I am working on a multi module Java project, it consists of multiple projects in Eclipse that depend on each other, now to add GUI Theme support I have created a Java project that does not contain any code, just the icons and pictures needed for the GUI, I made it a Java project so Maven will build it into a .jar. Now there's a main Application that loads the multiple projects into a main GUI, which gets its icons from resources within the actual modules at the moment. All I want to do is load all these resources from the external dummy .jar that is included in the classpath of the main application's .jar. Every method I have found so far does not seem to work. The .jar contains no actual java .classes, so there is no ClassLoader to reference. Are there any other methods to load the pictures without extracting the .jar?
Three things you might try:
Create a dummy class in the resource jar, which you could use to get the ClassLoader reference.
Use URLClassLoader, if you know, where your jar-File resides.
You always have the possibility to create your own ClassLoader by extending java.lang.ClassLoader.
Treat the external jar file as a zip-archive and read the images/resource something like this:
public ZipStuff(String filename) throws IOException
{
ZipFile zf = new ZipFile(filename);
Enumeration<? extends ZipEntry> entries = zf.entries();
while (entries.hasMoreElements()) {
ZipEntry ze = entries.nextElement();
ImageIcon ii = new ImageIcon(ImageIO.read(zf.getInputStream(ze)));
JLabel l = new JLabel(ii);
getContentPane().add(l);
}
setVisible(true);
pack();
}
In this case I have a jar file with one single image which I load into a JLabel. The ZipStuff-class is a JFrame.
You could get all of the resources (all jar files on classpath should work even one without classes):
Enumeration<URL> resources = null;
try {
resources = Thread.currentThread().getContextClassLoader().getResources(someResource);
} catch (Exception ex) {
//no op
}
if (resources == null || !resources.hasMoreElements()) {
resources = ClasspathReader.class.getClassLoader().getResources(someResource);
}
Then check to see if the current resource is a file. Files you can deal with direct as files.
But your question was about jar files so I wont go there.
while (resources.hasMoreElements()) {
URL resource = resources.nextElement();
if (resource.getProtocol().equals("file")) {
//if it is a file then we can handle it the normal way.
handleFile(resource, namespace);
continue;
}
At this point you should have only jar:file resources so...
Split up the string that looks like this:
jar:file:/Users/rick/.m2/repository/invoke/invoke/1.0-SNAPSHOT/invoke-1.0-SNAPSHOT.jar!/org/node/
into this
/Users/rick/.m2/repository/invoke/invoke/1.0-SNAPSHOT/invoke-1.0-SNAPSHOT.jar
and this
/org/node/
Here is the code to do the above devoid of pesky error checking. :)
String[] split = resource.toString().split(":");
String[] split2 = split[2].split("!");
String zipFileName = split2[0];
String sresource = split2[1];
System.out.printf("After split zip file name = %s," +
" \nresource in zip %s \n", zipFileName, sresource);
Now we have the zip file name so we can read it:
ZipFile zipFile = new ZipFile(zipFileName);
Now we can iterate through its entries:
Enumeration<? extends ZipEntry> entries = zipFile.entries();
while (entries.hasMoreElements()) {
ZipEntry entry = entries.nextElement();
/* If it is a directory, then skip it. */
if (entry.isDirectory()) {
continue;
}
String entryName = entry.getName();
System.out.printf("zip entry name %s \n", entryName);
See if it starts with the resource we are looking for.
if (!entryName.startsWith(someResource)) {
continue;
}
There were two tricks I did earlier to see if it was a directory
boolean isDir = !someResource.endsWith(".txt");
This only works because I am looking for resources that ends with .txt and I assume if it does not end with .txt that it is an dir /foo/dir and /foo/dir/ both work.
The other trick was this:
if (someResource.startsWith("/")) {
someResource = someResource.substring(1);
}
Classpath resource can never really start with a starting slash. Logically they do, but in reality you have to strip it. This is a known behavior of classpath resources. It works with a slash unless the resource is in a jar file. Bottom line, by stripping it, it always works.
