I've got a project to do with 2 other classmates.
We used Dropbox to share the project so we can write from our houses (Isn't a very good choice, but it worked and was easier than using GitHub)
My question is now about sharing the object stream.
I want to put the file of the stream in same dropbox shared directory of the code.
BUT, when i initialize the file
File f = new File(PATH);
i must use the path of my computer (C:\User**Alessandro**\Dropbox)
As you can see it is linked to Alessandro, and so to my computer.
This clearly won't work on another PC.
How can tell the compiler to just look in the same directory of the source code/.class files?
You can use Class#getResource(java.lang.String) to obtain a URL corresponding to the location of the file relative to the classpath of the Java program:
URL url = getClass().getResource("/path/to/the/file");
File file = new File(url.getPath());
Note here that / is the root of your classpath, which is the top of the directory containing your class files. So your resource should be placed inside the classpath somewhere in order for it to work on your friend's computer.
Don't use absolute paths. Use relative paths like ./myfile.txt. Start the program with the project directory as the current dir. (This is the default in Eclipse.) But then you have to ensure that this both works for development and for production use.
As an alternative you can create a properties file and define the path there. Your code then only refers to a property name and each developer can adjust the configuration file. See Properties documentation.
I am using Maven for a project. In one of my java class, I need to grab a property file. I cannot use absolute file location but relative path. I tried several ways but no luck.
The java is in
myProject/src/main/java/com/some/myConfiguration.java
The property file is in
myProject/src/main/com/some/resources/myProperties/myFeatures.properties
And in my java, I have some code like this
String filePath = "correct/path/to/myFeatures.properties";
File repositoryFile = new File(filePath);
So my problem is what I should give to filePath
Thank you for any advice.
put your myFeatures.properties property file inside src/main/resources as suggested by Maven's Standard Directory Layout
access it in your Java code using:
this.getClass().getClassLoader().getResourceAsStream("myFeatures.properties");
You can try this.getClass.getResource("/myProperties/myFeatures.properties") because the compiled path of resources is under the class path.
I can use:
getClass().getClassLoader().getResource("").getPath());
To dynamically get the path of the Class files in my project.
But I need to get the path dynamically for the home directory of the project. I can't do this statically, like this:
File file = new File("C:\Users\etc...");
Because the destination of the project can change from one computer to the next. I need to call a file in the project's home directory. How do I do that dynamically. It would be similar to how I did it in the beginning of this question, but it can not be just for Classes.
Any idea?
There's no such thing as a "java project" in Java. If you know where your class files are in relationship to whatever directory you're trying to locate, you can use the path you obtained in your first example:
Url classUrl = ...
File resourcePath = new File(classUrl.toUri());
File rootDir = resourcePath.getParentFile().getParentFile().getParentFile();
or
File rootDir = new File(resourcePath.getPath() +
"/../../../".replace("/", File.separatorChar));
Of course, this will only work as long as the files are actually on the file system; if they're in a jar, you'd need to locate the jar file instead.
But this is usually not something you want to do. It's a better practice to configure a sandbox for your application by passing them to your application as command-line arguments or environment variables. In some cases, it's also useful to use the pre-defined System properties such as user.home, user.dir, and java.io.tmpdir.
We are working on project.
Every colleague have different folder for the install
In my case the folder of my files is in
C:\\p4_3202\\CAR\\car.rt.appl\\dev\\car.components\\cars\\res\\car.rt.components.cars\\resources\\js;
for other colleague it could be in
C:\\my_3202\\CAR2\\car.rt.appl\\dev\\car.components\\cars\\res\\car.rt.components.cars\\resources\\js;
it is depends how you config your perforce.
I need to read files from my folder but i don't know the name of the folder ( as i explained it could be different )
File folderFile = new File(folder);
How i can find the location of my folder ? ( c:\p4\......test.js )
I tried with
System.getProperty("sun.java.command");
System.getProperty("user.home")
but it didn't give me the path of my folder
I would use a system property for each user. So all users tell where perforce is installed (might already exist a property for this, look at the docs).
This could then be read by your code like:
System.getenv().get("PROP");
On a unix/Linux system you can set the property in a shell/environment variable using:
export PROP=thepath
Windows was a long time ago for me but if I remember correctly its somewhere under System on control panel :)
Update:
http://www.itechtalk.com/thread3595.html
file.getAbsolutePath() or file.getAbsoluteFile()
String path = new java.io.File(".").getCanonicalPath();
If it are files for reading only, being stored inside the final produced jar, then use URL url = getClass().getResource("/.../...") or InputStream in = getClass().getResourceAsStream("/.../...").
That uses a path inside the jar/class path. Using only the jar would never do with File.
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;
}