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Is there a way to get the file path of the .java file executed or compiled?

In Python the global variable __file__ is the full path of the current file.
System.getProperty("user.dir"); seems to return the path of the current working directory.
I want to get the path of the current .java, .class or package file.
Then use this to get the path to an image.
My project file structure in Netbeans looks like this:
(source: toile-libre.org)
Update to use code suggested from my chosen best answer:
// read image data from picture in package
try {
InputStream instream = TesseractTest.class
.getResourceAsStream("eurotext.tif");
bufferedImage = ImageIO.read(instream);
}
catch (IOException e) {
System.out.println(e.getMessage());
}
This code is used in the usage example from tess4j.
My full code of the usage example is here.

If you want to load an image file stored right next to your class file, use Class::getResourceAsStream(String name).
In your case, that would be:
try (InputStream instream = TesseractTest.class.getResourceAsStream("eurotext.tif")) {
// read stream here
}
This assumes that your build system copies the .tif file to your build folder, which is commonly done by IDEs, but requires extra setup in build tools like Ant and Gradle.
If you package your program to a .jar file, the code will still work, again assuming your build system package the .tif file next to the .class file.

Is there a way to get the file path of the .java file executed or compiled?
For completeness, the literal answer to your question is "not easily and not always".
There is a round-about way to find the source filename for a class on the callstack via StackFrameElement.getFileName(). However, the filename won't always be available1 and it won't necessarily be correct2.
Indeed, it is quite likely that the source tree won't be present on the system where you are executing the code. So if you needed an image file that was stashed in the source tree, you would be out of luck.
1 - It depends on the Java compiler and compilation options that you use. And potentially on other things.
2 - For example, the source tree can be moved or removed after compilation.
Andreas has described the correct way to solve your problem. Make sure that the image file is in your application's JAR file, and access it using getResource or getResourceAsStream. If your application is using an API that requires a filename / pathname in the file system, you may need to extract the resource from the JAR to a temporary file, or something like that.

public class Main {
public static void main(String[] args) throws Exception {
System.out.println(getPackageParent(Main.class, false));
}
public static String getPackageParent(Class<?> cls, boolean include_last_dot)
throws Exception {
StringBuilder sb = new StringBuilder(cls.getPackage().getName());
if (sb.lastIndexOf(".") > 0)
if (include_last_dot)
return sb.delete(sb.lastIndexOf(".") + 1, sb.length())
.toString();
else
return sb.delete(sb.lastIndexOf("."), sb.length()).toString();
return sb.toString();
}
}

Replace specific file inside Zip archive without extracting the whole archive in Java

I'm trying to get a specific file inside a Zip Archive, extract it, Encrypt it, and then get it back inside the archive replacing the origial one.
here's what I've tried so far..
public static boolean encryptXML(File ZipArchive, String key) throws ZipException, IOException, Exception {
ZipFile zipFile = new ZipFile(ZipArchive);
List<FileHeader> fileHeaderList = zipFile.getFileHeaders();
for (FileHeader fh : fileHeaderList)
{
if (fh.getFileName().equals("META-INF/file.xml"))
{
Path tempdir = Files.createTempDirectory("Temp");
zipFile.extractFile(fh, tempdir.toString());
File XMLFile = new File(tempdir.toFile(), fh.getFileName());
// Encrypting XMLFile, Ignore this part
// Here, Replace the original XMLFile inside ZipArchive with the encrypted one <<<<<<<<
return true;
}
}
return false;
}
I stuck at the replacing part of the code is there anyway I can do this without having to extract the whole Zip Archive?
Any help is appreciated, thanks in advance.

Not sure if this will help you as you are using a different library but the solution in ZT Zip would be the following.
ZipUtil.unpackEntry(new File("/tmp/demo.zip"), "foo.txt", new File("foo.txt"));
// encrypt the foo.txt
ZipUtil.replaceEntry(new File("/tmp/demo.zip"), "foo.txt", new File("foo.txt"));
This will unpack the foo.txt file and then after you encrypt it you can replace the previous entry with the new one.

You may use the ZipFilesystem (as of Java 7) as explained in the Oracle documentation to read/write within a zip file as if it were its own file system.
However, on my machine, this unpacks and re-packs the zip file under the hood anyway (tested with 7 and 8). I am not sure if there is a way to reliably change zip files like you describe.

