When I attempt to access a File object in Java using Eclipse, the system sends a FileNotFoundException whenever the file is accessed. I have created the file manually and placed it within Eclipse's file browser. The file does have content, so the problem isn't that it's empty. I am initializing the File with
File file = new File(this.getClass().getResource("save.txt").toString());
and accessing it with
PrintWriter p = new PrintWriter(file);
which sends the error. The stacktrace flows back through nothing except a path of initializations that create this class. The exception is
java.io.FileNotFoundException: file:\C:\Users\Nathaniel\Downloads\ERPGE-Global%20Equestria%20Workspace\Swarm\bin\main\save.txt (The filename, directory name, or volume label syntax is incorrect)
What could be causing this error, and how can it be fixed?
You need to make sure that the resource is included in the project's build path.
You can do this in one of two ways:
Put it in a folder (e.g. "res") then add that folder to the build path. That way, all the files in the folder will be added to the build path.
Put it inside (one of) your source folders. Be aware that if you put it inside a package folder, you'll need to include the relative path to that folder (see below).
Personally I prefer option #1.
Edit: You might also find it useful to have sub-directories of the res folder. If you have a file such as res/images/button.png and you've added the res folder to the build path, you need to access the file with the string "images/button.png".
Edit 2: Fixed some misleading info.
The file needs to be placed in the parent project folder itself and not the source folder. If a file is created without a specified file path, this is where it would be stored.
Provide the actual path to the file and get the file Object. Then you can use it.
File file = new File("C:\Users\Nathaniel\Downloads\save.txt");
PrintWriter p = new PrintWriter(file);
EDIT:
Externalising the File Name:
Create a properties file filedetails.properties in the external folder/package.
fileName=C:\Users\Nathaniel\Downloads\save.txt
Accessing the resource bundle.
ResourceBundle b=ResourceBundle.getBundle("external.filedetails") ;
String fileName=b.getString("fileName");
File file=new File(fileName);
Related
I am implementing a Branch Predictor for one of my classes and I am trying to read files from my src folder in Eclipse but for some reason it is not able to open the files. I have done this before with the exact same process so I'm not sure what is different.
traceFile is set from the command line and if I print "input", it will print out the correct file path and I have confirmed it is there manually.
ClassLoader loader = BiModalPredictor.class.getClassLoader();
File input = new File(loader.getResource(traceFile).getFile());
Scanner fin = new Scanner(input);
Is there any insight as to why this might be happening? I've tried restarting Eclipse, refreshing the files, and I've also tested it on another program which worked. No idea why it can't find this file.
Resources on the classpath, i.e. available through the classloaders getResource method, will not be files on the file system when your application is deployed as a jar file, or deployed in general. Do not use File with such resources, instead use getResourceAsStream to access the resource content.
Besides, your code is wrong. getResource() returns a URL. If you want a File object from a URL, you should use new File(uri), where the URI is obtained by calling url.toURI().
File input = new File(loader.getResource(traceFile).toURI());
I'm struggling to reference a .dat file in my java project in Eclipse. My file structure looks like this:
I'm trying to reference 'GeoIPLite.dat' from my 'LookupCountry.java' file. When I say "Reference" I actually mean just getting a String object of the path and then sending that as a parameter to a class that is located under the 'Referenced Libraries' in a jar.
My code in LookupCountry.java is as follow:
try{
String dbFile = "../resources/GeoIPLite.dat"
// TRIED ALL THESE AS WELL:
//String dbFile = "/../resources/GeoIPLite.dat"
//String dbFile = "/resources/GeoIPLite.dat"
//String dbFile = "resources/GeoIPLite.dat"
//Some more code calling the jar file......
}catch(IOException e){
//Handles exception
}
So basically I get a IOException that gets thrown from the jar saying that the path to the .dat file is incorrect.
Any ideas??
Use the same package structure in your resource folder as you did in your src folder. Add your resource folder as a source folder in Eclipse (see screenshot). Place the .dat file, in the resource folder, relative to the class that is going to use it. You should see your resource files then along side your compiled classes.
In your code, reference the resource file like so:
this.getClass().getResource("foo.txt");
In Eclipse, for example, when you run a class, you configure it as to where the 'current folder' is. From the menu: Run > Run Configuration ... and select the one you are using.
Then click the 'Arguments' tab and check the working directory.
Your path should be relative to the location shown there. (In the clipped image it shows the default which is the base folder of the project itself.)
You would open the file as new File("resources/GeoIPLite.dat") with the default working directory.
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).
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");
In my Java app I need to get some files and directories.
This is the program structure:
./main.java
./package1/guiclass.java
./package1/resources/resourcesloader.java
./package1/resources/repository/modules/ -> this is the dir I need to get
./package1/resources/repository/SSL-Key/cert.jks -> this is the file I need to get
guiclass loads the resourcesloader class which will load my resources (directory and file).
As to the file, I tried
resourcesloader.class.getClass().getResource("repository/SSL-Key/cert.jks").toString()
in order to get the real path, but this way does not work.
I have no idea which path to use for the directory.
I had problems with using the getClass().getResource("filename.txt") method.