We need to get the actual fileName of the darn thing. The fileName part from the entry name.
where /foo/bar/foo/bee/bar.txt, and we want 'bar.txt' which is the fileName. Back inside of our while loop.
while (entries.hasMoreElements()) {
ZipEntry entry = entries.nextElement();
...
String entryName = entry.getName(); //entry is zipEntry
String fileName = entryName.substring(entryName.lastIndexOf("/") + 1);
/** See if the file starts with our namespace and ends with our extension. */
if (fileName.startsWith(namespace) && fileName.endsWith(".txt")) {
Next we see if these entry matches our criteria and if so read contents of the file to System.out.
try (Reader reader = new InputStreamReader(zipFile.getInputStream(entry))) {
StringBuilder builder = new StringBuilder();
int ch = 0;
while ((ch = reader.read()) != -1) {
builder.append((char) ch);
}
System.out.printf("zip fileName = %s\n\n####\n contents of file %s\n###\n",
entryName, builder);
} catch (Exception ex) {
ex.printStackTrace();//it is an example/proto :)
}
}
You can see the full example here: Sleepless in Pleasanton.
Related
I'm trying to access an XML file within a jar file, from a separate jar that's running as a desktop application. I can get the URL to the file I need, but when I pass that to a FileReader (as a String) I get a FileNotFoundException saying "The file name, directory name, or volume label syntax is incorrect."
As a point of reference, I have no trouble reading image resources from the same jar, passing the URL to an ImageIcon constructor. This seems to indicate that the method I'm using to get the URL is correct.
URL url = getClass().getResource("/xxx/xxx/xxx/services.xml");
ServicesLoader jsl = new ServicesLoader( url.toString() );
Inside the ServicesLoader class I have
XMLReader xr = XMLReaderFactory.createXMLReader();
xr.setContentHandler( this );
xr.setErrorHandler( this );
xr.parse( new InputSource( new FileReader( filename )));
What's wrong with using this technique to read the XML file?
Looks like you want to use java.lang.Class.getResourceAsStream(String), see
https://docs.oracle.com/javase/8/docs/api/java/lang/Class.html#getResourceAsStream-java.lang.String-
You don't say if this is a desktop or web app. I would use the getResourceAsStream() method from an appropriate ClassLoader if it's a desktop or the Context if it's a web app.
It looks as if you are using the URL.toString result as the argument to the FileReader constructor. URL.toString is a bit broken, and instead you should generally use url.toURI().toString(). In any case, the string is not a file path.
Instead, you should either:
Pass the URL to ServicesLoader and let it call openStream or similar.
Use Class.getResourceAsStream and just pass the stream over, possibly inside an InputSource. (Remember to check for nulls as the API is a bit messy.)
The problem was that I was going a step too far in calling the parse method of XMLReader. The parse method accepts an InputSource, so there was no reason to even use a FileReader. Changing the last line of the code above to
xr.parse( new InputSource( filename ));
works just fine.
I'd like to point out that one issues is what if the same resources are in multiple jar files.
Let's say you want to read /org/node/foo.txt, but not from one file, but from each and every jar file.
I have run into this same issue several times before.
I was hoping in JDK 7 that someone would write a classpath filesystem, but alas not yet.
Spring has the Resource class which allows you to load classpath resources quite nicely.
I wrote a little prototype to solve this very problem of reading resources form multiple jar files. The prototype does not handle every edge case, but it does handle looking for resources in directories that are in the jar files.
I have used Stack Overflow for quite sometime. This is the second answer that I remember answering a question so forgive me if I go too long (it is my nature).
This is a prototype resource reader. The prototype is devoid of robust error checking.
I have two prototype jar files that I have setup.
<pre>
<dependency>
<groupId>invoke</groupId>
<artifactId>invoke</artifactId>
<version>1.0-SNAPSHOT</version>
</dependency>
<dependency>
<groupId>node</groupId>
<artifactId>node</artifactId>
<version>1.0-SNAPSHOT</version>
</dependency>
The jar files each have a file under /org/node/ called resource.txt.
This is just a prototype of what a handler would look like with classpath://
I also have a resource.foo.txt in my local resources for this project.
