I have a huge JAR file, which I created with Maven Shade plugin.
I run this with java -jar foo.jar , and my UI opens. Now I want to execute *.exe file, which is also in that JAR file, how can I do this?
I tried putting the exe to my classpath and run it from there, but after trying I found out that classpath is actually the location, where my JAR is.
Any suggestions?
Found this thing here, but is this really the best solution? Seems like alot of work, I think I have different case here, because I can define the location of exe myself and the JAR is created by me.
run exe which is packaged inside jar
Why I need this?
I want to give user a single file executable, which he can run, but my program is using the *.exe. Should I put the exe next to my jar and there will be 2 files or there is solution for my requirements?
Copying the file to a temporary location and running it is the way to go. The answer you linked to does much more work that necessary, as you can get your exe file as an InputStream and copy it to a file with a utility like Apache Commons IO FileUtils.copy(in, out)
See How do I copy a text file from a jar into a file outside of the jar? for example.
It's not about the location, it's about the fact that you need to tell your OS to run the exe and, unfortunately, you can't do that by providing a location within a jar.
Related
I am trying to create an executable jar file that a user can just click on and execute the code. I have done some research but almost all of the examples are for single files. I currently have all my classes and images being used in my files in a folder. My images exceed the default java memory space and thus I have to run my code with -Xmx500m.
How can I create an executable jar that will go into my folder and run my program from there? Any help will be greatly appreciated!
Just use jar command to create the jar. You may need to link this type of file to Java so that it can be run by clicking on it in Windows. Something like this-
jar cvf test.jar <path to your classes / resources>
I am trying to create an executable jar file that a user can just click on and execute the code
Use Java Web Start to deploy it.
Put all the images and other media in Jar files mentioned in the launch file and they will be downloaded when needed. JWS also has facilities to adjust the memory allocated to an app.
I recommend using https://netbeans.org/ to build your project. Once built it will create the jar for you. You are also able to set compile settings such as the amount of ram you need.
See the following directions https://netbeans.org/kb/docs/java/quickstart.html#build
I have a Java application that was converted to an .exe with launch4j. For several reasons I need to get access to the class files of the application.
The first thing I tried was unpacking the exe with 7zip. That way I get a handful of class files, but definately not the whole application packed in the exe (the class files seem to belong to launch4j).
What other options are there?
You can use something like this to extract the JAR from the Launch4j executable:
http://www.nirsoft.net/utils/resources_extract.html
Typically the executable packages necessary JAR's as executable resources and then looks for a JDK on the path to execute.
Once you get the JAR file you should be able to decompile it.
You cannot get access to class files. Exes are created, So that nobody can get access to your code.
If somehow , you are able to get access to jar files then you can decompile them to get the code and make changes.
I have created a simple class with only inherent Java dependencies (java.io, etc).
I have set up my jar file and the bat file in the same folder. The BAT simply reads:
java -jar "MyApp.jar"
pause
I have been able to run it from several different locations on my computer. But when I sent it to a coworker as a zip file, he was unable to run it by double clicking the BAT file.
The command window came back with an error
could not find the main class: MyApp.MyApp. Program will exit.
I've poked around this site but most similar errors involve use on the same computer.
Yes the other computer has Java installed 6.29
Any help much appreciated.
Two options that I can think off the top of my head:
1) He might not have extracted them both to the same directory (or) after extraction, he might have moved around the JAR file to another location.
2) His classpath does not include the current directory. Your classpath has a '.' (indicating the current directory) while his doesn't. If that is the case, you can probably modify your command to include the '-cp' switch.
In order to run a jar that way, you need a META-INF folder inside it with a manifest file inside that. The manifest file needs a main-class line that points at your class with a main(). Your IDE probably added that, but maybe in the process of extracting things he unzipped the jar file too, or something "interesting" like that.
http://docs.oracle.com/javase/tutorial/deployment/jar/appman.html
I have a jar file which is used in html file as applet. I want to modify the content of the jar file and to rebuild the jar file so that the html will work fine with the new jar file. How can i do this??
