Develop Reference

windows - User folder with special character -> class.getResource returns null [duplicate]

The getResourceAsStream-method returns null whenever running the executable jar in a directory which ends with a exclamation mark.
For the following example, I have a Eclipse project the following directory structure:
src\ (Source Folder)
main\ (Package)
Main.java
res\ (Source Folder)
images\
Logo.png
I'm reading the Logo.png as follows:
public static void main(String[] args) throws IOException {
try (InputStream is = Main.class.getClassLoader().getResourceAsStream("images/Logo.png")) {
Image image = ImageIO.read(is);
System.out.println(image);
}
}
See the attachment for 2 test cases. First, the executable jar is started from the directory "D:\test123!##" without any problems. Secondly, the executable jar is started from the directory "D:\test123!##!!!", with problems.
Are directories ending with an exclamation mark not supported? Is the code wrong?
Thanks in advance.

Probably because of this bug or any of the many similar bugs in the Java bug database:
http://bugs.sun.com/view_bug.do?bug_id=4523159
The reason is that "!/" in a jar URL is interpreted as the separator between the JAR file name and the path within the JAR itself. If a directory name ends with !, the "!/" character sequence at the end of the directory is incorrectly interpreted. In your case, you are actually trying to access a resource with the following URL:
jar:file:///d:/test1231##!!!/test.jar!/images/Logo.png
The bug has been open for almost 12 years and is not likely to be fixed. Actually I don't know how it can be fixed without breaking other things. The problem is the design decision to use ! as a character with a special meaning (separator) in the URL scheme for JAR files:
jar:<URL for JAR file>!/<path within the JAR file>
Since the exclamation mark is an allowed character in URLs, it may occur both in the URL to the JAR file itself, as well as in the path within the JAR file, making it impossible in some cases to find the actual "!/" separator.

A simple work around for Windows is to use "\" instead of "/" in the path. That would mean the "!/" character sequence is found after the full path. For instance:
new URL("jar:file:\\d:\\test1231##!!!\\test.jar!/images/Logo.png");
My Code:
File jar = new File(jarPath + "/" + jarName);
URL url = new URL("jar:" + jar.toURI() + "!" + dataFilePath);
InputStream stream = null;
try {
stream = url.openStream();
} catch (FileNotFoundException e) {
// Windows fix
URL urlFix = new URL("jar:" + jar.toURI().toString().replace('/', '\\')
+ "!" + dataFilePath);
stream = urlFix.openStream();
}
I use toURI() because it handles things like spaces.
Fixes:
The fix itself would be for Java to check if the file exists and if not continue to the next separator (the "!/" part of the url) until the separators are exhausted, then throw the exception. So it would see that "d:\test1231##!!" throws a java.io.FileNotFoundException and would then try "d:\test1231##!!!\test.jar" which does exist. This way it does not matter if there are "!" in the file path or in the jar's files.
Alternatively the "!/" can be switched to something else that is an illegal file name or to something specific (like "jarpath:").
Alternatively make the jar's file path use another parameter.
Note:
It may be possible to override something, swap a handler, or change the code to open the file first then look inside the jar file later but I have not looked.

Convert Windows style path into unix path in java code

I am working in a java code that was designed to run on windows and contains a lot of references to files using windows style paths "System.getProperty("user.dir")\trash\blah". I am in charge to adapt it and deploy in linux. Is there an efficient way to convert all those paths(\) to unix style (/) like in "System.getProperty("user.dir")/trash/blah". Maybe, some configuration in java or linux to use \ as /.

My approach is to use the Path object to hold the path information, handle concatenate and relative path. Then, call Path's toString() to get the path String.
For converting the path separator, I preferred to use the apache common io library's FilenameUtils. It provides the three usefule functions:
String separatorsToSystem(String path);
String separatorsToUnix(String path);
String separatorsToWindows(String path)
Please look the code snippet, for relative path, toString, and separator changes:
private String getRelativePathString(String volume, Path path) {
Path volumePath = Paths.get(configuration.getPathForVolume(volume));
Path relativePath = volumePath.relativize(path);
return FilenameUtils.separatorsToUnix(relativePath.toString());
}

I reread your question and realize you likely don't need help writing paths. For what you're trying to do I am not able to find a solution. When I did this in a project recently I had to take time to convert all paths. Further, I made the assumption that working out of the "user.home" as a root directory was relatively sure to include write access for that user running my application. In any case, here are some path problems I addressed.
I rewrote the original Windows code like so:
String windowsPath = "C:\temp\directory"; //no permission or non-existing in osx or linux
String otherWindowsPath = System.getProperty("user.home") + "\Documents\AppFolder";
String multiPlatformPath = System.getProperty("user.home") + File.separator + "Documents" + File.separator + "AppFolder";
If you're going to be doing this in a lot of different places, perhaps write a utility class and override the toString() method to give you your unix path over and over again.
String otherWindowsPath = System.getProperty("user.home") + "\Documents\AppFolder";
otherWindowsPath.replace("\\", File.separator);

Write a script, replace all "\\" with a single forward slash, which Java will convert to the respected OS path.

filePath.getFileName() behaves differently

I tried the use of Path interface;
//get a path object with relative path
Path filePath = Paths.get("C:\\Test\\filename.txt");
System.out.println("The file name is: " + filePath.getFileName());
Path filePath2 = Paths.get("/home/shibu/Desktop/filename.txt");
System.out.println("The file name is: " + filePath2.getFileName());
The out put is like;
The file name is: C:\Test\filename.txt
The file name is: filename.txt
For the windows file it printed full path and for linux file it printed only file name.
Why this difference?