It picks them all up and prints them out.
package com.foo;
import java.io.File;
import java.io.FileReader;
import java.io.InputStreamReader;
import java.io.Reader;
import java.net.URI;
import java.net.URL;
import java.util.Enumeration;
import java.util.zip.ZipEntry;
import java.util.zip.ZipFile;
/**
* Prototype resource reader.
* This prototype is devoid of error checking.
*
*
* I have two prototype jar files that I have setup.
* <pre>
* <dependency>
* <groupId>invoke</groupId>
* <artifactId>invoke</artifactId>
* <version>1.0-SNAPSHOT</version>
* </dependency>
*
* <dependency>
* <groupId>node</groupId>
* <artifactId>node</artifactId>
* <version>1.0-SNAPSHOT</version>
* </dependency>
* </pre>
* The jar files each have a file under /org/node/ called resource.txt.
* <br />
* This is just a prototype of what a handler would look like with classpath://
* I also have a resource.foo.txt in my local resources for this project.
* <br />
*/
public class ClasspathReader {
public static void main(String[] args) throws Exception {
/* This project includes two jar files that each have a resource located
in /org/node/ called resource.txt.
*/
/*
Name space is just a device I am using to see if a file in a dir
starts with a name space. Think of namespace like a file extension
but it is the start of the file not the end.
*/
String namespace = "resource";
//someResource is classpath.
String someResource = args.length > 0 ? args[0] :
//"classpath:///org/node/resource.txt"; It works with files
"classpath:///org/node/"; //It also works with directories
URI someResourceURI = URI.create(someResource);
System.out.println("URI of resource = " + someResourceURI);
someResource = someResourceURI.getPath();
System.out.println("PATH of resource =" + someResource);
boolean isDir = !someResource.endsWith(".txt");
/** Classpath resource can never really start with a starting slash.
* Logically they do, but in reality you have to strip it.
* This is a known behavior of classpath resources.
* It works with a slash unless the resource is in a jar file.
* Bottom line, by stripping it, it always works.
*/
if (someResource.startsWith("/")) {
someResource = someResource.substring(1);
}
/* Use the ClassLoader to lookup all resources that have this name.
Look for all resources that match the location we are looking for. */
Enumeration resources = null;
/* Check the context classloader first. Always use this if available. */
try {
resources =
Thread.currentThread().getContextClassLoader().getResources(someResource);
} catch (Exception ex) {
ex.printStackTrace();
}
if (resources == null || !resources.hasMoreElements()) {
resources = ClasspathReader.class.getClassLoader().getResources(someResource);
}
//Now iterate over the URLs of the resources from the classpath
while (resources.hasMoreElements()) {
URL resource = resources.nextElement();
/* if the resource is a file, it just means that we can use normal mechanism
to scan the directory.
*/
if (resource.getProtocol().equals("file")) {
//if it is a file then we can handle it the normal way.
handleFile(resource, namespace);
continue;
}
System.out.println("Resource " + resource);
/*
Split up the string that looks like this:
jar:file:/Users/rick/.m2/repository/invoke/invoke/1.0-SNAPSHOT/invoke-1.0-SNAPSHOT.jar!/org/node/
into
this /Users/rick/.m2/repository/invoke/invoke/1.0-SNAPSHOT/invoke-1.0-SNAPSHOT.jar
and this
/org/node/
*/
String[] split = resource.toString().split(":");
String[] split2 = split[2].split("!");
String zipFileName = split2[0];
String sresource = split2[1];
System.out.printf("After split zip file name = %s," +
" \nresource in zip %s \n", zipFileName, sresource);
/* Open up the zip file. */
ZipFile zipFile = new ZipFile(zipFileName);
/* Iterate through the entries. */
Enumeration entries = zipFile.entries();
while (entries.hasMoreElements()) {
ZipEntry entry = entries.nextElement();
/* If it is a directory, then skip it. */
if (entry.isDirectory()) {
continue;
}
String entryName = entry.getName();
System.out.printf("zip entry name %s \n", entryName);
/* If it does not start with our someResource String
then it is not our resource so continue.
*/
if (!entryName.startsWith(someResource)) {
continue;
}
/* the fileName part from the entry name.
