I have a Spring boot project that also executes a node.js file at one point. My project structure is:
ROOT
- src/
- jsModules/
- script.js
At one point, I execute the script.js (with some parameters) from the java code using:
Process proc = Runtime.getRuntime().exec(executingScriptString);
Once I pack that into a fatJar i have to manually copy-paste the jsModules folder and the script.js to be in the folder root where the .jar is.
Is it maybe somehow possible to pack all the JS files into the fatJar so once I pack it up, I don't need to worry about JS files being copied.
I tried searching how to do this but didn't come across any answers.
Here is the answer:
Place the JS file inside the resources folder which makes it available on the classpath.
When executing the script make sure to adjust the path of the file. In my case it would be:
node src/main/resources/script.js.
Instead of the "/" symbol I'm using File.separator from java.io package and seems to be working correctly although on Windows it creates a string of node src\main\resources\pptrSoccer.js.
Hope it helps someone.
EDIT:
I tried using ResourceLoader (from Spring framework)
resourceLoader.getResource("classpath:script.js").getFile().getAbsolutePath();
It works while it's exploded (running from IDE) but when I pack to a fatJar it doesn't work.
Related
For a project, I created an executable jar using the maven shade plugin. So far it runs, as it should, double clicking it starts the application just like executing the main method in my IDE. My next task is to have a properties file (named connection.properties), which is needed for the application, outside of the jar (at best in the same directory that the jar is in). I have already successfully excluded it from the jar with a filter, but I don't know how make the jar use the file while it is in the same directory, but not in the jar itself.
Any help or comments would be appreciated very much.
if the jar is in the current directory:
Paths.get("").resolve('myfile')
if the jar is elsewhere, see That answer
You can do the following to find the file in the current path (where the jar is running from)
File propertyFile = new File(Paths.get("").toAbsolutePath().toString() + File.separatorChar + "connection.properties");
I have a maven web project in eclipse. I need to get the project's path, actually have to get list of files under src\main\resources\someFolder in project.
Tried String dataDir = "src\\main\\resources\\someFolder";, on running this directory structure is created inside eclipse folder like F:\softwares\eclipse\eclipse\src\main\resources\someFolder. Same when using / instead of \\.
Tried System.getProperty("user.dir")and new File(".").getAbsolutePath(), they return F:\softwares\eclipse\eclipse.
I need to access the project folder in my workspace F:\workspace\Project\src\main\resources\someFolder
But when created a core java app and used System.getProperty("user.dir")and new File(".").getAbsolutePath(), I am getting project path in workspace, F:\workspace\Project. This src\\main\\resources\\someFolder also works fine then.
Why this odd behavior from eclipse?
As mentioned here the directory user.dir is the place where the JVM is started. As web applications are mostly jar/war/ear packages placed somewhere within the folder of the server eclipse handles them in a different way because the behaviour of such a web application is different. You cannot expect to have file access from outside the jar/war/ear file. Within the jar/war/ear file everything from within src/main/resources will be available just by using getResourceAsStream as described in many other stackoverflow articles. This way you mustn't use src/main/resources/myfile.txt but myfile.txt.
Don't try to guess or use what the user.dir / JVM/server start folder is!
I've currently finished my project, but can't get it to work when it is exported. I use JAXB to read and write XML Files and also have dependencies on other external Folders, which are needed to use a POS-Printer.
I've managed to link my external XML Save-Files with absolute paths, but not with relative paths. So that worked, although not the way i wanted. Yet, using the external class folder for the printer didn't work at all.
This means, that in my Eclipse Project Build Path i've added a class folder, which contains all of these needed files (which are not only jars, so adding them one by one wouldnt work). So exporting my project to a jar either includes all the files into the jar itself, or doesnt include them at all.
Everything works perfectly in Eclipse, but not when i export it.
My folder structure looks like this:
src
/model
/view
/control
data
/articles.xml
/...
