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]
Related
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.
First of all i request people do not consider it as a duplicate question, please look into query.
I am copying the xml files from one folder to other folder, in the source folder, i have some files which have some content like "backingFile="$IDP_ROOT/metadata/iPAU-SP-metadata.xml" but while writing to the destination folder.
i am replacing the "$IDP_ROOT" with my current working directory. The entire copying of files is for deploying into tomcat
server. The copying is done only when server starts for the first time.
Problem: If i change the folder name from my root path in my machine after i run the server,
the entire process will be stopped because the destination folder files already contains the content which is with
existed files names or folder names.
So i want to change it to relative path instead absolute path. What is the best way to do it?
Please look at code below:
// Getting the current working directory
String currentdir = new File(".").getAbsoluteFile().getParent() + File.separator;
if(currentdir.indexOf("ControlPanel")!=-1){
rootPath=currentdir.substring(0, currentdir.indexOf("ControlPanel"));
}else{
rootPath=currentdir;
}
rootPath = rootPath.replace("\\", "/");
// target file in source folder is having "backingFile="$IDP_ROOT/metadata/iPAU-SP-metadata.xml"
String content = FileReaderUtil.readFile(targetFile,
rootPath + "Idp/conf");
content = updatePath(content, Install.rootPath
+ "IdP/IdPserver/metadata","$IDP_ROOT");
FileWriterUtil.writeToFile(Install.rootPath
+ "IdP/IdPserver/idp/conf", content,
targetFile);
// update method
public String updatePath(String content, String replaceString,String replaceKey) {
replaceKey = replaceKey!=null ? replaceKey : "$IDP_SERVER_ROOT";
replaceString= replaceString.replace("\\","/");
String updateContent = content.replace(replaceKey,
replaceString);
return updateContent;
}
I trying to check whether a file exists at given directory location.
File seacrhFile = new File("D:/input", contract.conf);
if (seacrhFile.exists()) {
returnFile = seacrhFile;
} else {
System.out.println("No such file exists");
}
reutrn returnFile;
This is working in D:/input directory scenario, but if I Change the directory location to src/test/resources/input folder then I am getting No such file exists, eventhough the file exists.
If you want to have access to
src/test/resources/input
you probably should use
System.getProperty("user.dir") + File.separator + "src/test/resources/input"
because the system-property "user.dir" points to the projectlocation.
If you just use "src/test/resources/input" you will get your mentioned exception, because the File Object don't "start" at the project location. So you have to specify it manually.
Nevertheless it's better to use the getResource-Method to retrieve different resources within your project, because if you run your project with the jar-File you need to tweak around to get "user.dir" to work correctly.
Just a basic example for the Classloader:
ClassLoader.getSystemClassLoader().getResource("test/resources/input");
This returns an URL-Object, with this object you can get the File-Object using ...
URL filePath = ClassLoader.getSystemClassLoader().getResource("test/resources/input");
File file = new File( filePath.toURI() );
Remove the src and try it again .
I'm uploading images using Spring and Hibernate. I'm saving images on the server as follows.
File savedFile = new File("E:/Project/SpringHibernet/MultiplexTicketBooking/web/images/" + itemName);
item.write(savedFile);
Where itemName is the image file name after parsing the request (enctype="multipart/form-data"). I however need to mention the relative path in the constructor of File. Something like the one shown below.
File savedFile = new File("MultiplexTicketBooking/web/images/" + itemName);
item.write(savedFile);
But it doesn't work throwing the FileNotFoundException. Is there a way to specify a relative path with File in Java?
Try printing the working directory from your program.
String curDir = System.getProperty("user.dir");
Gets you that directory. Then check if the directories MultiplexTicketBooking/web/images/ exist in that directory.
Can't count the number of times I've been mistaken about my current dir and spent some time looking for a file I wrote to...
It seems the server should offer functionality as might be seen in the methods getContextPath() or getRealPath(String). It would be common to build paths based on those types of server related and reproducible paths. Do not use something like user.dir which makes almost no sense in a server.