* where /foo/bar/foo/bee/bar.txt, bar.txt is the file
*/
String fileName = entryName.substring(entryName.lastIndexOf("/") + 1);
System.out.printf("fileName %s \n", fileName);
/* See if the file starts with our namespace and ends with our extension.
*/
if (fileName.startsWith(namespace) && fileName.endsWith(".txt")) {
/* If you found the file, print out
the contents fo the file to System.out.*/
try (Reader reader = new InputStreamReader(zipFile.getInputStream(entry))) {
StringBuilder builder = new StringBuilder();
int ch = 0;
while ((ch = reader.read()) != -1) {
builder.append((char) ch);
}
System.out.printf("zip fileName = %s\n\n####\n contents of file %s\n###\n", entryName, builder);
} catch (Exception ex) {
ex.printStackTrace();
}
}
//use the entry to see if it's the file '1.txt'
//Read from the byte using file.getInputStream(entry)
}
}
}
/**
* The file was on the file system not a zip file,
* this is here for completeness for this example.
* otherwise.
*
* #param resource
* #param namespace
* #throws Exception
*/
private static void handleFile(URL resource, String namespace) throws Exception {
System.out.println("Handle this resource as a file " + resource);
URI uri = resource.toURI();
File file = new File(uri.getPath());
if (file.isDirectory()) {
for (File childFile : file.listFiles()) {
if (childFile.isDirectory()) {
continue;
}
String fileName = childFile.getName();
if (fileName.startsWith(namespace) && fileName.endsWith("txt")) {
try (FileReader reader = new FileReader(childFile)) {
StringBuilder builder = new StringBuilder();
int ch = 0;
while ((ch = reader.read()) != -1) {
builder.append((char) ch);
}
System.out.printf("fileName = %s\n\n####\n contents of file %s\n###\n", childFile, builder);
} catch (Exception ex) {
ex.printStackTrace();
}
}
}
} else {
String fileName = file.getName();
if (fileName.startsWith(namespace) && fileName.endsWith("txt")) {
try (FileReader reader = new FileReader(file)) {
StringBuilder builder = new StringBuilder();
int ch = 0;
while ((ch = reader.read()) != -1) {
builder.append((char) ch);
}
System.out.printf("fileName = %s\n\n####\n contents of file %s\n###\n", fileName, builder);
} catch (Exception ex) {
ex.printStackTrace();
}
}
}
}
}
You can see a fuller example here with the sample output.
Here's a sample code on how to read a file properly inside a jar file (in this case, the current executing jar file)
Just change executable with the path of your jar file if it is not the current running one.
Then change the filePath to the path of the file you want to use inside the jar file. I.E. if your file is in
someJar.jar\img\test.gif
. Set the filePath to "img\test.gif"
File executable = new File(BrowserViewControl.class.getProtectionDomain().getCodeSource().getLocation().toURI());
JarFile jar = new JarFile(executable);
InputStream fileInputStreamReader = jar.getInputStream(jar.getJarEntry(filePath));
byte[] bytes = new byte[fileInputStreamReader.available()];
int sizeOrig = fileInputStreamReader.available();
int size = fileInputStreamReader.available();
int offset = 0;
while (size != 0){
fileInputStreamReader.read(bytes, offset, size);
offset = sizeOrig - fileInputStreamReader.available();
size = fileInputStreamReader.available();
}
Outside of your technique, why not use the standard Java JarFile class to get the references you want? From there most of your problems should go away.
If you use resources extensively, you might consider using
Commons VFS.
Also supports:
* Local Files
* FTP, SFTP
* HTTP and HTTPS
* Temporary Files "normal FS backed)
* Zip, Jar and Tar (uncompressed, tgz or tbz2)
* gzip and bzip2
* resources
* ram - "ramdrive"
* mime
There's also JBoss VFS - but it's not much documented.
I have 2 CSV files that I use to read data. The java program is exported as a runnable jar file. When you export it, you will figure out it doesn't export your resources with it.
I added a folder under project called data in eclipse. In that folder i stored my csv files.