I already tried unzipping using 7zip nad modified the source and created the new jar. But when i use it in html it shows some java.lang.Classnotfound error
You can unjar or rejar the classes and source files as you wish.
unjar
jar -xvf abc.jar
jar
jar cf abc.jar input-files
http://docs.oracle.com/javase/tutorial/deployment/jar/build.html
Make the changes in the code (.java files), recompile to get the .class files. Then simply replace the old .class files in the jar with the new ones. I usually use WinZip, but you can use whatever app that can handle .Zip files. It should just work.
I've faced cases where the launcher of the app uses some sort of verification and checks for this kind of changes. I had to use a new launch script. This doesn't seem to be your case though.
This is surely possible from the command line. Use the u option for jar
From the Java Tutorials:
jar uf jar-file input-file(s)
"Any files already in the archive having the same pathname as a file being added will be overwritten."
See Updating a JAR File
A brief test shows this quickly updates changes apart from trying to delete the file.
I haven't seen this answer on other threads about modifying jar files, and many, marked as duplicates, suggest there is no alternative but to remake the jar completely. Please correct if wrong.
JARs are just ZIP files, use whatever utility you like and edit away!
Disclaimer: When reverse engineering any code be sure that you are staying within the limits of the law and adhering to the license of that code.
Follow the instructions above to unpack the JAR.
Find the original source of the JAR (perhaps its on SourceForge) and download the source, modify the source, and rebuild your own JAR.
You can also decompile the class files in the JAR. This is a rather advanced process and has a lot of "gotchas".
I have a java class that uses a bat file to execute commands. However I developed it in Eclipse IDE. It works fine in there. But as I export it in a jar file, it fails to find the bat file that was included.(gives me an IOException)
The file structure in eclipse is as follows
:
Project1
---->src
------>com.myproj
-------->BatFileRead.java
----md.bat
----ul.bat
md.bat and ul.bat is same level as src directory. After jarring it src folder disappears.
Could someone help me with this.
Thanks
In order to execute the command, you'll have to extract the bat file afterwards. You can't run executables which are inside jar files. Basically you'll need to open the batch file entry in the jar file as an input stream, and copy the data to a FileOutputStream on disk. You won't be able to execute it until it's a proper standalone file on the file system.
If you're already trying to extract it, chances are you're using getResource or getResourceAsStream slightly incorrectly. This is easy to do, because it depends whether you're calling ClassLoader.getResourceAsStream or Class.getResourceAsStream. The first only ever uses absolute paths (implicitly) and the second can use either absolute or relative paths. For example, in your case you'd want:
BatFileRead.class.getResourceAsStream("/md.bat")
or
BatFileRead.class.getClassLoader().getResourceAsStream("md.bat")
Have you checked that the bat files are definitely ending up in the jar file? Just list the contents with
jar tvf file.jar
to see what's in there.
Well this can be very dangerous. Be sure to use gloves when dealing with the BAT. They bite and quite painfull. Also try getting jar that has a big enough opening, although the bats will fit almost through any hole.
Good luck, and don't try this at home. This is done by professionals.
Try copying file.jar to file.zip, and opening file.zip (e.g. just double clicking, or using e.g. 7-Zip). Can you find your .bat files inside? If the .bat file isn't there, you have to add the .bat file to your Eclipse project, and rebuild in Eclipse. If the .bat file is there, but you still get an IOException opening it from you .java application, then please post the code snippet triggering the IOException, and please give a full listing of file.zip (get the listing by extracting file.zip to C:\testdir, and running dir /s C:\testdir in the command line).
Please note that Jon Skeet is right that although it is possible to open any file in the .jar file as an InputStream using java.lang.Class.getResourceAsStream and java.lang.ClassLoader.getResourceAsStream (see tutorial or find them in the Java API docs), but in order to execute a .bat file inside, you have to extract the .jar file first.