I'm trying to set the system property "java.security.policy" programmatically.
It works, as long as the path to the security policy file has no spaces.
File myFileReference = new File("C:\folder_name\security.policy")
System.setProperty("java.security.policy", myFileReference.getAbsolutePath());
System.setSecurityManager(new RMISecurityManager());
If there are spaces int the file path, they get escaped with a %20, like this
"C:\folder%20name\security.policy".
The code above executes fine, but then all security checks fail. I assume setProperty doesn't really find the file.
On Windows, writing the file name without that escaping for spaces works.
System.setProperty("java.security.policy", "C:\\some folder\\wideopen.policy");
So, the problem seems to be that %20 space escaping. I could replace it using a regex, but maybe that would make it work just on Windows, and fails somewhere else.
Also, I don't want to hard-code the file path like that.
I looked at the Java doc for a File function that returns a "System.setProperty" compatible path name that works on any platform. I also tried things like toURI().toString(), to no avail.
Is there an elegant way to get a working file path String from a File reference in 1 line of code?
EDIT:
This was simplified code, I construct the file like this
URL policyURL = Class.class.getResource("/sub local folder/wideopen.policy");
new File(policyURL.getFile())
I needed a relative path, so I used that little getResource trick, which happens to return an URL, with the nasty %20 escapings.
I can use an URLDecoder to strip them away now that I know what the problem is.
But is there a less error prone way?
I solved using the URLDecoder class. Here is a complete example of what I was trying to do, and the solution that made it work.
//here is the trick
URL policyFileURL = Class.class.getResource("/server/model/easy.policy");
String policyFilePath = "";
try {
policyFilePath = URLDecoder.decode(policyFileURL.getFile(), "UTF-8");
}
catch (UnsupportedEncodingException e) {}
//here what I wanted to do (now it works)
server.activate(port, new File(policyFilePath));
Related
I want to load a font in a SWT. My ttf file is in the resources/fonts directory of my Maven project. I try to load it like this:
URL fontURL = MyClass.class.getResource("/fonts/myfont.ttf");
boolean fontLoaded = display.loadFont(fontURL.getPath());
But the resulting boolean is always false. I tried to prompt the result of fontURL.getPath(), and it is something like /C:/Users/myuser/Documents/.... If I copy this result in a String, remove the first / and try to call display.loadFont() with it, it works.
Another weird thing is that this is not the only resource I load this way. For example, this is how I load the icon of the window:
URL iconURL = MyClass.class.getResource("/images/myicon.png");
Image icon = new Image(display, iconURL.getPath());
shell.setImage(icon);
And it works fine. The only file posing problem is the font file. Does anybody know why ?
The reason for / at the beginning is that getPath of the URL class returns the URL path defined by RFC 2396 (see javadocs).
As for why it's working for the Image constructor and not for loadFont() method, the answer can be found in the implementation.
The constructor uses FileInputStream which internally normalizes the path, whereas loadFont() has a native implementation for loading which does not support such path.
Since in both cases a file path is expected, what you want to do is normalize the path yourself using either File constructor or Paths.get(url.toURI()).toString() method.
I'm trying to get this folder
File imagesOrg = new File(getClass().getResource("/stock").getPath());
I've printed it out in console with
System.out.println(imagesOrg.getAbsolutePath());
and there is a space within my path so it was changed to %20 and because of that the rest of my code doesn't work which is:
for(final File child : imagesOrg.listFiles()) {
System.out.println(child.getName());
}
If I put the whole path in new File with a space instead of %20 it works fine is there an easy solution to this?
I recommand URL decoder.
You can use like this..
String result = java.net.URLDecoder.decode(url, "UTF-8");
I am trying to d file = new file (location) while location of file is absolute path with somethign like this : \\test\hold\REPO/TEST/Letter/123.pdf
I am getting file not found exception while files are there in this path. what could be going wrong? can i have path with both forward and backward slash?
