I developed a small program that takes as input some configuration parameters from a .cfg file and produces an output .txt file based on the values taken from the .cfg file.
While the program runs perfectly in eclipse, I receive a NullPointerException error when I create a JAR file of this program and try to run it. From my understanding I have to make the JAR access its internal files or try to receive the needed information (in this case the .cfg file) externally, e.g. create a resource folder next to the JAR file.
I have searched many related questions asked here but I got even more confused whether there is an optimal way to produce a JAR file that can access input files and produces output files. Should I modify my code to achieve this or there is another way?
For the record, I use FileReader and FileWriter to access and produce the files.
If your .cfg file is outside of the JAR file, it should work just as in Eclipse.
If you want to access it from inside of the JAR archive, then you should use a class loader to load it, instead of FileReader...
Related
I am trying to access the text file in java and the code I have works in other IDEs but in VS code it keeps saying File Not Found - (produced from my exception handling)
I can't find anything about how to access text files with VScode. Is this possible please and does anyone know how?
I have tried to create the file in VScode with the new text file option and also by loading a pre-existing file and get the same result each time
Place the .txt file in the project root directory so that the code reads to the file location. As shown below
If the file is not with the project, use absolute paths in the code.
I'm trying to read a text file located in src/main/resources/test/file.txt. I'm trying to get the path of the file using String path = getClass().getResource("/text/file.txt").getFile(); but when I try to read it I get a FileNotFoundException. I tried putting many different paths, all of which failed. How can I go about doing this?
The idea of putting something into the src/main/resources tree is that it will be copied into the JAR file that you build from your project. It will then be available to your application via the Class methods getResource(String) and getResourceAsStream(String) methods.
When you are running in your application in the development environment, it is certainly possible to use FileInputStream etcetera to access the resource. But this won't work in production. In production, the resources will then be inside your app's JAR file. FileInputStream cannot open a JAR file and its contents by name.
When you do this:
getClass().getResource("/text/file.txt");
you get a URL for the resource, which will look something like this:
jar:file:/path/to/your.jar!/text/file.txt"
It is not possible to turn that into a pathname the FileInputStream will understand. Whatever you try will give you a FileNotFoundException ... or something that is not the resource you want to read.
So what to do?
You have a few options, depending on your application's requirements.
You can use getResourceAsStream and use the resulting input stream directly.
You can copy the contents of getResourceAsStream to a temporary file, and then use the pathname of the temporary file.
You can create an application specific directory (e.g. in the user's home directory) and extract the file you need from the JAR into the directory. You might do this the first time the application runs.
You could open the JAR file as a JarFile and use that API to open an InputStream for the resource. But this assumes that that the resources are in a JAR ... and on some platforms (e.g. Windows) you may encounter problems with file locking. (And it would be a bad idea to attempt to update the resource in the JAR.)
Try giving complete path of the file from the disk.
C:\Users\MyUser\Desktop\file name with extension
I made a small Java program for academic purposes, its main focus is to read some .txt files and present the information to the user. These files are present in the resources folder, under the src folder.
The program runs as intended when launched from Eclipse.
Using the Launch4j app I was able to successfully create an exe which runs fine and does what's intended, up until I try to read the .txt files I have in the resources folder, which appears not to be able to reach.
I'm guessing that when I launch the exe the run time path would change to where the exe was created, so I created the program in a desktop folder and specified this path in the program, but that doesn't seem to solve the situation.
As an alternative, I moved the .txt files out of the program and once again created the exe in a desktop folder with said .txt files, linked the program to this path and once again it didn't work.
The command used to get the .txt files is:
Files.readAllLines(Paths.get(doc)).get(line)
And doc is simply the path to the intended .txt file.
It's worth noting that I have no previous experience in Java and throughout the development of the program I tried my best to use commands I'd fully understand and to keep it as simple as possible. I hope the solution can be along these lines! I'm very confident this must be a rookie mistake, but I can't seem to find the solution to this specific problem anywhere.
The paths to files in Eclipse are different than the paths to files in an .exe or JAR file.
I will let this other user explain it because I am lazy :p
Rather than trying to address the resource as a File just ask the
ClassLoader to return an InputStream for the resource instead via
getResourceAsStream:
InputStream in = getClass().getResourceAsStream("/file.txt");
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
As long as the file.txt resource is available on the classpath then
this approach will work the same way regardless of whether the
file.txt resource is in a classes/ directory or inside a jar.
The URI is not hierarchical occurs because the URI for a resource
within a jar file is going to look something like this:
file:/example.jar!/file.txt. You cannot read the entries within a jar
(a zip file) like it was a plain old File.
This is explained well by the answers to:
How do I read a resource file from a Java jar file?
Java Jar file: use resource errors: URI is not hierarchical
The original post is here, all credit to its author.
Fixing your URL should let you read from that file when you are using the .exe.
EDITED FOR CORRECTION. Thanks #VGR (see comments) for correcting my mistake.
I have a Scala project that I want to export as a jar file so I can run it on another machine(I know how to export as a jar file). My main function reads in an existing input.json file and writes to a output.json file. The path where the the input.json is hard coded in my main and the path where the output.json will be written is hard coded as well. My goal is to export a jar file and pass in multiple input.json files to this jar file and my desire is the output.json file to be written to some directory. Basically, I want to have a large pipeline that feeds many different input.json files to this jar file and outputs all the output.json to some directory. I'm not too sure if this can be done, and if so how exactly?
Try to use the path from the configuration and from the environment variables and if the environment variable is not present it will pick up the path from the Configuration.
Change the environmental variable according to your needs and there you go.!
But in the case you want a whole directory to be taken as input containing multiple json files. Then in that case you have to implement a hack !
Read the directory path from the configuration file or environmental variable, and read it as directory, extract the list of names and perform operations on them inside a map !
I hope I answered your query!
I am using CLIPSJNI.
What I have is:
Environment clips = new Environment();
clips.load("main.clp");
where main.clp is put in the same level as src and bin folder.
This runs fine in Eclipse. However when I export to JAR. It cannot work.
I understand that there are some problems with the path when we export to JAR.
So I've seen people suggesting using this.getClass().getResourceStream() but this is not the case. Because what I need is the name of the file, not its content.
Any suggestions on how to fix this?
The issue is that the load is being done within the native library on the C side which is being passed a file name as an argument. The C code has no concept of a JAR file or how to extract files embedded within one. I think what you would need to do is always place your .clp files within the JAR file and then have a routine which extracts the data from the JAR file and saves it to a file. You can then load it using the load method and delete the file once done.