When I need to reference those files I do it like this...
private static final String ZIP_FILE_LOCATION_PRIMARY = "free-zipcode-database-Primary.csv";
private static final String ZIP_FILE_LOCATION = "free-zipcode-database.csv";
private static String getFileLocation(){
String loc = new File("").getAbsolutePath() + File.separatorChar +
"data" + File.separatorChar;
if (usePrimaryZipCodesOnly()){
loc = loc.concat(ZIP_FILE_LOCATION_PRIMARY);
} else {
loc = loc.concat(ZIP_FILE_LOCATION);
}
return loc;
}
Then when you put the jar in a location so it can be ran via commandline, make sure that you add the data folder with the resources into the same location as the jar file.
In Groovy, I am reading a file from inside a JAR, and after some processing, I wish to delete this JAR, but once accessed through the URL, it doesn't seem to let me.
Example:
File jarFile = new File('jarFile.jar')
URL url = jarFile.toURI().toURL()
URL intUrl = new URL("jar:$url!/internalFile.json")
println intUrl.text // reads text correctly
jarFile.delete() // returns false, cannot delete
The Javadoc of the getText() command says the connection is closed at the end of the call, and this JAR normally isn't on classpath. Is there any way to make this code work?
Try setting the sun.zip.disableMemoryMapping system property:
java -Dsun.zip.disableMemoryMapping=true ....
(or however you set system properties when invoking Gradle). ZipFile (which backs jar: URLs) uses memory mapping by default, and this may be causing Windows to think that the file in question is still open. If this is not an option then you could try using the commons-compress ZipFile implementation instead of the java.util.zip one:
#Grab(group='org.apache.commons', module='commons-compress', version='1.4.1')
import org.apache.commons.compress.archivers.zip.*
File jarFile = new File('jarFile.jar')
ZipFile f = new ZipFile(jarFile)
ZipArchiveEntry json = f.getEntry('internalFile.json')
if(json) {
f.getInputStream(json)?.withStream {
println it.getText('UTF-8')
}
}
f.close()
jarFile.delete()
You could try this:
import java.util.zip.ZipFile
File jarFile = new File( 'jarFile.jar' )
String text = new ZipFile( jarFile ).with { zf ->
String result = zf.entries().findResult { ze ->
if( ze.name == 'internalFile.json' ) {
zf.getInputStream( ze ).withReader {
it.text
}
}
}
zf.close()
result
}
println text
jarFile.delete()
To avoid (what I suspect is) the classloader locking the jar file
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);
Im using the following Code to try an package a directory with all its subdirectories and files into one jar-file.
private void writeFile(JarOutputStream jos, File f) throws IOException {
byte buffer[] = new byte[BUFFER_SIZE];
if (f == null || !f.exists())
return; //
JarEntry jarAdd = new JarEntry(f.getName());
jarAdd.setTime(f.lastModified());
jos.putNextEntry(jarAdd);
if(f.isDirectory()){
File[] files = f.listFiles();
for(int i = 0; i < files.length ; i++){
writeFile(jos,files[i]);
}
}
else{
FileInputStream in = new FileInputStream(f);
while (true) {
int nRead = in.read(buffer, 0, buffer.length);
if (nRead <= 0)
break;
jos.write(buffer, 0, nRead);
}
in.close();
}
}
However FileInputStream does not work on directories, so i assume adding the JarEntry will suffice.
Since I do have another problem, I am not able to see if this assumption is correct.
This problem is that many entries made are duplicate.
When trying to add the structure...
activation
/--activation
/-- somefile.txt
...this code fails, because activation is added twice to the jar.
Since those are different directories nested within each other I do not see why this should not work in some way.
It seems that adding the entry using file.getName() will get entry-ids that do not
differentiate between the directories those files are in.
However when I use
file.getAbsolutePath()
for example, the jar structure is messed up with
directories like "C_" and the contained files are not readable anymore.
Can anyone recommend a way to package directory structures in a jar file correctly?
Note that a JarEntry representing a directory (folder) must end with a slash ("/") in order to be treated as such; otherwise it will be considered a plain file.
