For my Java application, I use some input files inside src folder. But after I create the jar file and use it, it gives an error saying cannot find the file.
How to add an input file when creating jar file in NetBeans?
You need to use getResourceAsStream(String name), to get a stream to the file:
InputStream is = MyClass.class.getResourceAsStream("fileName");
This will return an input stream to the file fileName which is located inside a directory that MyClass.class is in.
Related
I'm trying to save a YAML formatted file packaged in the JAR to the user's disk.
The file is named config.yml and sits directly under one of the project's source folders. I can confirm that the file is packaged in the JAR with WinRAR.
I am using the following code to get the file from the JAR:
File file = new File(this.getClassLoader().getResource("config.yml").getPath());
And I am using this code to copy the file to a directory:
FileUtils.copyFileToDirectory(file, this.getDataFolder());
The getDataFolder() method is implemented by a reliable 3rd party API.
However, when I use a file instance defined with the same getDataFolder() File instance and the path "config.yml":
new File(this.getDataFolder(), "config.yml")
The console logs a FileNotFoundException.
The file path given in the stack trace is this: "file:\C:\Users\Evan\Desk
top\Test%20Server\plugins\lottocrates-1.0.jar!\config.yml" which seems correct. I tried opening the file with run prompt, which I was able to open the JAR with, but not the config.yml file.
i have a small application which checks for values from a file and display the result in a jframe.
A file contain list of word to check. this file is placed in project folder "testing" and the main source testing.java file is present in location "testing\src\testing"
input file : c:\document..\netbeans\testing\
java file : c:\document..\netbeans\testing\src\testing\
when i place the input file inside folder "c:\document..\netbeans\testing\src\testing\
" the input file is not taken as input, it works only when kept on folder "c:\document..\netbeans\testing\"
so when a jar file is created it has not included the input file in that, even i manually input that is not taking the input file in and working.
some path setting issue? what can be done to solve this issue?
any help pls??
Once you create the jar, the file becomes an embedded resource. If you try to read it as a File it will no long be the same file system path as you originally use in the program. It must now be read from the class path.
To read the file from the class path, you will want to use getClass().getResourceAsStream(), which return an InputStream. If your file is in the same location (package) as your class file, then you should use
InputStream is = getClass().getResourceAsStream("input.txt");
Then you can read from the InputStream
BufferedReader reader = new BufferedReader (new InputStreamReader(is));
This generally happens, when you don't use absolute path...!
As when you run your program from IDE(Netbeans) then the HOME_FOLDER is your ProjectFolder. Relative to which you would have given the file_path(that has to be accessed in your program).
But after building, jar is present in ProjectFolder/dist. When you run the jar file the HomeFolder is not ProjectFolder rather it is ProjectFolder/dist.
So, to make it successful, to need to copy all files and folders from ProjectFolder/dist to ProjectFolder.
Then run the jar.. Hope it will fix the issue
Try putting double backslashes in your file paths. Like this:
c:\\document..\\netbeans\\testing\\src\\testing\\
This is the format that java normally requires it to be in
How would you read a file into a program that's compiled into a jar next to it through its local directory? The type read would be a simple .txt file.
It depends on what the usage of the program is. Do you know how the jar is supposed to be executed? When you try to run it, does it spit out a "usage: somejar firstarg secondarg" type message?
Also, if its a jar that you've compiled and you know how it should be executed, then you may have forgot to set its main class or manifest.
Check this: http://www.mkyong.com/java/how-to-make-an-executable-jar-file/
If you want to read a file that exists within an external .jar file, you will need to unzip the .jar file first in your code and then retrieve the file. You can do this using Java's zip APIs. See this answer if this is the case: Easiest way to unpack a jar in java
If you want to read a file that is in the same .jar file that your code is executing, you can get the file as a resource. See this answer: Get a resource using getResource()
If the file is simply in the exact same directory as the executable .jar, create a new file like so:
File input = new File("myfile.txt");
I have a scanner that's trying to read a file named info.data in the src folder.I get Exception in thread "main" java.io.FileNotFoundException: info.data (The system cannot find the file specified). What's the address I should put in the scanner?
If the input file is always part of your application (i.e. you also put this into the .jar file later) you should use getResourceAsStream() in order to read its contents.
InputStream in = getClass().getResourceAsStream(filename);
Scanner scanner = new Scanner(in);
In netbeans, the src folder isn't the destination of the compiled classes, so if you are using a relative path, the location your program launches is not going to be the src folder.
That means you typically should "extend" your build to copy a non-source file into the build path if you want it to operate in the manner you imply. Many files already copy over to the build path (like properties files), but if you are including a data file that doesn't have a rule for being place in the build path, you need to add the rule yourself.
Try putting the path to it.
File f = new File("C:\\path\\src\\info.data");
I want to manipulate a file in my java program.The file to read must be paralled to my src folder.
What should I give as file path?
An elaborated example might help. From your question, what I get is,
Source Path : /home/user/project1/src/
File Path : /home/user/project1/src/
If this is the case, then once you build the project, the file path is not going to remain the same. So if you say that relative path for the file to open remains the same in built code, then you can use Class.getResourceAsStream(String path) which returns you the InputStream for given file. You can then construct the File object using it.
Refer this for details.
You should have a File object representing your src folder, and then create a new File object using that:
File textFile = new File(srcFolder, relativePath);
How you determine srcFolder really depends on the context.
EDIT: If you're just trying to read a file which is present at build time, you should include it in your built jar file and use either ClassLoader.getResourceAsStream or Class.getResourceAsStream to load it at execution time.
For example, if you have this structure:
src\
com\
xyz\
Foo.class
data\
input.txt
Then you could use Foo.class.getResourceAsStream("/data/input.txt") or Foo.class.getClassLoader.getResourceAsStream("data/input.txt"). Both will give you an InputStream you can use to load the data.