I am getting some strange behavior when trying to convert between Files and URLs, particularly when a file/path has spaces in its name. Is there any safe way to convert between the two?
My program has a file saving functionality where the actual "Save" operation is delegated to an outside library that requires a URL as a parameter. However, I also want the user to be able to pick which file to save to. The issue is that when converting between File and URL (using URI), spaces show up as "%20" and mess up various operations. Consider the following code:
//...user has selected file
File userFile = myFileChooser.getSelectedFile();
URL userURL = userFile.toURI().toURL();
System.out.println(userFile.getPath());
System.out.println(userURL);
File myFile = new File(userURL.getFile());
System.out.println(myFile.equals(userFile);
This will return false (due to the "%20" symbols), and is causing significant issues in my program because Files and URLs are handed off and often operations have to be performed with them (like getting parent/subdirectories). Is there a way to make File/URL handling safe for paths with whitespace?
P.S. Everything works fine if my paths have no spaces in them (and the paths look equal), but that is a user restriction I cannot impose.
The problem is that you use URL to construct the second file:
File myFile = new File(userURL.getFile());
If you stick to the URI, you are better off:
URI userURI = userFile.toURI();
URL userURL = userURI.toURL();
...
File myFile = new File(userURI);
or
File myFile = new File( userURL.toURI() );
Both ways worked for me, when testing file names with blanks.
Use instead..
System.out.println(myFile.toURI().toURL().equals(userURL);
That should return true.
Related
I'm loading some network path in my java code. It is not taking the same format as present in the configuration file, missing one slash.
Example:
String path = "//abckatte.com/abc/test";
File fileobj = new File(path);
Whenever I saw the fileobj in log message it is displaying as /abckatte.com/abc/test. One slash is missing.
I tried with appending two more slash like.
String path = "////abckatte.com/abc/test";
then also it is not working.
You could make use of Apache Commons VFS 2, as it provides access to several file systems. Chek it out here, in Local files file://///somehost/someshare/afile.txt.
This is such a simple question, I'm sure the answer is out there and I'm simply not searching with the proper lingo. I'm new to Java, using Java 8, and want to learn how to properly handle this, rather than rigging it together.
The application takes in arguments via command line.
$ MyApp /home/user/thefiletheywant.me
I have tried the following:
// Missing Scheme, I know I can just force ("file:" + args[0]) but is that proper?
URI fileIn = new URI(args[0]);
// I've learned this is the same thing as above
URI fileIn = URI.create(args[0]);
I've seen examples that take the string, check with File.Separator to verify it is "/" and if not, replace it, then simply tack on "file:" in front. Which, again, seems sloppy.
What if the user added "http:"?
What if the user specifies a full path or a path relative to the directory they are currently in?
Do the builtin functions verify the path is proper? I'm aware of file.isFile() and file.exists(), which I can check myself easy enough.
If I knew exactly where the file was every time, of course the URI.create would be fine. But for future education, I want to know how to properly handle this very simple scenario. Please forgive me if in my searches I've simply somehow missed what I suspect is an easy solution.
You could just create a File object in java, which is OS-independent (Windows uses backslash for instance), check if it exists, and use the handy toURI() method on it, to create a valid URI object.
File myFile = new File(args[0]);
URI fileUri = null;
if(myFile.exists()) {
fileUri = myFile.toURI();
}
A user can submit a subfolder/filename to download.
The subfolder/filename will then be used to serve a file from a predertemined folder.
In the end, I am doing new File(folder, "subfolder/filename").
But before I do that, I also check that !"subfolder/filename".contains("..")
But is this enough? Is there possibly a scenario where two dots (..) may not come after each other, but still be interpreted as two dots when passed to new File(...) ?
Are there any other way a user can navigate back and reach content outside this folder?
Do you need to do something else to secure such a subfolder/filename access from folder?
One can get the absolute paths, from the OS, so a bit slow.
String folderPath = folder.getCanonicalPath() + File.separator;
File file = new File(folder, "subfolder/filename");
String path = file.getCanonicalPath();
if (!path.startsWith(folderPath)) {
log(Level.ERROR, "Security breach attempt: ...");
return;
}
A simple check would probably do too:
Pattern BREACH = Pattern.compile("\\.[\\\\]*\\.");
if (BREACH.matcher(path).find()) { ... }
Mind when you use version control or other "protected" files/folders, then names of files or folders starting with a dot are illegal too.