Bingo!
I'm able to do it that way
ZipParameters parameters = new ZipParameters();
parameters.setIncludeRootFolder(true);
zipFile.removeFile(fh);
zipFile.addFolder(new File(tempdir.toFile(), "META-INF"), parameters);

java.io.IOException source exists but not in a directory

I am a real beginner at Java programming so I hope I'm not wasting anyone's time. I tried my best to research this but couldn't come up with a solution.
I am following the Lynda video series "Java Essential Training" and it's been very good so far. I am currently learning how to copy the contents of a text file onto a new text file. However, the video shows a alternative method by downloading commons IO from Apache commons and adding the .jar file to the project.
In the video the jar file was added to build path. My version of eclipse seemed to do it automatically as "Referenced Libraries" popped up, and when I tried to add it eclipse said it was already there.
I followed the video exactly. The code looks like this
package com.lynda.files;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.IOException;
import org.apache.commons.io.FileUtils;
public class Main {
public static void main(String[] args) {
try {
File f1 = new File("loremipsum.txt");
File f2 = new File("target.txt");
FileUtils.copyDirectory(f1, f2);
System.out.println("File copied!");
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
}
When I ran the code I got the message in console
java.io.IOException: Source 'loremipsum.txt' exists but is not a directory
at org.apache.commons.io.FileUtils.copyDirectory(FileUtils.java:1371)
at org.apache.commons.io.FileUtils.copyDirectory(FileUtils.java:1261)
at org.apache.commons.io.FileUtils.copyDirectory(FileUtils.java:1230)
at com.lynda.files.Main.main(Main.java:16)
In the code it says the FileUtils imported but eclipse tells me "The source attachment does not contain the source for file FileUtils.class". I tried to change the attached source but it gave me the error "Could not write to file BlahBlahBlah.classpath (Access is denied)
Hopefully I didn't drone on about something obvious and simple. I thought it best to be as clear as possible in case someone else has a similar problem.
Edit
I feel so stupid. Thank you for your help. I clicked on "copyDirectory" instead of "copyFile". Next time, instead of panicking, googling every line of error and asking people for help, I'll take the time to go through each line and think about what it does. Thanks to all of you for your help and patience.

See (http://commons.apache.org/proper/commons-io/javadocs/api-2.4/org/apache/commons/io/FileUtils.html#copyFile%28java.io.File,%20java.io.File%29)
Use FileUtils.copyFile(f1, f2); instead of FileUtils.copyDirectory(f1, f2);

The source and target File parameters of copyDirectory must be directories, but you are suppling text files.
public static void copyDirectory(File srcDir,File destDir)
throws IOException
Copies a whole directory to a new location preserving the file dates.
This method copies the specified directory and all its child directories and files to the specified destination. The destination is the new location and name of the directory.
The destination directory is created if it does not exist. If the destination directory did exist, then this method merges the source with the destination, with the source taking precedence.
Note: This method tries to preserve the files' last modified date/times using File.setLastModified(long), however it is not guaranteed that those operations will succeed. If the modification operation fails, no indication is provided.
Parameters:
srcDir - an existing directory to copy, must not be null
destDir - the new directory, must not be null
Throws:
NullPointerException - if source or destination is null
IOException - if source or destination is invalid
IOException - if an IO error occurs during copying
Since:
1.1
(Source)

I found this that may be of help to you:
copyFile(File srcFile, File destFile) Copies a file to a new location preserving the file date.
static void copyFile(File srcFile, File destFile, boolean preserveFileDate) Copies a file to a new location.
static long copyFile(File input, OutputStream output) Copy bytes from a File to an OutputStream.
static void copyFileToDirectory(File srcFile, File destDir) Copies a file to a directory preserving the file date.
static void copyFileToDirectory(File srcFile, File destDir, boolean preserveFileDate) Copies a file to a directory optionally preserving the file date.
Source
Although it's from the Apache site, it does talk about the Java classes.

Please read the error message again:
Source 'loremipsum.txt' exists but is not a directory
This is not exactly what you wrote in your subject. Indeed the file 'loremipsum.txt' exists but it not a directory. It is regular file. However you try to call FileUtils.copyDirectory() and pass this regular file to this method. But this method is not ready to work with files. It supports directories only. This is exactly what is written in error message.
EDIT
Now the question is why do you call method that definitely for intended for directories with parameters that are definitely files?

