resource path "<PathRelativeToThisClassFile>/<ResourceDirectory>" E.g. if your class path is com.abc.package.MyClass and your resoure files are within src/com/abc/package/resources/:
URL url = MyClass.class.getResource("resources/");
if (url == null) {
// error - missing folder
} else {
File dir = new File(url.toURI());
for (File nextFile : dir.listFiles()) {
// Do something with nextFile
}
}
You can also use
URL url = MyClass.class.getResource("/com/abc/package/resources/");
The following code returns the wanted "folder" as Path regardless of if it is inside a jar or not.
private Path getFolderPath() throws URISyntaxException, IOException {
URI uri = getClass().getClassLoader().getResource("folder").toURI();
if ("jar".equals(uri.getScheme())) {
FileSystem fileSystem = FileSystems.newFileSystem(uri, Collections.emptyMap(), null);
return fileSystem.getPath("path/to/folder/inside/jar");
} else {
return Paths.get(uri);
}
}
Requires java 7+.
I know this is many years ago . But just for other people come across this topic.
What you could do is to use getResourceAsStream() method with the directory path, and the input Stream will have all the files name from that dir. After that you can concat the dir path with each file name and call getResourceAsStream for each file in a loop.
I had the same problem at hands while i was attempting to load some hadoop configurations from resources packed in the jar... on both the IDE and on jar (release version).
I found java.nio.file.DirectoryStream to work the best to iterate over directory contents over both local filesystem and jar.
String fooFolder = "/foo/folder";
....
ClassLoader classLoader = foofClass.class.getClassLoader();
try {
uri = classLoader.getResource(fooFolder).toURI();
} catch (URISyntaxException e) {
throw new FooException(e.getMessage());
} catch (NullPointerException e){
throw new FooException(e.getMessage());
}
if(uri == null){
throw new FooException("something is wrong directory or files missing");
}
/** i want to know if i am inside the jar or working on the IDE*/
if(uri.getScheme().contains("jar")){
/** jar case */
try{
URL jar = FooClass.class.getProtectionDomain().getCodeSource().getLocation();
//jar.toString() begins with file:
//i want to trim it out...
Path jarFile = Paths.get(jar.toString().substring("file:".length()));
FileSystem fs = FileSystems.newFileSystem(jarFile, null);
DirectoryStream<Path> directoryStream = Files.newDirectoryStream(fs.getPath(fooFolder));
for(Path p: directoryStream){
InputStream is = FooClass.class.getResourceAsStream(p.toString()) ;
performFooOverInputStream(is);
/** your logic here **/
}
}catch(IOException e) {
throw new FooException(e.getMessage());
}
}
else{
/** IDE case */
Path path = Paths.get(uri);
try {
DirectoryStream<Path> directoryStream = Files.newDirectoryStream(path);
for(Path p : directoryStream){
InputStream is = new FileInputStream(p.toFile());
performFooOverInputStream(is);
}
} catch (IOException _e) {
throw new FooException(_e.getMessage());
}
}
Another solution, you can do it using ResourceLoader like this:
import org.springframework.core.io.Resource;
import org.apache.commons.io.FileUtils;
#Autowire
private ResourceLoader resourceLoader;
...
Resource resource = resourceLoader.getResource("classpath:/path/to/you/dir");
File file = resource.getFile();
Iterator<File> fi = FileUtils.iterateFiles(file, null, true);
while(fi.hasNext()) {
load(fi.next())
}
If you are using Spring you can use org.springframework.core.io.support.PathMatchingResourcePatternResolver and deal with Resource objects rather than files. This works when running inside and outside of a Jar file.
PathMatchingResourcePatternResolver r = new PathMatchingResourcePatternResolver();
Resource[] resources = r.getResources("/myfolder/*");
Then you can access the data using getInputStream and the filename from getFilename.
Note that it will still fail if you try to use the getFile while running from a Jar.
As the other answers point out, once the resources are inside a jar file, things get really ugly. In our case, this solution:
https://stackoverflow.com/a/13227570/516188
works very well in the tests (since when the tests are run the code is not packed in a jar file), but doesn't work when the app actually runs normally. So what I've done is... I hardcode the list of the files in the app, but I have a test which reads the actual list from disk (can do it since that works in tests) and fails if the actual list doesn't match with the list the app returns.
That way I have simple code in my app (no tricks), and I'm sure I didn't forget to add a new entry in the list thanks to the test.
Below code gets .yaml files from a custom resource directory.
ClassLoader classLoader = this.getClass().getClassLoader();
URI uri = classLoader.getResource(directoryPath).toURI();
if("jar".equalsIgnoreCase(uri.getScheme())){
Pattern pattern = Pattern.compile("^.+" +"/classes/" + directoryPath + "/.+.yaml$");
log.debug("pattern {} ", pattern.pattern());
ApplicationHome home = new ApplicationHome(SomeApplication.class);
JarFile file = new JarFile(home.getSource());
Enumeration<JarEntry> jarEntries = file.entries() ;
while(jarEntries.hasMoreElements()){
JarEntry entry = jarEntries.nextElement();
Matcher matcher = pattern.matcher(entry.getName());
if(matcher.find()){
InputStream in =
file.getInputStream(entry);
//work on the stream
}
}
}else{
//When Spring boot application executed through Non-Jar strategy like through IDE or as a War.
String path = uri.getPath();
File[] files = new File(path).listFiles();
for(File file: files){
if(file != null){
try {
InputStream is = new FileInputStream(file);
//work on stream
} catch (Exception e) {
log.error("Exception while parsing file yaml file {} : {} " , file.getAbsolutePath(), e.getMessage());
}
}else{
log.warn("File Object is null while parsing yaml file");
}
}
}
Took me 2-3 days to get this working, in order to have the same url that work for both Jar or in local, the url (or path) needs to be a relative path from the repository root.
..meaning, the location of your file or folder from your src folder.
could be "/main/resources/your-folder/" or "/client/notes/somefile.md"
Whatever it is, in order for your JAR file to find it, the url must be a relative path from the repository root.
it must be "src/main/resources/your-folder/" or "src/client/notes/somefile.md"
Now you get the drill, and luckily for Intellij Idea users, you can get the correct path with a right-click on the folder or file -> copy Path/Reference.. -> Path From Repository Root (this is it)
Last, paste it and do your thing.
Simple ... use OSGi. In OSGi you can iterate over your Bundle's entries with findEntries and findPaths.
Inside my jar file I had a folder called Upload, this folder had three other text files inside it and I needed to have an exactly the same folder and files outside of the jar file, I used the code below:
URL inputUrl = getClass().getResource("/upload/blabla1.txt");
File dest1 = new File("upload/blabla1.txt");
FileUtils.copyURLToFile(inputUrl, dest1);
URL inputUrl2 = getClass().getResource("/upload/blabla2.txt");
File dest2 = new File("upload/blabla2.txt");
FileUtils.copyURLToFile(inputUrl2, dest2);
URL inputUrl3 = getClass().getResource("/upload/blabla3.txt");
File dest3 = new File("upload/Bblabla3.txt");
FileUtils.copyURLToFile(inputUrl3, dest3);