JarEntry dir = new JarEntry("activation/");
The name of the entry should be a relative path from the root of your tree. If you have
a
-- b
------c.txt
the entries should be a/, a/b/ and a/b/c.txt. Add a prefix argument to you recursive method (which is the empty string initially), and each time a directory is met, call the method with the current prefix + the directory name appended + /. And always prepend the prefix to the file name to compose the entry name.
I'm trying to access an XML file within a jar file, from a separate jar that's running as a desktop application. I can get the URL to the file I need, but when I pass that to a FileReader (as a String) I get a FileNotFoundException saying "The file name, directory name, or volume label syntax is incorrect."
As a point of reference, I have no trouble reading image resources from the same jar, passing the URL to an ImageIcon constructor. This seems to indicate that the method I'm using to get the URL is correct.
URL url = getClass().getResource("/xxx/xxx/xxx/services.xml");
ServicesLoader jsl = new ServicesLoader( url.toString() );
Inside the ServicesLoader class I have
XMLReader xr = XMLReaderFactory.createXMLReader();
xr.setContentHandler( this );
xr.setErrorHandler( this );
xr.parse( new InputSource( new FileReader( filename )));
What's wrong with using this technique to read the XML file?
Looks like you want to use java.lang.Class.getResourceAsStream(String), see
https://docs.oracle.com/javase/8/docs/api/java/lang/Class.html#getResourceAsStream-java.lang.String-
You don't say if this is a desktop or web app. I would use the getResourceAsStream() method from an appropriate ClassLoader if it's a desktop or the Context if it's a web app.
It looks as if you are using the URL.toString result as the argument to the FileReader constructor. URL.toString is a bit broken, and instead you should generally use url.toURI().toString(). In any case, the string is not a file path.
Instead, you should either:
Pass the URL to ServicesLoader and let it call openStream or similar.
Use Class.getResourceAsStream and just pass the stream over, possibly inside an InputSource. (Remember to check for nulls as the API is a bit messy.)
The problem was that I was going a step too far in calling the parse method of XMLReader. The parse method accepts an InputSource, so there was no reason to even use a FileReader. Changing the last line of the code above to
xr.parse( new InputSource( filename ));
works just fine.
I'd like to point out that one issues is what if the same resources are in multiple jar files.
Let's say you want to read /org/node/foo.txt, but not from one file, but from each and every jar file.
I have run into this same issue several times before.
I was hoping in JDK 7 that someone would write a classpath filesystem, but alas not yet.
Spring has the Resource class which allows you to load classpath resources quite nicely.
I wrote a little prototype to solve this very problem of reading resources form multiple jar files. The prototype does not handle every edge case, but it does handle looking for resources in directories that are in the jar files.
I have used Stack Overflow for quite sometime. This is the second answer that I remember answering a question so forgive me if I go too long (it is my nature).
This is a prototype resource reader. The prototype is devoid of robust error checking.
I have two prototype jar files that I have setup.
<pre>
<dependency>
<groupId>invoke</groupId>
<artifactId>invoke</artifactId>
<version>1.0-SNAPSHOT</version>
</dependency>
<dependency>
<groupId>node</groupId>
<artifactId>node</artifactId>
<version>1.0-SNAPSHOT</version>
</dependency>
The jar files each have a file under /org/node/ called resource.txt.
This is just a prototype of what a handler would look like with classpath://
I also have a resource.foo.txt in my local resources for this project.
It picks them all up and prints them out.
package com.foo;
import java.io.File;
import java.io.FileReader;
import java.io.InputStreamReader;
import java.io.Reader;
import java.net.URI;
import java.net.URL;
import java.util.Enumeration;
import java.util.zip.ZipEntry;
import java.util.zip.ZipFile;
/**
* Prototype resource reader.
* This prototype is devoid of error checking.
*
*
* I have two prototype jar files that I have setup.
* <pre>
* <dependency>
* <groupId>invoke</groupId>
* <artifactId>invoke</artifactId>
* <version>1.0-SNAPSHOT</version>
* </dependency>
*
* <dependency>
* <groupId>node</groupId>
* <artifactId>node</artifactId>
* <version>1.0-SNAPSHOT</version>
* </dependency>
* </pre>
* The jar files each have a file under /org/node/ called resource.txt.