Rename file in zip with zip4j

I'm using zip4j 1.3.1 to compress files in my application. Now I'm trying to rename the file inside the zip without having to rename the file itself. There seems to be a method for that, but it's not working.
My code looks like:
public static void zipFile(File dstPath, File srcFile, String optionalName) throws ZipException {
ZipFile zipFile = new ZipFile(dstPath);
ZipParameters parameters = new ZipParameters();
parameters.setCompressionMethod(Zip4jConstants.COMP_DEFLATE);
parameters.setCompressionLevel(Zip4jConstants.DEFLATE_LEVEL_MAXIMUM);
parameters.setEncryptFiles(true);
parameters.setEncryptionMethod(Zip4jConstants.ENC_METHOD_STANDARD);
parameters.setPassword(RutasDAO.getPassZip());
//here I'm setting the name in the zip, but it's not working
parameters.setFileNameInZip(optionalName);
zipFile.addFile(srcFile, parameters);
}
The file inside the zip has the same name as the file I'm passing in srcFile.
Maybe it's not yet implemented? do you know any alternative, method or library?
Thanks.
Edit:
I've come up with a solution creating a new file with the desired name, then using it as source file:
File renamedFile = new File(tmpDirectory + optionalFileName);
copyFile(srcFile, renamedFile);

today I met the same problem. After debugging zip4j, the solution is very simple.
parameters.setSourceExternalStream(true);
No other changes happened, but the file appeared in archive with the new name. I hope this setting will help you too.
Best wishes

decompress .gz file in batch

I have 100 of .gz files which I need to de-compress.
I have couple of questions
a) I am using the code given at http://www.roseindia.net/java/beginners/JavaUncompress.shtml to decompress the .gz file. Its working fine.
Quest:- is there a way to get the file name of the zipped file. I know that Zip class of Java gives of enumeration of entery file to work upon. This can give me the filename, size etc stored in .zip file. But, do we have the same for .gz files or does the file name is same as filename.gz with .gz removed.
b) is there another elegant way to decompress .gz file by calling the utility function in the java code. Like calling 7-zip application from your java class. Then, I don't have to worry about input/output stream.
Thanks in advance.
Kapil

a) Zip is an archive format, while gzip is not. So an entry iterator does not make much sense unless (for example) your gz-files are compressed tar files. What you want is probably:
File outFile = new File(infile.getParent(), infile.getName().replaceAll("\\.gz$", ""));
b) Do you only want to uncompress the files? If not you may be ok with using GZIPInputStream and read the files directly, i.e. without intermediate decompression.
But ok. Let's say you really only want to uncompress the files. If so, you could probably use this:
public static File unGzip(File infile, boolean deleteGzipfileOnSuccess) throws IOException {
GZIPInputStream gin = new GZIPInputStream(new FileInputStream(infile));
FileOutputStream fos = null;
try {
File outFile = new File(infile.getParent(), infile.getName().replaceAll("\\.gz$", ""));
fos = new FileOutputStream(outFile);
byte[] buf = new byte[100000];
int len;
while ((len = gin.read(buf)) > 0) {
fos.write(buf, 0, len);
}
fos.close();
if (deleteGzipfileOnSuccess) {
infile.delete();
}
return outFile;
} finally {
if (gin != null) {
gin.close();
}
if (fos != null) {
fos.close();
}
}
}

Regarding A, the gunzip command creates an uncompressed file with the original name minus the .gz suffix. See the man page.
Regarding B, Do you need gunzip specifically, or will another compression algorithm do? There's a java port of the LZMA compression algorithm used by 7zip to create .7z files, but it will not handle .gz files.

If you have a fixed number of files to decompress once, why don't you use existing tools for that?
As Paul Morie noticed, gunzip can do that:
for i in *.gz; do gunzip $i; done
And it would automatically name them, stripping .gz$
On windows, try winrar, probably, or gunzip from http://unxutils.sf.net

GZip is normally used only on single files, so it generally does not contain information about individual files. To bundle multiple files into one compressed archive, they are first combined into an uncompressed Tar file (with info about individual contents), and then compressed as a single file. This combination is called a Tarball.
There are libraries to extract the individual file info from a Tar, just as with ZipEntries. One example. You will first have to extract the .gz file into a temporary file in order to use it, or at least feed the GZipInputStream into the Tar library.
You may also call 7-Zip from the command line using Java. 7-Zip command-line syntax is here: 7-Zip Command Line Syntax. Example of calling the command shell from Java: Executing shell commands in Java. You will have to call 7-Zip twice: once to extract the Tar from the .tar.gz or .tgz file, and again to extract the individual files from the Tar.
Or, you could just do the easy thing and write a brief shell script or batch file to do your decompression. There's no reason to hammer a square peg in a round hole -- this is what batch files are made for. As a bonus, you can also feed them parameters, reducing the complexity of a java command line execution considerably, while still letting java control execution.

Have you tried
gunzip *.gz

.gz files (gzipped) can store the filename of a compressed file. So for example FuBar.doc can be saved inside myDocument.gz and with appropriate uncompression, the file can be restored to the filename FuBar.doc. Unfortunately, java.util.zip.GZIPInputStream does not support any way of reading the filename even if it is stored inside the archive.