You can execute something like
cd ./\.\.
In Unix it will change directory to parent. May be You can resolve file and when check if it under right parent?
UPD: looks like in java You cannot use \.\. pattern http://goo.gl/4Rszg5 still it does not mean what check for ".." is sufficient. Better check canonical path
I have some text configuration file that need to be read by my program. My current code is:
protected File getConfigFile() {
URL url = getClass().getResource("wof.txt");
return new File(url.getFile().replaceAll("%20", " "));
}
This works when I run it locally in eclipse, though I did have to do that hack to deal with the space in the path name. The config file is in the same package as the method above. However, when I export the application as a jar I am having problems with it. The jar exists on a shared, mapped network drive Z:. When I run the application from command line I get this error:
java.io.FileNotFoundException: file:\Z:\apps\jar\apps.jar!\vp\fsm\configs\wof.txt
How can I get this working? I just want to tell java to read a file in the same directory as the current class.
Thanks,
Jonah
When the file is inside a jar, you can't use the File class to represent it, since it is a jar: URI. Instead, the URL class itself already gives you with openStream() the possibility to read the contents.
Or you can shortcut this by using getResourceAsStream() instead of getResource().
To get a BufferedReader (which is easier to use, as it has a readLine() method), use the usual stream-wrapping:
InputStream configStream = getClass().getResourceAsStream("wof.txt");
BufferedReader configReader = new BufferedReader(new InputStreamReader(configStream, "UTF-8"));
Instead of "UTF-8" use the encoding actually used by the file (i.e. which you used in the editor).
Another point: Even if you only have file: URIs, you should not do the URL to File-conversion yourself, instead use new File(url.toURI()). This works for other problematic characters as well.
The default output of File.toURL() is
file:/c:/foo/bar
These don't appear to work on windows, and need to be changed to
file:///c:/foo/bar
Does the format
file:/foo/bar
work correctly on Unix (I don't have a Unix machine to test on)? Is there a library that can take care of generating a URL from a File that is in the correct format for the current environment?
I've considered using a regex to fix the problem, something like:
fileUrl.replaceFirst("^file:/", "file:///")
However, this isn't quite right, because it will convert a correct URL like:
file:///c:/foo/bar
to:
file://///c:/foo/bar
Update
I'm using Java 1.4 and in this version File.toURL() is not deprecated and both File.toURL().toString() and File.toURI().toString() generate the same (incorrect) URL on windows
The File(String) expects a pathname, not an URL. If you want to construct a File based on a String which actually represents an URL, then you'll need to convert this String back to URL first and make use of File(URI) to construct the File based on URL#toURI().
String urlAsString = "file:/c:/foo/bar";
URL url = new URL(urlAsString);
File file = new File(url.toURI());
Update: since you're on Java 1.4 and URL#toURI() is actually a Java 1.5 method (sorry, overlooked that bit), better use URL#getPath() instead which returns the pathname, so that you can use File(String).
String urlAsString = "file:/c:/foo/bar";
URL url = new URL(urlAsString);
File file = new File(url.getPath());
The File.toURL() method is deprecated - it is recommended that you use the toURI() method instead. If you use that instead, does your problem go away?
Edit:
I understand: you are using Java 4. However, your question did not explain what you were trying to do. If, as you state in the comments, you are attempting to simply read a file, use a FileReader to do so (or a FileInputStream if the file is a binary format).
What do you actually mean with "Does the format file:/c:/foo/bar work correctly on Unix"?
Some examples from Unix.
File file = new File("/tmp/foo.txt"); // this file exists
System.out.println(file.toURI()); // "file:/tmp/foo.txt"
However, you cannot e.g. do this:
File file = new File("file:/tmp/foo.txt");
System.out.println(file.exists()); // false
(If you need a URL instance, do file.toURI().toURL() as the Javadoc says.)
Edit: how about the following, does it help?
URL url = new URL("file:/tmp/foo.txt");
System.out.println(url.getFile()); // "/tmp/foo.txt"
File file = new File(url.getFile());
System.out.println(file.exists()); // true
(Basically very close to BalusC's example which used new File(url.toURI()).)