Update
ServletContext sc=request.getSession().getServletContext();
File savedFile = new File(sc.getRealPath("images")+"\\" + itemName);
Rather than use "\\" I'd tend to replace that with the following which will cause the correct file separator to be used for each platform. Retain cross-platform compatibility for when the client decides to swap the MS/ISS based server out for a Linux/Tomcat stack. ;)
File savedFile = new File(sc.getRealPath("images"), itemName); //note the ','
See File(String,String) for details.
You could get the path of your project using the following -
File file = new File("");
System.out.println("" + file.getAbsolutePath());
So you could have a constants or a properties file where you could define your path which is MultiplexTicketBooking/web/images/ after the relative path.
You could append your path with the path you get from file.getAbsolutePath() and that will be the real path of the file. - file.getAbsolutePath() + MultiplexTicketBooking/web/images/.
Make sure the folders after the Project path i.e. MultiplexTicketBooking/web/images/ exist.
You can specify the path both absolute and relative with File. The FileNotFoundException can be thrown because the folder might be there. Try using the mkdirs() method first in to create the folder structure you need in order to save your file where you're trying to save it.
If a file exists in the same directory where a Java application is running and I create a File object for that file the Java File methods for the path of the file include the filename as well. Code and output are below.
If this was a bug in the JDK version I'm using someone would surely have seen it by now.
Why do File.getAbsolutePath() and File.getCanonicalPath() include the file name? The Javadocs indicate that the directory name should be returned.
import java.io.File;
import java.io.IOException;
public class DirectoryFromFile {
private void getDirectoryOfFile(String fileName) throws IOException{
File f = new File(fileName );
System.out.println("exists(): " + f.exists());
System.out.println("getPath(): " + f.getPath());
System.out.println("getAbsolutePath(): " + f.getAbsolutePath());
System.out.println("getParent(): " + f.getParent() );
System.out.println("getCanonicalPath(): " + f.getCanonicalPath() );
System.out.println("getAbsoluteFile().getCanonicalPath(): " + f.getAbsoluteFile().getCanonicalPath() );
String dirname = f.getCanonicalPath();
System.out.println("dirname: " + dirname);
File dir = new File(dirname);
System.out.println("dir: " + dir.getAbsolutePath());
if (dirname.endsWith(fileName))
dirname = dirname.substring(0, dirname.length() - fileName.length());
System.out.println("dirname: " + dirname);
File dir2 = new File(dirname);
System.out.println("dir2: " + dir2.getAbsolutePath());
}
public static void main(String[] args) {
DirectoryFromFile dff = new DirectoryFromFile();
try {
dff.getDirectoryOfFile("test.txt");
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
Here' the output:
exists(): true
getPath(): test.txt
getAbsolutePath(): C:\dean\src\java\directorytest\directory.from.file\test.txt
getParent(): null
getCanonicalPath(): C:\dean\src\java\directorytest\directory.from.file\test.txt
getAbsoluteFile().getCanonicalPath(): C:\dean\src\java\directorytest\directory.from.file\test.txt
dirname: C:\dean\src\java\directorytest\directory.from.file\test.txt
dir: C:\dean\src\java\directorytest\directory.from.file\test.txt
dirname: C:\dean\src\java\directorytest\directory.from.file\
dir2: C:\dean\src\java\directorytest\directory.from.file
So far the only way I've found to get the directory in this case is to manually parse off the file name.
Does the File class have a way to get the directory name in this case (where a File that exists in the current directory is created without specifying a directory)?
Why do File.getAbsolutePath() and File.getCanonicalPath() include the
file name? The Javadocs indicate that the directory name should be
returned.
No, they don't. If you'd care to point out why you think they do, someone can probably identify the mistake in your reasoning. Also, if you specify exactly what you'd like to see for output given some particular input, we can help you out there, too. Your question title seems strange, too, since your problem seems to be that it is returning the full path to a file.
