We Keep Coding

Java parse File-path with different working directory

I have a file which contain several paths, like . (relative) or /Users/...../ (absolut). I need to parse the paths that are relative to the directory of the file that contains the paths and not the working-directory and create correct File-instances. I can not change the working directory of the Java-Program, since this would alter the behaviour of other components and i also have to parse several files. I don't think public File(String parent, String child)does what i want, but i may be wrong. The documentation is quite confusing.
Example:
file xy located under /system/exampleProgram/config.config has the following content:
.
/Users/Name/file
./extensions
i want to resolve these to:
/system/exampleProgram/
/Users/Name/file
/system/exampleProgram/file/

So, I am going to assume that you have access to the path of the file you opened (either via File.getAbsolutePath() if it was a File descriptor or via a regex or something)...
Then to translate your relative paths into absolute paths, you can create new File descriptions with your opened file, like so:
File f = new File(myOpenedFilePath);
File g = new File(f, "./extensions");
String absolutePath = g.getCanonicalPath();
When you create a file with a File object and a String, Java treats the String as a path relative to the File given as a first argument. getCanonicalPath will get rid of all the redundant . and .. and such.
Edit: as Leander explained in the comments, the best way to determine whether the path is relative or not (and thus whether it should be transformed or not) is to use file.isAbsolute().

Sounds like you probably want something like
File fileContainingPaths = new File(pathToFileContainingPaths);
String directoryOfFileContainingPaths =
fileContainingPaths.getCanonicalFile().getParent();
BufferedReader r = new BufferedReader(new FileReader(fileContainingPaths));
String path;
while ((path = r.readLine()) != null) {
if (path.startsWith(File.separator)) {
System.out.println(path);
} else {
System.out.println(directoryOfFileContainingPaths + File.separator + path);
}
}
r.close();
Don't forget the getCanonicalFile(). (You might also consider using getAbsoluteFile()).

Create directory at given path in Java - Path with space

I have my java code like below-
string folderName = "d:\my folder path\ActualFolderName";
File folder = new File( folderName );
folder.mkdirs();
So here as given directory path has space in it. folder created is d:\my, not the one I am expecting.
Is there any special way to handle space in file/folder paths.

You should us \\ for path in java. Try this code
String folderName = "D:\\my folder path\\ActualFolderName";
File folder = new File( folderName );
folder.mkdirs();
Or use front-slashes / so your application will be OS independent.
String folderName = "D:/my folder path1/ActualFolderName";

Unless you are running a really old version of Java, use the Path API from JDK7:
Path p = Paths.get("d:", "my folder path", "ActualFolderName");
File f = p.toFile();
It will take care of file separators and spaces for you automatically, regardless of OS.

Following alternatives should work in Windows:
String folderName = "d:\\my\\ folder\\ path\\ActualFolderName";
String folderName = "\"d:\\my folder path\\ActualFolderName\"";

You need to escape your path (use \\ in your path instead of \) and you also need to use String, with an uppercase S, as the code you posted does not compile. Try this instead, which should work:
String folderName = "D:\\my folder path\\ActualFolderName";
new File(folderName).mkdirs();
If you are getting your folder name from user input (ie.not hardcoded in your code), you don't need to escape, but you should ensure that it is really what you expect it is (print it out in your code before creating the File to verify).
If your are still having problems, you might want to try using the system file separator character, which you can get with System.getProperty(file.separator) or accesing the equivalent field in the File class. Also check this question.

You need to escape path seprator:
String folderName = "D:\\my folder path\\ActualFolderName";
File file = new File(folderName);
if (!file.exists()) {
file.mkdirs();
}

First of all, the String path you have is incorrect anyway as the backslash must be escaped with another backslash, otherwise \m is interpreted as a special character.
How about using a file URI?
String folderName = "d:\\my folder path\\ActualFolderName";
URI folderUri = new URI("file:///" + folderName.replaceAll(" ", "%20"));
File folder = new File(folderUri);
folder.mkdirs();

Why can't java find my file?

