Develop Reference

How to access version of my project in build.gradle from a Java class

I'm quite new to Gradle so the answer might be simple, so I apologize if the answer is simple: I have a testing tool that needs to fetch it's version and compare it to the version of the application it is testing. However , the version of my tool is in my build.graddle as
version '1.0'
I tried different way to access it ( such as) :
task generateSources {
File outDir
outDir = file("$buildDir/classes/test/tests/Version.java")
doFirst {
outDir.exists() || outDir.mkdirs()
new File(outDir).write("public class Version { public static final String VERSION = \"$project.version\"; }")
}
}
compileJava.dependsOn generateSources
compileJava.source generateSources.outputs.files, sourceSets.main.java
I found this piece of code to output the version to another java file, but I fail to see how I'd be able to retrieve that info afterwards ( I mean, my tests are defined in src and I would need to point to a file that doesn't exist at compilation -- correct me if I'm wrong here).
Any idea on how I could accomplish this task?

First of all, you are trying to create java source file in your build/classes (it should contain compiled classes, not sources) directory, but you have to do it in your sources, otherwise it won't be compiled. And if you need this new class to be vailable not for tests, then use src/main/java, not src/test/java/
But anyway, I suppose for your case it's much easier to use some properties file for that and replace some token within it during build. That will allow you to make some static logic to get this property value and use it yet before running the build. So all you need is:
1- to have some properties file in your resources src/main/resources (for example app.properties), where should version variable be stored, with it's value like APP_VERSION_TOKEN
version=%APP_VERSION_TOKEN%
2- configure you Gradle processResources to replace tokens, something like this:
processResources {
filesMatching('**/app.properties') {
filter {
it.replace('%APP_VERSION_TOKEN%', version)
}
}
}
3- make some method to read this file and return the value of the property and use it where you need.
And that's all. For unit tests you can have another file with the same name under src/test/resource with the unchanging value you need for testing.

this.getClass().getResource("").getPath() returns an incorrect path

I am currently making a small simple Java program for my Computer Science Final, which needs to get the path of the current running class. The class files are in the C:\2013\game\ folder.
To get this path, I call this code segment in my main class constructor:
public game(){
String testPath = this.getClass().getResource("").getPath();
//Rest of game
}
However, this command instead returns this String: "/" despite the correct output being "C:/2013/game"
Additionally, I attempted to rectify this by using this code:
public game(){
String testPath = this.getClass().getClassLoader().getResource("").getPath();
}
This returns a NullPointerException, which originates from the fact that getClassLoader() returns null, despite working on my Eclipse IDE. Any Ideas?

If you want to load a file in the same path as the code then I suggest you put it in the same root folder as the code and not the same path as the class.
Reason : class can be inside a jar, data file can be put in same jar but its more difficult to edit and update then.
Also suggest you see the preferences class suggested in comments : http://www.javacodegeeks.com/2011/09/use-javautilprefspreferences-instead-of.html though in some cases I think its okay to have your own data/ excel/csv/ java.util.Properties file
Not sure about why it is working in eclipse but I would suggest you focus on running it from a command prompt/ terminal as that is the 'real mode' when it goes live
You could just ask for your class
String s = getClass().getName();
int i = s.lastIndexOf(".");
if(i > -1) s = s.substring(i + 1);
s = s + ".class";
System.out.println("name " +s);
Object testPath = this.getClass().getResource(s);
System.out.println(testPath);
This will give you
name TstPath.class
file:/java/Projects/tests3b/build/classes/s/TstPath.class
Which is my eclipse build path ...
need to parse this to get the path where the class was loaded.
Remember:
App could be started from elsewhere
class can be in jar then path will be different (will point to a jar and file inside that
classpaths can be many at runtime and point 1
a class might be made at runtime via network/ Proxy / injection etc and thus not have a file source, so this is not a generic solution.
think what you want to acheive at a higher level and post that question. meaning why do you want this path?
do you want the app path :-
File f = new File("./");
f.getCanonicalPath();//...
So an app can be started from folder c:\app1\run\
The jar could be at c:\app1\libsMain\myapp.jar
and a helper jar could be at c:\commonlibs\set1
So this will only tell you where the JVM found your class, that may or maynot be what you need.
if inside a jar will give you some thing like this in unix or windows
jar:file:c:\app\my.jar!/s/TstPath.class
If package is s and class is TstPath, you can be sure this will work as the class has to be there ...
now to parse this you can look for your class name and remove / or \ till you get path you want. String lastIndexOf will help

You can use :
URL classURL = getClass().getProtectionDomain().getCodeSource().getLocation();

The call to getResource([String]) requires a path relative to the folder that contains the class it is being called from. So, if you have the following, anything you pass into MyClass.class.getResource([path]); must be a valid path relative to the com/putable/ package folder and it must point to a real file:
package com.putable;
public class MyClass{}
Using the empty string simply isn't valid, because there can never be a file name that equals the empty string. But, you could do getResource(getClass().getSimpleName()). Just remove the file name from the end of the path returned by that call and you will have the class directory you want.

