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Access file without knowing absolute path, only knowing file name

I'm trying to use a file in my code but I don't want to have specify the absolute file path, only the file name, for example "fileName.txt".
I want to do this so I have the ability to use this code on different laptops where the file may be stored in different folders.
The code below is what I'm using at the moment but I receive a NoSuchFileException when I ran it.
FileSystem fs FileSystems.getDefault();
Path fileIn = Paths.get("fileName.txt");
Any ideas how to overcome this problem so I can find the file without knowing its absolute path?

Ideas on how to find the file without knowing its absolute path:
Instruct the user of the app to place the file in the working directory.
Instruct the user of the app to give the path to the file as a program argument, then change the program to use the argument.
Have the program read a configuration file, found using options 1 or 2, then instruct the user of the app to give the path to the file in the configuration file.
Prompt the user for the file name.
(Not recommended) Scan the entire file system for the file, making sure there is only one file with the given name. Optional: If more than one file is found, prompt the user for which file to use.

if you don't ask the user for the complete path, and you don't have a specific folder that it must be in, then your only choice is to search for it.
Start with a rootmost path. Learn to use the File class. Then search all the children. This implementation only returned the first file found with that name.
public File findFile(File folder, String fileName) {
File fullPath = new File(folder,fileName);
if (fullPath.exists()) {
return fullPath;
}
for (File child : folder.listFiles()) {
if (child.isDirectory()) {
File possible = findFile(child,fileName);
if (possible!=null) {
return possible;
}
}
}
return null;
}
Then start this by calling either the root of the file system, or the configured rootmost path that you want to search
File userFile = findFile( new File("/"), fileName );
the best option, however, is to make the user input the entire path. There are nice file system browsing tools for most environments that will do this for the user.

Why Groovy and Java return a dot instead the file name?

In a piece of some Gradle code I want to get a current folder name without splitting the path. But all attempts that I tried:
Paths.get(".").getFileName()
new File(".").getName()
new File(".").name
, return a dot "." instead of the name. Is there some function that gives the name, not another string by which the folder could be addressed?
What is interesting, if I use:
String currentDirPath = new File(".").absolutePath
println currentDirPath
currentDirPath = currentDirPath.substring(0,currentDirPath.lastIndexOf("\\"))
println currentDirPath
String currentDir = currentDirPath.substring(currentDirPath.lastIndexOf("\\")+1)
, it is seen, that the path string looks as:
C:\Users\543829657\workspace\dev.appl.ib.cbl\application\.
So, it is simply incorrect to take the last substring after
'\'. But all those three functions take not the name of the name of the really actual folder, but the last "."!

Gradle is build upon Groovy, which is a JVM language, just like Java. So to get the current working directory, you can simply use the same ways you would use in Java. As an example, the following code will give you the name of the working directory, not the full path (check Frans answer).
new File('').absoluteFile.name
However, please mention that, in Gradle, you should not create or access files from the current working directory, but from the project (project.projectDir) or build (project.buildDir) directories, since you might accidentally build projects from lower directories, e.g. because Gradle checks for settings.gradle scripts in parent directories.

Is not
new File("").absolutePath
what you are looking for?

We should not really retrieve the absolute path that way.
In Gradle, you can use project.projectDir to get the project path or rootProject if its a multiproject or if you want to get a path of the file project.file('yourfile')

Java - How to get the correct absolute path of a file within a project

I have an XML file in a folder within my Java project, and I'd like to get its absolute path, so I can load it as a File in order to parse it(DOM). Instead of using an absolute/relative path, I want to specify only the file name, and get the absolute path after that. I tried to do this in a few different ways, but there is always a folder name missing from the path I get.
I get:
C:\Users\user\workspace\projectName\Input.xml<br>
instead of:
C:\Users\user\workspace\projectName\\**Folder1**\\Input.xml
-
File input = new File(project.getFile("Input.xml").getLocation().toString());`
File input = new File(project.getFile("Input.xml").getRawLocation().makeAbsolute().toString());
File input = new File(project.getFile("Input.xml").getLocationURI().getRawPath().toString());
File input = new File(project.getFile("Input.xml").getFullPath().toFile().getAbsolutePath());
How can I get the correct path, that includes that Folder1?

Reading your question (your project are in workspace directory) I suppose you're talking of a project in Eclipse.
Well the default directory where your app run into Eclipse is right the base dir of your project.
So if you run something like this in your main:
Files.newDirectoryStream(Paths.get("."))
.forEach(path -> {
System.out.println(path);
System.out.println(path.toFile().getAbsolutePath());
});
You should see all the files and directory that are in your project.
So if what you want is just the absolute path to your project run:
System.out.println(Paths.get(".").toFile().getAbsolutePath());
If you want open the resource Input.xml specifying only the name, I suggest to move all the files you need in a directory and run a method like this:
public static File getFileByName(String name, String path) throws IOException {
ArrayList<File> files = new ArrayList<>();
Files.newDirectoryStream(Paths.get(path))
.forEach(p -> {
if (p.getFileName()
.equals(name))
files.add(p.toFile());
});
return files.size() > 0 ? files.get(0) : null;
}

How can I get real path for file in my WebContent folder?

