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()).
Related
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 /[^/]+/[^/]+$
How do we actually discard last file from java string and just get the file path directory?
Random Path input by user:
C:/my folder/tree/apple.exe
Desired output:
C:/my folder/tree/
closest solution i found is from here . The answer from this forum only display last string acquired not the rest of it. I want to display the rest of the string.
The easiest and most failsafe (read: cross-platform) solution is to create a File object from the path.
Like this:
File myFile = new File( "C:/my folder/tree/apple.exe" );
// Now get the path
String myDir = myFile.getParent();
Try that:
String path = "C:/my folder/tree/apple.exe";
path = path.substring(0, path.lastIndexOf("/")+1);
I am looking at code new FileInputStream("config.properties").
I have the same file "config.properties" in multiple places in my project(doing windows file search) and I am now confused as to which one does this function call refer to. How do i get to know the absolute path of file?
I found this on the internet but this location doesnt look like the right answer.
"ClassName".class.getProtectionDomain().getCodeSource().getLocation().getPath() but this doesnt look it. Can you please correct it if I am wrong
You can use File:
File f = new File("config.properties");
System.out.println(f.getAbsolutePath());
The path returned will be deduced from the current working directory.
File f = new File("config.properties");
String dirPath = file.getParentFile().getAbsolutePath()
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...
I have this issue of accessing a file in one of the parent directories.
To explain, consider the following dir structure:-
C:/Workspace/Appl/src/org/abc/bm/TestFile.xml
C:/Workspace/Appl/src/org/abc/bm/tests/CheckTest.java
In the CheckTest.java I want to create a File instance for the TestFile.xml
public class Check {
public void checkMethod() {
File f = new File({filePath value I want to determine}, "TestFile.xml");
}
}
I tried a few things with getAbsolutePath() and the getParent() etc but was getting a bit complicated and frankly I think I messed it up.
The reason I don't want to use "C:/Workspace/Appl/src/org/abc/bm" while creating the File instance is because the C:/Workspace/Appl is not fixed and in all circumstances will be different at runtime and basically I don't want to hard-code.
What could be the easiest and cleaner way to achieve this ?
Thank you.
You should load it from Classpath in this case.
In your CheckTest.java, try
FileInputStream fileIs = new FileInputStream(CheckTest.class.getClassLoader().getResourceAsStream("org/abc/bm/TestFile.xml");
Use System.getProperty to get the base dir or you set the base.dir during application launch
java -Dbase.dir=c:\User\pkg
System.getProperty("base.dir");
and use
System.getProperty("file.separator");
What could be the easiest and cleaner way to achieve this ?
For accessing static resources use:
URL urlToResource = this.getClasS().getResource("path/to/the.resource");
If the resource is expected to change, write it to a sub-directory of user.home, where it is easy to locate later.
First of all, you can't get a reference to the source file path on runtime.
But, you can access the resrources included at your classpath (where you complied .class files will be).
Normally, your compiler will copy the xml file included at your srouce directory into the build directory, so at last, you could end up having something like this:
C:/Workspace/Appl/classes/org/abc/bm/TestFile.xml
C:/Workspace/Appl/classes/org/abc/bm/tests/CheckTest.class
Then, with your classpath pointing to the compiled classes root dir, you get the resources from this directory, using the ClassLoader.getResource method (or the equivalent Class.getResource() method).
public class Check {
public void checkMethod() {
java.net.URL fileURL=this.getClass().getResource("/org/abc/bm/tests/TestFile.xml");
File f=new File( fileURL.toURI());
}
}
One could do this:
String pathOfTheCurrentClass = this.getClass().getResource(".").getPath();
File file = new File(pathOfTheCurrentClass + "/..", "Testfile.xml");
or
String pathOfTheCurrentClass = this.getClass().getResource(".").getPath();
File filePath = new File(pathOfTheCurrentClass);
File file = new File(filePath.getParent(), "Testfile.xml");
But as Tomas Naros points out this gives you the file located in the build path.
Did you try
URL some=Test.class.getClass().getClassLoader().getResource("org/abc/bm/TestFile.xml");
File file = new File(some.getFile());