ClassLoader loader = Test.class.getClassLoader();
System.out.println(loader.getResource("Test.class"));
also
Test.class.getProtectionDomain().getCodeSource().getLocation().getPath());

Try this.
import java.io.File;
public class TT {
/**
* #param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
String path = TT.class.getResource("").getPath();
File file = new File(path);
System.out.println(file.getAbsolutePath());
}
}

Try use this code
public game()
{
String className = this.getClass().getSimpleName();
String testPath = this.getClass().getResource(className+".class");
System.out.println("Current Running Location is :"+testPath);
}
visit the link for more information
Find where java class is loaded from

Print out absolute path for a file in your classpath i.e. build/resources/main/someFileInClassPath.txt Disclaimer, this is similar to another solution on this page that used TT.class..., but this did not work for me instead TT..getClassLoader()... did work for me.
import java.io.File;
public class TT {
/**
* #param args
*/
public static void main(String[] args) {
String path = TT.getClassLoader().getResource("someFileInClassPath.txt").getPath();
File file = new File(path);
System.out.println(file.getAbsolutePath());
}
}

Because you used class.getResource(filePath).getpath() in a *.jar file. So the path includes "!". If you want to get content of file in *.jar file, use the following code:
MyClass.class.getResourceAsStream("/path/fileName")

Make JAR as a standalone executable

Is there a way to convert JAR lib into JAR standalone?
I need to find a standalone java executable that convert PDF into TIFF and I've found these JARs: http://www.icefaces.org/JForum/posts/list/17504.page
Any ideas?

Easiest might be to create another Jar with a Main() entry point, and then just use the java.exe executable to run it:
e.g.
> java.exe -cp MyJarMain.jar;MyPDFJar.jar com.mydomain.MyMain myPDF.pdf
Where MyMain is a class with a Main static method.
You'll need something with a main entry point to pass in and interpret some command line arguments (myPDF.pdf in my made-up example)

You could do an assembly (are you using maven?) and make sure the Main-Class entry in the manifest.mf points to the main class.

Since there is no main-Method, you have to write one, or write a whole new class to call the class/method TiffConver.convertPDF .
The question is, how you're going to use it. From the command line, you need no executable jar. From the Gui, maybe you want to pass a file to be converted by drag and drop? Then you should take the parameter(s) passed to main as Input-PDF-Names (if they end in .pdf) and pass the names iteratively to TiffConverter, for "a.pdf b.pdf" =>
TiffConver.convertPDF ("a.pdf", "a.tiff");
TiffConver.convertPDF ("b.pdf", "b.tiff");
TiffCoverter will silently overwrite existing tiffs, so check that before or change the code there - this is clearly bad habit, and look out for more such things - I didn't.
/*
* Remove target file if exists
*/
File f = new File(tif);
if (f.exists()) {
f.delete();
}
Maybe you wan't to write a swing-wrapper, which let's you choose Files interactively to be converted. This would be a nice idee, if no filename is given.
If the user passes "a.pdf xy.tiff" you could rename the converted file to xy, as additional feature.
Without a main-class, however, a standalone jar would be magic.
However, building a native executale is almost always a bad idea. You loose portability, you don't profit from security- and performance improvements to the JVM or fixed bugs. For multiple programs you need always an independend bugfix, which you might have to manage yourself, if you don't have a package-management as most linux distros have.
after clearing some questions:
public static void main (String [] args) {
if (args.length == 1 && args[0].endsWith (".pdf")) {
String target = args[0].replaceAll (".pdf$", ".tif");
convertPDF (args[0], target);
}
}
This method you put into TiffConvert. It will allow you to convert a simple pdf-File, and generate a tif-File with the same basename but ending in .tif, silently overwriting an existing one of the same name.
I guess you now need to know how to start it?