Simple: On Linux, the only illegal characters in a file name are / and the 0 byte. Everything else, including \, line feed and escape sequences, are valid.
That means C:\Test\filename.txt is a valid file name on Linux. The Java runtime doesn't attempt to be smart and guess that this might be a Windows path.
Note that this is different when you use /: This is a valid path delimiter on Windows when using Java. So the path a/foo.txt is a relative path both on Linux and Windows.
This means you can open files on Windows using Paths.get("/C:/Test/filename.txt");, for example.

Getting FileNotFoundException when using System.getProperty("user.dir")

I am attempting to access a file within my current working directory.
The error I am getting is
[java] java.io.FileNotFoundException: /u/user/Documents/DataComProject1\confA.txt
The line which is causing this, I presume is:
bufferedReader = new BufferedReader(new FileReader(System.getProperty("user.dir") + "/" + fileName));
Whenever I print the directory I'm attempting to use with FileReader() I get:
/u/user/Documents/DataComProject1/confA.txt
I believe the problem has to do with the the backslash before the text file name being in a different direction. Upon looking in the directory I can see the file is there.

You could use the Path library instead of creating the path yourself:
Path p = Paths.get(System.getProperty("user.dir"))
.resolve(filename);
File f = p.toFile();

Don't write "/" to separate path elements, use this instead to get the correct path separator that's appropriate for your platform:
File.separator

It looks like your fileName includes an embedded backslash: it's Documents/DataComProject1\confA.txt. Since the backslash is a valid character in a file name, DataComProject1\confA.txt is assumed to be name of the file, not a name of a file in a directory.
To fix the file name you have to change the embedded \\ into the correct separator character:
fileName = fileName.replace('\\', File.separator);
bufferedReader = new BufferedReader(new FileReader(System.getProperty("user.dir") + "/" + fileName));
It would be correct to use the slash / as the directory separator on every system, but here I use File.separator because not using a hardcoded value makes the intent of the code clearer.

Java File.renameTo(File) not working

I'm trying to list a directory's contents, and rename certain files.
public void run(String dirName) {
try {
File parDir = new File(dirName);
File[] dirContents = parDir.listFiles();
// Rename if necessary
for(File f : dirContents) {
System.out.println("f is:\n" + f.toString());
String name = f.getName();
String subbedName = name.replaceAll("\uFFFD", "_");
System.out.println("\n" + "name = " + name + ", subbedName = " + subbedName + "\n");
if(!name.equals(subbedName)) {
File newFile = new File(f.getParentFile(), subbedName);
System.out.println("newFile is:\n" + newFile.toString());
if(!f.renameTo(newFile))
System.out.println("Tried to change file name but couldn't.");
}
}
}
catch(Exception exc1) {
System.out.println("Something happened while listing and renaming directory contents: " + exc1.getMessage());
}
}
When I run this, I get "Tried to change file name but couldn't." I don't believe that Java is considering these files to be "open", so I don't think that's the reason. I've even ran chmod 777 myDir where myDir is the value of the dirName string passed into the run method.
What am I missing here? Why won't Java rename these file(s)? These are CentOS machines.
Edit: Added printouts for both f and newFile, which is as follows:
f is:
/root/path/to/mydir/test�.txt
newFile is:
/root/path/to/mydir/test_.txt

You need to create your new File object with the full pathname of those files. So
String name = f.getName(); // gets the name without the directory
should likely be:
String name = f.getAbsolutePath();
(your search/replace may need to change)

The problem is that f.getName() returns the last name component of the path that is represented by f. You then massage this String and turn it back into a File. But the File now represents a path relative to the current directory, not the directory containing the original path.
As a result your code is actually attempting to rename the files from dirName into the application's current directory. That could fail because files already exist in the current directory with those names, or because the dirName and the current directory are in different file systems. (You cannot rename a file from one filesystem to another ... you have to copy it.)
Please note that a File in Java represents a pathname, not a file or a folder. In your code, the f objects are the pathnames for file system objects (either files or folders) in the directory denoted by the String dirname. Each of these f objects will have a directory part.
There is more than one way to fix your code; for example
change name = f.getName() to name = f.toString()
change new File(subbedName) to new File(f.getParentFile(), subbedName)
I have an alternative / additional theory.
The pathname of the file containing the \uFFFD character is coming out as "mojibake"; i.e. the kind of garbled text that you get when you display encoded text using the wrong encoding. And since we are seeing 3 characters of garbled text, I suspect that it is attempting to display the UTF-8 rendering of \uFFFD as Latin-1.
So my theory is that the same think is happening when the File.renameTo method is converting f to the form that it is going to provide to the system call. For some reason that is no clear to me, Java could be using the wrong encoding, and as a result producing a "name" for the original file that doesn't match the name of the file in the file system. That would be sufficient to cause the rename to fail.
Possibly related questions / links:
File name charset problem in java
http://bugs.sun.com/bugdatabase/view_bug.do?bug_id=4733494 (Note that Sun decided this was not a Java bug, and most of the "me too" comments on the bug report are from people who do not understand the explanation ...)

f.getName(); only returns the name of the folder, not the full path. So subbedName becomes a relative path file. Try something with f.getCanonicalPath() instead.