String separator = System.getProperty("file.separator");
So location can be rewritten to
location=separator+"test"+separator+"hold"+separator +"REPO"+separator "TEST"+separator+"Letter"+separator+"123.pdf";
In this case no need to think about underlying OS
You need two slashes in a string literal to represent one in the filename. Try
"\\\\test\\hold\\REPO/TEST/Letter/123.pdf"
or better still
"//test/hold/REPO/TEST/Letter/123.pdf"
There is never a need to use a backslash in a filename in Java.
Try adding quotes around the filename.
You can write your code without single backward slashes.if you want to use back-word slashes use \\ instead of \ .A single backward slashes create a problem if you put it inside the String literal.So you can write you code in multiple ways to avoid your exceptions.
1) File f=new File("\\test/hold/REPO/TEST/Letter/123.pdf");
2) File f=new File("\\test\\hold\\REPO/TEST/Letter/123.pdf");
3) File f=new File("\\test\\hold\\REPO\\TEST\\Letter\\123.pdf");
4) File f=new File("/test/hold/REPO/TEST/Letter/123.pdf");
you can use :-
InputStream input = new URL("\\test\hold\REPO/TEST/Letter/123.pdf").openStream();
or
File file = new File(location);
where location=\\test\hold\REPO/TEST/Letter/123.pdf; and Check Using SOP statement whether URL is Call properly or not . hope it will help you to fine better solution
I'm trying to develop a server for mplayer using Java but I can't open files that have spaces in name (e.g. "File with space.mp3").
I'm following this tutorial here. The problem is, every time I try to open a file with spaces in name the getInputStream() read only the string before the space, generating a "file not found" error.
The path are correct in command, I tried even different formats (e.g. "File\ with\ space.mp3", "$PATH/File with space.mp3", etc), but nothing works.
What can I do to get data properly from getInputStream? How to avoid getInputStream to block when it founds a space in the String?
Ps. I use a linux system and the codes are the same as the link above (ctrl+c , ctrl+v).
thanks for the help.
The problem is the use of Runtime#exec. It thinks that the space in the file is another parameter.
Process mplayerProcess = Runtime.getRuntime().exec("/path/to/mplayer -slave -quiet -idle file/to/play.avi");
Instead, you should use ProcessBuilder which allows you to specify each parameter as a separate String eliminating the need to mess about with quotes.
ProcessBuilder pb = new ProcessBuilder("/path/to/mplayer", "-slave", "-quiet", "-idle", "file/to/play.avi");
// Other configuration options...
Process p = pb.start();
I am getting this error when I try to open a file:
java.io.FileNotFoundException: D:\Portable%20Programs\Android%20Development\workspace3\XXX-desktop\bin\World_X.fr (The system cannot find the path specified)
at java.io.FileInputStream.open(Native Method)
at java.io.FileInputStream.<init>(Unknown Source)
at java.util.Scanner.<init>(Unknown Source)
The file is existing in the directory but I am still getting this error. However when I copy the same file in the Eclipse workspace Project src folder, no such Exception is returned (though this method also creates the World_X.fr file in the bin folder).
What I am actually trying to do is get the absolute location of the .jar file through this:
fileLocation = new String(Main.class.getProtectionDomain().getCodeSource().getLocation().getPath());
And then I am appending "World_X.fr" to the fileLocation string but this is not working. Please help me in this regard.
The preferred way to convert a file: URL into an actual File is this:
File file = new File(url.toURI());
This takes care of all checks and quoting/escaping.
Using getPath() instead will leave these odd bits up to you.
You need to unescape the %20 to spaces. e.g.:
fileLocation = new String(
Main.class.getProtectionDomain().getCodeSource().getLocation().getPath())
.replaceAll("%20", " ");
Here is the solution for that , this will work only after JDK1.5 ,
try { f = new File("somePath".toURI().getPath()); } catch(Exception e) {}
The confirmed solution is quite old and even though it works for this particular case it is far more convenient to use URLDecoder, because %20 is only one encoded character, but in your path there can be whole lot of different encoded characters.
fileLocation = URLDecoder.decode(Main.class.getProtectionDomain().getCodeSource().getLocation().getPath(), "UTF-8");
Try leaving out the %20, and use normal spaces instead. Also, you're using backslashes, in your code if you're using backslashes make sure you escape them first.