Java System.getProperty("user.dir") on Mac OS X

I have an application bundle on Mac OS X 10.4 on the desktop. My application looks for a folder named "resources" in which files to be displayed are kept (kept in the same location as the runnable JAR). I know there is a folder named "Resources" within the app bundle too, sorry if thats confusing, but I never programmed on a Mac and didnt know this would be the same name.
In Windows, when I call System.getProperty("user.dir") I get the location at which the runnable JAR file is located. Exactly what I wanted.
Why then when I run an application bundle is the getProperty returning "/"? Thats all. I expected it to return something like "/Users/user_name/Desktop"...which is where my app bundle is located.

I instead used the system property "user.home" instead of "user.dir". This way i do not have to worry about where the JVM is looking. I have the application bundle refernce my jar file directly using a bash script as the executeable called by the info.plist file. i can always place the files to be displayed by the app on the users home because that location will always return a path.

That's because "user.dir" indicates the current user directory in effect when the JVM is run; in Windows, this is often the location of the JAR unless you specify otherwise. In OSX there may well be no concept of a current dir, but more likely it just has a different default.
Though I have never specifically tested this code under OSX, you can try this to locate the directory from which any class was loaded:
static public File getClassLocation(Class cls, boolean trmjar) {
ClassLoader clsldr; // class loader
URL urlobj; // url object
String exturl; // external form of URL
String lwrurl; // lowercase external form of URL
File rtnfil; // return file
if((clsldr=cls.getClassLoader())==null) { clsldr=ClassLoader.getSystemClassLoader(); }
if((urlobj=clsldr.getResource(cls.getName().replace('.','/')+".class"))==null) {
return null;
}
exturl=urlobj.toExternalForm();
lwrurl=exturl.toLowerCase();
while(lwrurl.startsWith("jar:") || lwrurl.startsWith("file:/")) {
if(lwrurl.startsWith("jar:")) {
if(lwrurl.indexOf("!/")!=-1) { exturl=exturl.substring(4,(exturl.indexOf("!/"))); } // strip encapsulating "jar:" and "!/..." from JAR url
else { exturl=exturl.substring(4 ); } // strip encapsulating "jar:"
}
if(lwrurl.startsWith("file:/")) {
exturl=exturl.substring(6); // strip encapsulating "file:/"
if(!exturl.startsWith("/")) { exturl=("/"+exturl); }
while(exturl.length()>1 && exturl.charAt(1)=='/') { exturl=exturl.substring(1); }
}
lwrurl=exturl.toLowerCase();
}
exturl=java.net.URLDecoder.decode(exturl,"UTF-8");
rtnfil=new File(exturl);
if(lwrurl.endsWith(".class") || (trmjar && lwrurl.endsWith(".jar"))) { rtnfil=rtnfil.getParentFile(); }
if(rtnfil.exists()) { rtnfil=rtnfil.getAbsoluteFile(); }
return rtnfil;
}
it's worked reliably for me for years under Windows for all versions of Java since Java 1.

Launching java classes via windows drag-and-drop

I have a java class file with a main method. In Windows, I would like to be able to drag files onto a desktop icon/short/etc that would call supply the filenames to my main method. Basically, I want to allow users to drag-and-drop files at program execution instead of having type them on the command line.
Any thoughts?

To build on daub815's answer, in Windows, you can use a batch file to pass
arguments to another command. In this case, we'll use the java launcher to
launch your class with the main method.
I did a quick Google search on how to do write a batch file to take multiple arguments,
and found a page with a batch file to pass arguments to another command. Adapting from
the example, here is what you can do:
#ECHO OFF
:Loop
IF "%1" == "" GOTO Done
java YourClass %1
SHIFT
GOTO Loop
:Done
Save the above file as a batch file (with a ".bat" extension), and then you can drag-and-drop
files onto it, and it will be passed as arguments.
Also, you can call the batch file from the command line and pass arguments as well.
Edit: It appears that the batch file will not work with quoted arguments which contain spaces. Using a workaround presented in the site I've linked to will split the spaces contained in the quoted full path of the file into separate arguments, so that won't work either. If anyone has a good idea how to fix this, please either edit this entry, or post another answer. I will make this a community wiki.