I need to get real path for file in my WebContent directory, so that framework that I use can access that file. It only takes String file as attribute, so I need to get the real path to this file in WebContent directory.
I use Spring Framework, so solution should be possible to make in Spring.

If you need this in a servlet then use getServletContext().getRealPath("/filepathInContext")!

getServletContext().getRealPath("") - This way will not work if content is being made available from a .war archive. getServletContext() will be null.
In this case we can use another way to get real path. This is example of getting a path to a properties file C:/Program Files/Tomcat 6/webapps/myapp/WEB-INF/classes/somefile.properties:
// URL returned "/C:/Program%20Files/Tomcat%206.0/webapps/myapp/WEB-INF/classes/"
URL r = this.getClass().getResource("/");
// path decoded "/C:/Program Files/Tomcat 6.0/webapps/myapp/WEB-INF/classes/"
String decoded = URLDecoder.decode(r.getFile(), "UTF-8");
if (decoded.startsWith("/")) {
// path "C:/Program Files/Tomcat 6.0/webapps/myapp/WEB-INF/classes/"
decoded = decoded.replaceFirst("/", "");
}
File f = new File(decoded, "somefile.properties");

you must tell java to change the path from your pc into your java project so
if you use spring use :
#Autowired
ServletContext c;
String UPLOAD_FOLDEdR=c.getRealPath("/images");
but if you use servlets just use
String UPLOAD_FOLDEdR = ServletContext.getRealPath("/images");
so the path will be /webapp/images/ :)

In situations like these I tend to extract the content I need as a resource (MyClass.getClass().getResourceAsStream()), write it as a file to a temporary location and use this file for the other call.
This way I don't have to bother with content that is only contained in jars or is located somewhere depending on the web container I'm currently using.

Include the request as a parameter. Spring will then pass the request object when it calls the mapped method
#RequestMapping .....
public String myMethod(HttpServletRequest request) {
String realPath = request.getRealPath("/somefile.txt");
...

You could use the Spring Resource interface (and especially the ServletContextResource): http://static.springsource.org/spring/docs/current/javadoc-api/org/springframework/core/io/Resource.html

This approach uses the resource loader to get the absolute path to a file in your app, and then goes up a few folders to the app's root folder. No servlet context required! This should work if you have a "web.xml" in your WEB-INF folder. Note that you may want to consider using this solely for development, as this type of configuration is usually best stored externally from the app.
public String getAppPath()
{
java.net.URL r = this.getClass().getClassLoader().getResource("web.xml");
String filePath = r.getFile();
String result = new File(new File(new File(filePath).getParent()).getParent()).getParent();
if (! filePath.contains("WEB-INF"))
{
// Assume we need to add the "WebContent" folder if using Jetty.
result = FilenameUtils.concat(result, "WebContent");
}
return result;
}

my solve for: ..../webapps/mydir/ (..../webapps/ROOT/../mydir/)
String dir = request.getSession().getServletContext().getRealPath("/")+"/../mydir";
Files.createDirectories(Paths.get(dir));

try to use this when you want to use arff.txt in your development and production level too
String path=getServletContext().getRealPath("/WEB-INF/files/arff.txt");

How can I get relative path of the folders in my android project?

How can I get the relative path of the folders in my project using code?
I've created a new folder in my project and I want its relative path so no matter where the app is, the path will be correct.
I'm trying to do it in my class which extends android.app.Activity.
Perhaps something similar to "get file path from asset".

Make use of the classpath.
ClassLoader classLoader = Thread.currentThread().getContextClassLoader();
URL url = classLoader.getResource("path/to/folder");
File file = new File(url.toURI());
// ...

Are you looking for the root folder of the application? Then I would use
String path = getClass().getClassLoader().getResource(".").getPath();
to actually "find out where I am".

File relativeFile = new File(getClass().getResource("/icons/forIcon.png").toURI());
myJFrame.setIconImage(tk.getImage(relativeFile.getAbsolutePath()));

With this I found my project path:
new File("").getAbsolutePath();
this return "c:\Projects\SampleProject"

You can check this sample code to understand how you can access the relative path using the java sample code
import java.io.File;
public class MainClass {
public static void main(String[] args) {
File relative = new File("html/javafaq/index.html");
System.out.println("relative: ");
System.out.println(relative.getName());
System.out.println(relative.getPath());
}
}
Here getPath will display the relative path of the file.

In Android, application-level meta data is accessed through the Context reference, which an activity is a descendant of.
For example, you can get the source directory via the getApplicationInfo().sourceDir property.
There are methods for other folders as well (assets directory, data dir, database dir, etc.).

Generally we want to add images, txt, doc and etc files inside our Java project and specific folder such as /images.
I found in search that in JAVA, we can get path from Root to folder which we specify as,
String myStorageFolder= "/images"; // this is folder name in where I want to store files.
String getImageFolderPath= request.getServletContext().getRealPath(myStorageFolder);
Here, request is object of HttpServletRequest. It will get the whole path from Root to /images folder. You will get output like,
C:\Users\STARK\Workspaces\MyEclipse.metadata.me_tcat7\webapps\JavaProject\images

With System.getProperty("user.dir") you get the "Base of non-absolute paths" look at
Java Library Description