* <br />
* This is just a prototype of what a handler would look like with classpath://
* I also have a resource.foo.txt in my local resources for this project.
* <br />
*/
public class ClasspathReader {
public static void main(String[] args) throws Exception {
/* This project includes two jar files that each have a resource located
in /org/node/ called resource.txt.
*/
/*
Name space is just a device I am using to see if a file in a dir
starts with a name space. Think of namespace like a file extension
but it is the start of the file not the end.
*/
String namespace = "resource";
//someResource is classpath.
String someResource = args.length > 0 ? args[0] :
//"classpath:///org/node/resource.txt"; It works with files
"classpath:///org/node/"; //It also works with directories
URI someResourceURI = URI.create(someResource);
System.out.println("URI of resource = " + someResourceURI);
someResource = someResourceURI.getPath();
System.out.println("PATH of resource =" + someResource);
boolean isDir = !someResource.endsWith(".txt");
/** Classpath resource can never really start with a starting slash.
* Logically they do, but in reality you have to strip it.
* This is a known behavior of classpath resources.
* It works with a slash unless the resource is in a jar file.
* Bottom line, by stripping it, it always works.
*/
if (someResource.startsWith("/")) {
someResource = someResource.substring(1);
}
/* Use the ClassLoader to lookup all resources that have this name.
Look for all resources that match the location we are looking for. */
Enumeration resources = null;
/* Check the context classloader first. Always use this if available. */
try {
resources =
Thread.currentThread().getContextClassLoader().getResources(someResource);
} catch (Exception ex) {
ex.printStackTrace();
}
if (resources == null || !resources.hasMoreElements()) {
resources = ClasspathReader.class.getClassLoader().getResources(someResource);
}
//Now iterate over the URLs of the resources from the classpath
while (resources.hasMoreElements()) {
URL resource = resources.nextElement();
/* if the resource is a file, it just means that we can use normal mechanism
to scan the directory.
*/
if (resource.getProtocol().equals("file")) {
//if it is a file then we can handle it the normal way.
handleFile(resource, namespace);
continue;
}
System.out.println("Resource " + resource);
/*
Split up the string that looks like this:
jar:file:/Users/rick/.m2/repository/invoke/invoke/1.0-SNAPSHOT/invoke-1.0-SNAPSHOT.jar!/org/node/
into
this /Users/rick/.m2/repository/invoke/invoke/1.0-SNAPSHOT/invoke-1.0-SNAPSHOT.jar
and this
/org/node/
*/
String[] split = resource.toString().split(":");
String[] split2 = split[2].split("!");
String zipFileName = split2[0];
String sresource = split2[1];
System.out.printf("After split zip file name = %s," +
" \nresource in zip %s \n", zipFileName, sresource);
/* Open up the zip file. */
ZipFile zipFile = new ZipFile(zipFileName);
/* Iterate through the entries. */
Enumeration entries = zipFile.entries();
while (entries.hasMoreElements()) {
ZipEntry entry = entries.nextElement();
/* If it is a directory, then skip it. */
if (entry.isDirectory()) {
continue;
}
String entryName = entry.getName();
System.out.printf("zip entry name %s \n", entryName);
/* If it does not start with our someResource String
then it is not our resource so continue.
*/
if (!entryName.startsWith(someResource)) {
continue;
}
/* the fileName part from the entry name.
* where /foo/bar/foo/bee/bar.txt, bar.txt is the file
*/
String fileName = entryName.substring(entryName.lastIndexOf("/") + 1);
System.out.printf("fileName %s \n", fileName);
/* See if the file starts with our namespace and ends with our extension.
