I'm currently reading a txt file in Java, which is located in a package with the scanner object.
To receive the file location I use a quick and dirty method:
File currentDirectory = new File(new File(".").getAbsolutePath());
String location = currentDirectory.getAbsolutePath().replace(".", "")+"\\corefiles\\src\\filereadingexample\\";
Is there a better way of doing so?
I'd love to improve my code.
Greetings
J
Use a URL
URL resource = getClass().getResource("/path/to/text/file.txt");
As the path, since your text file is inside a package, use the package structure. For example, assume your file(myfile.txt) is inside
com.myproject.files
package, your path should be
"/com/myproject/files/myfile.txt"
(mind the leading slash. It's necessary)
Now you can create a File using the URL
new File(resource.getFile());
The "resource.getFile()" returns the absolute path to the file also.
Hope this helps
Get current diretory path
String currentDirPath = System.getProperty("user.dir");
String ohterPackages = currentDirPath + File.separator + "filereadingexample\\fileName.txt";
User home directory
String homePath = System.getProperty("user.home");
Related
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();
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);
After reading that is it possible to create a relative filepath name using "../" I tried it out.
I have a relative path for a file set like this:
String dir = ".." + File.separator + "web" + File.separator + "main";
But when I try setting the file with the code below, I get a FileNotFoundException.
File nFile= new File(dir + File.separator + "new.txt");
Why is this?
nFile prints: "C:\dev\app\build\..\web\main"
and
("") file prints "C:\dev\app\build"
According to your outputs, after you enter build you go up 1 time with .. back to app and expect web to be there (in the same level as build). Make sure that the directory C:\dev\app\web\main exists.
You could use exists() to check whether the directory dir exist, if not create it using mkdirs()
Sample code:
File parent = new File(dir);
if(! parent.exists()) {
parents.mkdirs();
}
File nFile = new File(parent, "new.txt");
Note that it is possible that the file denoted by parent may already exist but is not a directory, in witch case it would not be possible to use it a s parent. The above code does not handle this case.
Why wont you take the Env-Varable "user.dir"?
It returns you the path, in which the application was started from.
System.getProperty(user.dir)+File.separator+"main"+File.separator+[and so on]
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...