JavaPOS <--- needed folder with all its files
/jpos.xml
/xerxers.jar
/swt-..-.dll
I've tried:
InputStreams is = getClass().getResourceAsStream(url);
absolute paths
manipulating the manifest file and/or jar structure
runnable and non runnable jars with nearly every combination of options
putting the files inside the library "by hand"
changing the build path of the project
My Question is:
How do i get my jar-file to know where these files are?
EDIT:
Do you think Maven or an Ant file could solve my problems? I don't have any experience with those.
The Problem was, that i had more than one JRE installed and that the one eclipse was using, had all the dll files, but the other ones didnt have it. So i had to add them manually, because reinstalling the drivers of the printer didnt change anything. Gotta fix that somehow, but right now it works and that is all i wanted.
Turns out i didn't even need that Folder, just needed one file out of it and the missing dlls.
I've been working on a processing application using ControlP5 and Twitter4j. I want to have my project run from a single jar file from any operating system. Basically I want to package up my application. My application uses images. I've been browsing for more than an hour, but I cant find how to do this. Any suggestions?
using
processing 2
twitter4j3
Thanks in advance!
I dont know if you can directly do it from the Processing IDE however, if export your sketch to a Java applet then locate the .java the the sketch folder you can use this in conjunction with Eclipse to export to a jar file.
So, I know that this post is very old but if you are still looking for a solution, or to other people that see this thread, it's relatively simple.
Export the project
In the folder with the exported project (something like application.windows64), navigate to lib and find core.jar and project name.jar (you need to have file name extensions visible)
Rename the files to .zip files
Extract core.jar to whatever folder
Extract project name.jar into the same folder (make sure you don't do it into a subfolder)
Click yes if it asks if it wants you to replace a file (if it doesn't you extracted the files incorrectly)
Delete core.jar and project name.jar
If the project uses images, move them into the same folder as all the other files
Select all of the files in the folder, right click, hover over send to and select compressed (zipped) folder
Rename the .zip file to name of project.jar
This might be old, but i still find other posts about it on processing forums.
This is the best way to run processing project as a jar file.
When exporting application, you will always end up with a lib folder inside exported application(whether for Linux and Windows). For windows, open command prompt(or power shell), you can use right-click+shift and then click on open power shell here.
After that you can run the following command.
java -classpath lib\* DisplayDepthStream
Now DisplayDepthStream is the name of sketch file.
To explain the command, -classpath lib\* tells java to add everything under lib directory to the class path. And DisplayDepthStream is the name of my main class.
Hope this helps~!
Chears
I have a Java project which uses a third party application. I have the license file (.lic format) stored in the resources folder. Upon running the Ant script, it will copy this file into the /lib/jar directory as it rolls up the project into a Jar file to use on the server. This is where I will need to access the file when running the system live. Here is how the folder structure looks
MyProject
src
package
AccessingClass.java
resources
File.lic
lib
jar
File.lic (upon copy from Ant)
I am not sure the best way to do this so any suggestions other than how I have been trying will probably be helpful. The 3rd party project has a method in a class like License.setLicense(), which can either take a String to the location or an InputStream of the file.
I have been playing around with feeding it an InputStream, but always get a null value when calling getClass().getResourceAsStream(). Here is everything I have tried:
getClass().getResourceAsStream("../../../lib/jar/File.lic");
getClass().getResourceAsStream("/File.lic");
And as a backup I also tried (for local builds I figure I would try the resource folder):
getClass().getResourceAsStream("../../../resources/File.lic");
getClass().getResourceAsStream("/File.lic");
Is there a better method to perform this action? Or would someone be able to tell me why what I am trying is failing? Thanks ahead of time.
Are you running this code standalone or in IDE env looks like classpath issue. If you are running at command prompt you have to set classpath to lib dir if in ide make sure you resources dir is in classpath.
First, you need to ensure that the JAR is added in your class path.
Below should work.
InputStream inputStream =
getClass().getClassLoader().getResourceAsStream("/resources/File.lic");
Assuming File.lic is placed in root folder of the jar.