*/
if (fileName.startsWith(namespace) && fileName.endsWith(".txt")) {
/* If you found the file, print out
the contents fo the file to System.out.*/
try (Reader reader = new InputStreamReader(zipFile.getInputStream(entry))) {
StringBuilder builder = new StringBuilder();
int ch = 0;
while ((ch = reader.read()) != -1) {
builder.append((char) ch);
}
System.out.printf("zip fileName = %s\n\n####\n contents of file %s\n###\n", entryName, builder);
} catch (Exception ex) {
ex.printStackTrace();
}
}
//use the entry to see if it's the file '1.txt'
//Read from the byte using file.getInputStream(entry)
}
}
}
/**
* The file was on the file system not a zip file,
* this is here for completeness for this example.
* otherwise.
*
* #param resource
* #param namespace
* #throws Exception
*/
private static void handleFile(URL resource, String namespace) throws Exception {
System.out.println("Handle this resource as a file " + resource);
URI uri = resource.toURI();
File file = new File(uri.getPath());
if (file.isDirectory()) {
for (File childFile : file.listFiles()) {
if (childFile.isDirectory()) {
continue;
}
String fileName = childFile.getName();
if (fileName.startsWith(namespace) && fileName.endsWith("txt")) {
try (FileReader reader = new FileReader(childFile)) {
StringBuilder builder = new StringBuilder();
int ch = 0;
while ((ch = reader.read()) != -1) {
builder.append((char) ch);
}
System.out.printf("fileName = %s\n\n####\n contents of file %s\n###\n", childFile, builder);
} catch (Exception ex) {
ex.printStackTrace();
}
}
}
} else {
String fileName = file.getName();
if (fileName.startsWith(namespace) && fileName.endsWith("txt")) {
try (FileReader reader = new FileReader(file)) {
StringBuilder builder = new StringBuilder();
int ch = 0;
while ((ch = reader.read()) != -1) {
builder.append((char) ch);
}
System.out.printf("fileName = %s\n\n####\n contents of file %s\n###\n", fileName, builder);
} catch (Exception ex) {
ex.printStackTrace();
}
}
}
}
}
You can see a fuller example here with the sample output.
Here's a sample code on how to read a file properly inside a jar file (in this case, the current executing jar file)
Just change executable with the path of your jar file if it is not the current running one.
Then change the filePath to the path of the file you want to use inside the jar file. I.E. if your file is in
someJar.jar\img\test.gif
. Set the filePath to "img\test.gif"
File executable = new File(BrowserViewControl.class.getProtectionDomain().getCodeSource().getLocation().toURI());
JarFile jar = new JarFile(executable);
InputStream fileInputStreamReader = jar.getInputStream(jar.getJarEntry(filePath));
byte[] bytes = new byte[fileInputStreamReader.available()];
int sizeOrig = fileInputStreamReader.available();
int size = fileInputStreamReader.available();
int offset = 0;
while (size != 0){
fileInputStreamReader.read(bytes, offset, size);
offset = sizeOrig - fileInputStreamReader.available();
size = fileInputStreamReader.available();
}
Outside of your technique, why not use the standard Java JarFile class to get the references you want? From there most of your problems should go away.
If you use resources extensively, you might consider using
Commons VFS.
Also supports:
* Local Files
* FTP, SFTP
* HTTP and HTTPS
* Temporary Files "normal FS backed)
* Zip, Jar and Tar (uncompressed, tgz or tbz2)
* gzip and bzip2
* resources
* ram - "ramdrive"
* mime
There's also JBoss VFS - but it's not much documented.
I have 2 CSV files that I use to read data. The java program is exported as a runnable jar file. When you export it, you will figure out it doesn't export your resources with it.
I added a folder under project called data in eclipse. In that folder i stored my csv files.
When I need to reference those files I do it like this...
private static final String ZIP_FILE_LOCATION_PRIMARY = "free-zipcode-database-Primary.csv";
private static final String ZIP_FILE_LOCATION = "free-zipcode-database.csv";
private static String getFileLocation(){
String loc = new File("").getAbsolutePath() + File.separatorChar +
"data" + File.separatorChar;
if (usePrimaryZipCodesOnly()){
loc = loc.concat(ZIP_FILE_LOCATION_PRIMARY);
} else {
loc = loc.concat(ZIP_FILE_LOCATION);
}
return loc;
}
Then when you put the jar in a location so it can be ran via commandline, make sure that you add the data folder with the resources into the same location as the jar file.