Upon reading the Java docs instructions, if your resource is not in the same package as the class you are trying to access the resource from, then you have to give it relative path starting with '/'. The recommended strategy is to put your resource files under a "resources" folder in the root directory. So for example if you have the structure:
src/main/com/mycompany/myapp
then you can add a resources folder as recommended by maven in:
src/main/resources
furthermore you can add subfolders in the resources folder
src/main/resources/textfiles
and say that your file is called myfile.txt so you have
src/main/resources/textfiles/myfile.txt
Now here is where the stupid path problem comes in. Say you have a class in your com.mycompany.myapp package, and you want to access the myfile.txt file from your resource folder. Some say you need to give the:
"/main/resources/textfiles/myfile.txt" path
or
"/resources/textfiles/myfile.txt"
both of these are wrong. After I ran mvn clean compile, the files and folders are copied in the:
myapp/target/classes
folder. But the resources folder is not there, just the folders in the resources folder. So you have:
myapp/target/classes/textfiles/myfile.txt
myapp/target/classes/com/mycompany/myapp/*
so the correct path to give to the getClass().getResource("") method is:
"/textfiles/myfile.txt"
here it is:
getClass().getResource("/textfiles/myfile.txt")
This will no longer return null, but will return your class.
It is strange to me, that the "resources" folder is not copied as well, but only the subfolders and files directly in the "resources" folder. It would seem logical to me that the "resources" folder would also be found under `"myapp/target/classes"
Supply the path relative to the classloader, not the class you're getting the loader from. For instance:
resourcesloader.class.getClassLoader().getResource("package1/resources/repository/SSL-Key/cert.jks").toString();
In the hopes of providing additional information for those who don't pick this up as quickly as others, I'd like to provide my scenario as it has a slightly different setup. My project was setup with the following directory structure (using Eclipse):
Project/
src/ // application source code
org/
myproject/
MyClass.java
test/ // unit tests
res/ // resources
images/ // PNG images for icons
my-image.png
xml/ // XSD files for validating XML files with JAXB
my-schema.xsd
conf/ // default .conf file for Log4j
log4j.conf
lib/ // libraries added to build-path via project settings
I was having issues loading my resources from the res directory. I wanted all my resources separate from my source code (simply for managment/organization purposes). So, what I had to do was add the res directory to the build-path and then access the resource via:
static final ClassLoader loader = MyClass.class.getClassLoader();
// in some function
loader.getResource("images/my-image.png");
loader.getResource("xml/my-schema.xsd");
loader.getResource("conf/log4j.conf");
NOTE: The / is omitted from the beginning of the resource string because I am using ClassLoader.getResource(String) instead of Class.getResource(String).
When you use 'getResource' on a Class, a relative path is resolved based on the package the Class is in. When you use 'getResource' on a ClassLoader, a relative path is resolved based on the root folder.
If you use an absolute path, both 'getResource' methods will start at the root folder.
#GianCarlo:
You can try calling System property user.dir that will give you root of your java project and then do append this path to your relative path for example:
String root = System.getProperty("user.dir");
String filepath = "/path/to/yourfile.txt"; // in case of Windows: "\\path \\to\\yourfile.txt
String abspath = root+filepath;
// using above path read your file into byte []
File file = new File(abspath);
FileInputStream fis = new FileInputStream(file);
byte []filebytes = new byte[(int)file.length()];
fis.read(filebytes);
For those using eclipse + maven. Say you try to access the file images/pic.jpg in src/main/resources. Doing it this way :
ClassLoader loader = MyClass.class.getClassLoader();
File file = new File(loader.getResource("images/pic.jpg").getFile());
is perfectly correct, but may result in a null pointer exception. Seems like eclipse doesn't recognize the folders in the maven directory structure as source folders right away. By removing and the src/main/resources folder from the project's source folders list and putting it back (project>properties>java build path> source>remove/add Folder), I was able to solve this.
resourcesloader.class.getClass()
Can be broken down to:
Class<resourcesloader> clazz = resourceloader.class;
Class<Class> classClass = clazz.getClass();
Which means you're trying to load the resource using a bootstrap class.
Instead you probably want something like:
resourcesloader.class.getResource("repository/SSL-Key/cert.jks").toString()
If only javac warned about calling static methods on non-static contexts...
Doe the following work?
resourcesloader.class.getClass().getResource("/package1/resources/repository/SSL-Key/cert.jks")
Is there a reason you can't specify the full path including the package?
Going with the two answers as mentioned above. The first one
resourcesloader.class.getClassLoader().getResource("package1/resources/repository/SSL-Key/cert.jks").toString();
resourcesloader.class.getResource("repository/SSL-Key/cert.jks").toString()
Should be one and same thing?
In Order to obtain real path to the file you can try this:
URL fileUrl = Resourceloader.class.getResource("resources/repository/SSL-Key/cert.jks");
String pathToClass = fileUrl.getPath;
Resourceloader is classname here.
"resources/repository/SSL-Key/cert.jks" is relative path to the file. If you had your guiclass in ./package1/java with rest of folder structure remaining, you would take "../resources/repository/SSL-Key/cert.jks" as relative path because of rules defining relative path.
This way you can read your file with BufferedReader. DO NOT USE THE STRING to identify the path to the file, because if you have spaces or some characters from not english alphabet in your path, you will get problems and the file will not be found.
BufferedReader bufferedReader = new BufferedReader(
new InputStreamReader(fileUrl.openStream()));
I made a small modification on #jonathan.cone's one liner ( by adding .getFile() ) to avoid null pointer exception, and setting the path to data directory. Here's what worked for me :
String realmID = new java.util.Scanner(new java.io.File(RandomDataGenerator.class.getClassLoader().getResource("data/aa-qa-id.csv").getFile().toString())).next();
Use this:
resourcesloader.class.getClassLoader().getResource("/path/to/file").**getPath();**
One of the stable way to work across all OS would be toget System.getProperty("user.dir")
String filePath = System.getProperty("user.dir") + "/path/to/file.extension";
Path path = Paths.get(filePath);
if (Files.exists(path)) {
return true;
}