PhiLho's answer works perfectly if you pack the classes in an executable JAR file (it's how you're meant to do it anyway) and make a .reg file that looks like the one below. Then just double-click that .reg file to merge it into the registry and you're good to go. This lets you both double-click a JAR file to run it, and starting it by Drag & Drop.
Do remember to change the path to where your Java executable is installed.
Windows Registry Editor Version 5.00
[HKEY_CLASSES_ROOT\.jar]
#="jarfile"
[HKEY_CLASSES_ROOT\jarfile\DefaultIcon]
#="C:\\Java\\jdk1.7.0\\bin\\java.exe,1"
[HKEY_CLASSES_ROOT\jarfile\shell\open]
#="Run Java Program"
[HKEY_CLASSES_ROOT\jarfile\shell\open\command]
#="\"C:\\Java\\jdk1.7.0\\bin\\java.exe\" -jar \"%1\" %*"
[HKEY_CLASSES_ROOT\jarfile\shellex\DropHandler]
#="{86C86720-42A0-1069-A2E8-08002B30309D}"

OK, I made it work... The base knowledge is to use DropHandler UUID in the registry. I made a base setting, as follow:
Windows Registry Editor Version 5.00
[HKEY_CLASSES_ROOT\.class]
#="JavaClass"
[HKEY_CLASSES_ROOT\JavaClass\DefaultIcon]
#="C:\\Java\\jdk1.6.0_05\\bin\\java.exe,1"
[HKEY_CLASSES_ROOT\JavaClass\shell\open]
#="Run Java class"
[HKEY_CLASSES_ROOT\JavaClass\shell\open\command]
#="\"C:\\Java\\jdk1.6.0_05\\bin\\java.exe\" \"%1\" %*"
[HKEY_CLASSES_ROOT\JavaClass\shellex\DropHandler]
#="{86C86720-42A0-1069-A2E8-08002B30309D}"
and... it didn't work!
I just forgot that java.exe wants a class name, not a file name! But I see no way to do that in the registry.
Fortunately, there is a workaround, which still need a script file if we want to be generic, to work on any/all class files (with static main function, of course!). Not batch, I avoid them when I can. I chose to use WSH, as it should be available on any modern Windows system. I also chose to make a JS script, it could have been a VB script as well.
So I made the following script (LaunchJavaClass.js):
if (WScript.Arguments.count() == 0)
{
WScript.StdOut.Write("No parameters");
WScript.Quit(1);
}
var className = WScript.Arguments.Item(0);
//~ WScript.StdOut.Write(className + "\n");
var m = className.match(/^(.*)\\(.+?)\.class$/);
if (m == null)
{
WScript.StdOut.Write("Not a class file");
WScript.Quit(1);
}
var classPath = m[1];
className = m[2];
//~ WScript.StdOut.Write(classPath + " >>> " + className + "\n");
var params = new Array();
for (i = 1; i < WScript.Arguments.count(); i++)
{
params[params.length] = WScript.Arguments.Item(i);
}
var cmd = "cmd /c cd /D " + classPath +
" & C:/Java/jdk1.6.0_05/bin/java.exe " +
className + " " + params.join(" ");
//~ WScript.StdOut.Write(cmd + "\n");
var shell = WScript.CreateObject("WScript.Shell");
//~ var exec = shell.Exec(cmd); // Can be used to get stdout
shell.Run(cmd, 0);
I left some output, not useful in this context, but usable for debugging (run with cscript).
Of course, the path to the JRE must be adjusted.
And I changed the command in the registry, as follow:
[HKEY_CLASSES_ROOT\JavaClass\shell\open\command]
#="\wscript -b "D:\\_PhiLhoSoft\\WSH\\LaunchJavaClass.js\" %1 %*"
Of course, adjust path, and keep the above other lines.
Now, if I drag'n'drop some files to a .class file, it gets the short file paths as arguments of the main() function.
import java.io.*;
class TestDnD
{
public static void main(String[] args)
{
Writer output = null;
try
{
output = new BufferedWriter(new FileWriter(new File("LogFile.txt")));
for (String arg : args)
{
output.write(arg + "\n");
}
}
catch (IOException ioe)
{
ioe.printStackTrace();
return;
}
finally
{
try { output.close(); } catch (IOException e) {}
}
}
}
I think the first version of the .reg file can be used for something else, eg. to drag'n'drop on .jar files (adapting it, of course).
This technique has limited use: we rarely make one-class programs in Java! But it looked like a good and interesting challenge, so I didn't resist to solve it. Note: you can add stuff like -Djava.ext.dirs="some path;another path" if you ever need to use external libraries (in jar files).

Adding onto Adiel A. If you create a batch file, which launches your a Java window using Swing. You would have the user drop the files onto that window. You could then be able to root through those dropped files.

So there's no way to have windows itself pass the args into main() via drag and drop?