Edit: I think I understand the source of your confusion. A File represents a file system path in a platform-agnostic way. It can be a path to a file or to a directory. It also always represents the same path, though not necessarily the same absolute path. This is a very fine distinction but a very important one. A File object representing a relative path is always relative. Given a File representing a relative path, you can get the current corresponding absolute path using getAbsolutePath(). This doesn't, however, alter the fact that the File represents a relative path. Further invocations of getAbsolutePath() on the same File object may return different values. Consider, for example:
// A relative file
File foo = new File("foo.txt");
// Resolve relative file against CWD
System.out.println(foo.getAbsolutePath());
// Output: D:\dev\projects\testbed\foo.txt
System.setProperty("user.dir", "C:\\somewhere");
// Resolve relative file against new CWD
System.out.println(foo.getAbsolutePath());
// Output: C:\somewhere\foo.txt
// Get an absolute file
File absoluteFoo = foo.getAbsoluteFile();
// Show absolute path
System.out.println(absoluteFoo.getAbsolutePath());
// Output: C:\somewhere\foo.txt
System.setProperty("user.dir", "D:\\somewhere-else");
// An absolute path doesn't change when the CWD changes
System.out.println(absoluteFoo.getAbsolutePath());
// Output: C:\somewhere\foo.txt
It should be clear now that the path a File represents is only that: a path. Further, a path can be composed of zero or more parts, and calling getParent() on any File gives back the path of that File with the last path element removed unless there isn't a "last path element" to remove. Thus the expected result of new File("foo").getParent() is null since the relative path "foo" has no parent.
From the example and explanation above, you should be able to see that the way to get the containing directory when you've created relative-path File object is with
String absoluteParentDirPath = someRelativeFile.getAbsoluteFile().getParent();
with the caveat that the "absolute path" depends on your environment at the time.
Additional note: Since File is Serializable, you could write a relative-path file to disk or send it across a network. That File, when deserialized in another JVM, will still represent a relative path and will be resolved against whatever the current working directory of that JVM happens to be.
The behaviour is expected. The documentation does not mention that the filename is not included.
Perhaps you are confused by the difference between getAbsolutePath() and getAbsoluteFile(). It's that the latter returns a File instance.
I'm not sure why you think the Javadoc says that it returns the directory name.
Here is the Javadoc --
An abstract representation of file and directory pathnames.
User interfaces and operating systems use system-dependent pathname strings to name files and directories. This class presents an abstract, system-independent view of hierarchical pathnames. An abstract pathname has two components:
An optional system-dependent prefix string, such as a disk-drive specifier, "/" for the UNIX root directory, or "\\" for a Microsoft Windows UNC pathname, and
A sequence of zero or more string names.
The first name in an abstract pathname may be a directory name or, in the case of Microsoft Windows UNC pathnames, a hostname. Each subsequent name in an abstract pathname denotes a directory; the last name may denote either a directory or a file. The empty abstract pathname has no prefix and an empty name sequence.
http://download.oracle.com/javase/6/docs/api/java/io/File.html#getAbsolutePath%28%29
Returns the absolute pathname string of this abstract pathname.
In addition to the existing answers with regards to getAbsolutePath and getCanonicalPath, please also note, that File.getParent() does not mean "parent directory" it merely refers to the parent file object that was used to create the file.
For example, if the file object can be created as such:
File dir = new File("/path/to/a/directory");
File f1 = new File(dir, "x.txt");
File f2 = new File(dir, "../another/y.txt");
File f3 = new File("z.txt");
f1 would refer to /path/to/a/directory/x.txt, it's parent is dir (/path/to/a/directory)
f2 would refer to /path/to/a/directory/../another/y.txt, it's canonical path would be /path/to/a/another/y.txt, but it's parent is still the reference to dir (/path/to/a/directory)
f3 would refer to z.txt in the current directory. It does not have a parent file object, so f3.getParent() or f3.getParentFile() would return null.
path is the full path
if you only want the directory you need to call file.getParent()