I'm trying to see how many of these text files exist, but even with them there, the program always says the numFiles = 0. I have the files in a folder called Levels within the src folder. Thanks
int numFiles = 0;
for(int i = 0; i < 24; i++){
File file = new File("/Levels/level" + (i+1) + ".txt");
if(file.exists()){
numFiles++;
}
}
System.out.println(numFiles);

Edited
I overlooked that DirectoryStream doesn't support count()
You could go with an absolute path and make use of Stream API and lambdas. Like so:
String dirString = "..." //absolute Path
Path dir = Paths.get(dirString);
int numFiles = dir.getNameCount();
System.out.println(numFiles);
One advantage is that you can rename the files at will as long as they stay in the same directory. If you only want to work with specific files you can use filter() like so:
Files.newDirectoryStream(dir).filter(Predicate);
or add the filter directly when creating the DirectoryStream like so:
Files.newDirectoryStream(dir, RegEx);
To do something with each File you can use the consumer forEach() or have a look at Stream JavaDoc for other consumers/intermediate operations. Also double check if the DirectoryStream supports the Stream operation you want to use.

Your path is incorrect - if you are referring to an absolute location only then start with a /.
Also if you are using an editor remember your Java files are in src but but you don't run Java File you run class files and the class files may be in your bin/build directory most likely - check if the text file are in the build or bin directory.

Your path is incorrect, if you are referring to a local file(like something in your project folder) use
File file = new File("Levels/level" + (i+1) + ".txt");
the slash you used in front of the name makes it look in the root of the drive, not the local directory.

Jump one folder up in a URL in Java

I have a URL that I get as below:
String jarFilePath = getClass().getProtectionDomain().getCodeSource().getLocation()
This would get me the complete path to the jar file. Now how do I jump one folder up and append some other path to it? For example., if the jarFilePath is something like:
c:/path/to/jar/file.jar
I want to jump one folder up and append another relative path like below:
c:/path/to/resources/path/to/resources/
Where the folders resources and jar are at the same directory level in the file system.

File f = new File("C:/path/to/jar/file.jar");
File dest = new File(f.getParentFile().getParentFile(), "resources/path/to/resources");

Just use the File-Object, that makes a lot of things easier:
import java.io.File;
String pathname = "c:/path/to/jar/file.jar";
File f = new File(pathname);
String p = f.getParent();

Try and use a File object:
File jarFile = new File(jarFilePath);
File newFolder = new File( jarFile.getParentFile().getParentFile(), "resources/path/to/resources");
If you want to use the path as a string, try using Apache Commons IO's FilenameUtils:
String resourcesPath = FilenameUtils.normalize( FilenameUtils.getPath(jarFilePath) + "/../resources/path/to/resources");

You have several ways.
You can split the path into it's elements and rebuild it until array.length -3 (-1 would be filename, -2 the last folder)
You could simple remove the file and append another ../ (which just means: "Go one directory back") (that would be something like this then: c:/path/to/jar/../resources/path/to/resources/)
You gould use a regex to get rid of the last folder and file. something like /[^/]+/[^/]+$

Get the filePath from Filename using Java

Is there a easy way to get the filePath provided I know the Filename?

You can use the Path api:
Path p = Paths.get(yourFileNameUri);
Path folder = p.getParent();

Look at the methods in the java.io.File class:
File file = new File("yourfileName");
String path = file.getAbsolutePath();

I'm not sure I understand you completely, but if you wish to get the absolute file path provided that you know the relative file name, you can always do this:
System.out.println("File path: " + new File("Your file name").getAbsolutePath());
The File class has several more methods you might find useful.

Correct solution with "File" class to get the directory - the "path" of the file:
String path = new File("C:\\Temp\\your directory\\yourfile.txt").getParent();
which will return:
path = "C:\\Temp\\your directory"

You may use:
FileSystems.getDefault().getPath(new String()).toAbsolutePath();
or
FileSystems.getDefault().getPath(new String("./")).toAbsolutePath().getParent()
This will give you the root folder path without using the name of the file. You can then drill down to where you want to go.
Example: /src/main/java...