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.
Related
I've created a file inside a project package using this code:
File xmlFile = new File("src/com/company/project/xml/tags.xml");
I am able to read the file while running from eclipse. However, after creating .jar, I'm unable to read the file. So I want to put absolute path while reading the file from the project package. How it can be done? Help and suggestions are appreciated.
In most cases, IDE's will include no Java files in the resulting Jar. Most IDE's will also include the src directory in the classpath when you run/debug the program from within them.
As a general rule of thumb, never include src in any path, src will simply not exist once the program is built.
Instead you need to make use of Class#getResource or Class#getResourceAsStream, depending on your needs. You should remember, you should never treat an "embedded" resource as a File, as in most cases it won't be, it'll be a stream of bytes in a zip file.
Something like...
URL xmlFile = getClass().getResource("/com/company/project/xml/tags.xml");
will return a URL reference to the resource. Remember, if you need a InputStream, you'll have to Class#getResourceAsStream.
If you want the resource to be writable, then you will need to find a different location to store it, as embedded resources are read only
Try with getClass().getResource()
new File(getClass().getResource("src/com/company/project/xml/tags.xml").toURI());
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
I have an application that creates a temporary mp3-file and puts it in a directory like C:\
File tempfile = File.createTempFile("something", ".mp3", new File("C:\\));
I'm able to read it by just using that same tempfile again.
Everything works fine in the Eclipse IDE.
But when I export my project for as a Runnable jar, my files are still being made correctly (I can play them with some normal music player like iTunes) but I can't seem to read them anymore in my application.
I found out that I need to use something like getClass().getResource("/relative/path/in/jar.mp3") for using resource files that are in the jar. But this doesn't seem to work if I want to select a file from a certain location in my file system like C:\something.mp3
Can somebody help me on this one?
It seems you dont have file name of the temp files . When you was running your program in eclipse that instance was creating a processing files, but after you made a runable you are not able to read those file that instance in eclipse created, You runable file can create its own temp file and can process them,
To make temp files globe put there (path + name ) entries in some db or property file
For example of you will create a temp file from the blow code
File tempfile = File.createTempFile("out", ".txt", new File("D:\\"));
FileWriter fstream = new FileWriter(tempfile);//write in file
out = new BufferedWriter(fstream);
the out will not be out.txt file it will be
out6654748541383250156.txt // it mean a randum number will be append with file
and you code in runable jar is no able to find these temp files
getClass().getResource() only reads resources that are on your classpath. The path that is passed to getResource() is, in fact, a path relative to any paths on your current classpath. This sounds a bit confusing, so I'll give an example:
If your classpath includes a directory C:\development\resources, you would be able to load any file under this directory using getResource(). For example, there is a file C:\development\resources\mp3\song.mp3. You could load this file by calling
getClass().getResource("mp3/song.mp3");
Bottom line: if you want to read files using getResource(), you will need those files to be on your classpath.
For loading from both privileged JARs and the file system, I have had to use two different mechanisms:
getClass().getClassLoader().getResource(path), and if that returns null,
new File(path).toURI().toURL();
You could turn this into a ResourceResolver strategy that uses the classpath method and one or more file methods (perhaps using different base paths).
Suppose I have a Java class that needs to access a file with absolute path
/home/gem/projects/bar/resources/test.csv:
package com.example
class Foo {
String filePath = ????? // path to test.csv
String lines = FileInputStream(new File(filePath).readAllLines();
}
Where the path to Foo.java is /home/gem/projects/bar/src/com/example.
Of course I cannot specify absolute path to the resource file. This is because jar file will be distributed as library for any clients to use in their own environments.
Assume the resource file like test.csv is always in the same path relative to project root. When a jar is created containing Foo.class, this jar also contains test.csv in the same relative path ( relative to project root).
What is the way to specify relative path that would work no matter where the project bar is moved to? Also how can I create a jar file (which can be in any location) so that the path to the resource file test.csv would still be correct.
To keep things simple, I have used invalid Java API ( readAllLines() which reads all the lines and return a string containing entire file content. Also not using try/catch).
Assume csv file can be read as well as written to.
I hope this makes it clear now.
Put the test.csv file into the src folder and use this:
Foo.class.getResourceAsStream("/test.csv")
To get an InputStream for the file. This will work wherever the project is moved, including packaged as a JAR file.
Example:
ProjectX\src\Test.java
ProjectX\resources\config.properties
If you have the above structure and you want to use your config.properties file, this is how you do it:
InputStream input = new FileInputStream("./resources/config.projects");
In this example you don't have to worry about packaging your source into jar file. You can still modify your resources folder anytime.
Use getResource(), as shown here.
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");