Develop Reference

How do I tell Java to open a file using the system's default text editor, if it doesn't have a known extension?

I have a file named error.log.1.
I would like to tell Java to use the system editor to open this file.
If it were named error.log, then the following would work:
Desktop.getDesktop().edit(new File("error.log") );
However, since it is not a recognized file extension, it doesn't open. Instead I get an error:
Exception in thread "main" java.io.IOException: Failed to open error.log.1.
Error message: No application is associated with the specified file for this operation.
at sun.awt.windows.WDesktopPeer.ShellExecute(Unknown Source)
at sun.awt.windows.WDesktopPeer.open(Unknown Source)
at java.awt.Desktop.open(Unknown Source)
at net.joshuad.hypnos.workbench.EditorTest.main(EditorTest.java:9)

I am not sure what "System Editor" is, but if it is a specific application you want to open, you'll need to run that application and pass the log filename as an argument.
You will need to determine the path to the application and then you can use the Runtime.getRuntime().exec() method to open the file.
For example, if you wanted to open the log file using Notepad, you could do so like this:
Runtime.getRuntime().exec("C:\\Windows\\System32\\notepad.exe error.log.1");
This, of course, assumes the application accepts a filename as a parameter. You'll need to look into what commandline syntax this System Editor has.

There are two ways to handle this problem:
renaming your file and then open it as you mentioned;
explicitly using an command (the editor) to open the file;
In the second case, it will be something as:
private static void openByCommand(String filePath){
try {
Process process = new ProcessBuilder("gedit", filePath)
.directory(new File("/home/hearen")) // set up your working directory;
.start();
int exitCode = process.waitFor();
System.out.println(exitCode);
} catch (IOException | InterruptedException ignored) {
ignored.printStackTrace();
}
}

As suggested by Zephyr, a good solution would be to specify the path to the default editor (Write some code to be platform specific). Or better yet, you could package your own cross-platform text editor, something like jEdit could be a good option.
However, if you really want to use Desktop.getDesktop().edit(path), then a hack is to simply check the file extension, and if it is unknown then add to ".log" or ".txt" to the end.
A bit like this:
File originalName = new File("path/to/my/file/error.log.1");
File appendedName = new File(originalName.getAbsolutePath()+".log");
boolean success = originalName.renameTo(appendedName);
if (success) {
Desktop.getDesktop().edit(appendedName);
}
//Change it back when you are done:
appendedName.renameTo(originalName);
Obviously this will not work if multiple sources/applications are reading from the original file at the same time (You could make a copy instead of renaming it), but it may fit your use case.

Modify the feet to meters program to read from, write to a file. The input file named “feet.txt” and the output file “meters.txt” [duplicate]

I have an assignment for my CS class where it says to read a file with several test scores and asks me to sum and average them. While summing and averaging is easy, I am having problems with the file reading. The instructor said to use this syntax
Scanner scores = new Scanner(new File("scores.dat"));
However, this throws a FileNotFoundException, but I have checked over and over again to see if the file exists in the current folder, and after that, I figured that it had to do something with the permissions. I changed the permissions for read and write for everyone, but it still did not work and it still keeps throwing the error. Does anyone have any idea why this may be occurring?
EDIT: It was actually pointing to a directory up, however, I have fixed that problem. Now file.exists() returns true, but when I try to put it in the Scanner, it throws the FileNotFoundException
Here is all my code
import java.util.Scanner;
import java.io.*;
public class readInt{
public static void main(String args[]){
File file = new File("lines.txt");
System.out.println(file.exists());
Scanner scan = new Scanner(file);
}
}

There are a number situation where a FileNotFoundException may be thrown at runtime.
The named file does not exist. This could be for a number of reasons including:
The pathname is simply wrong
The pathname looks correct but is actually wrong because it contains non-printing characters (or homoglyphs) that you did not notice
The pathname is relative, and it doesn't resolve correctly relative to the actual current directory of the running application. This typically happens because the application's current directory is not what you are expecting or assuming.
The path to the file is is broken; e.g. a directory name of the path is incorrect, a symbolic link on the path is broken, or there is a permission problem with one of the path components.
The named file is actually a directory.
The named file cannot be opened for reading for some reason.
The good news that, the problem will inevitably be one of the above. It is just a matter of working out which. Here are some things that you can try:
Calling file.exists() will tell you if any file system object exists with the given name / pathname.
Calling file.isDirectory() will test if it is a directory.
Calling file.canRead() will test if it is a readable file.
This line will tell you what the current directory is:
System.out.println(new File(".").getAbsolutePath());
This line will print out the pathname in a way that makes it easier to spot things like unexpected leading or trailing whitespace:
System.out.println("The path is '" + path + "'");
Look for unexpected spaces, line breaks, etc in the output.
It turns out that your example code has a compilation error.
I ran your code without taking care of the complaint from Netbeans, only to get the following exception message:
Exception in thread "main" java.lang.RuntimeException: Uncompilable
source code - unreported exception java.io.FileNotFoundException; must
be caught or declared to be thrown
If you change your code to the following, it will fix that problem.
public static void main(String[] args) throws FileNotFoundException {
File file = new File("scores.dat");
System.out.println(file.exists());
Scanner scan = new Scanner(file);
}
Explanation: the Scanner(File) constructor is declared as throwing the FileNotFoundException exception. (It happens the scanner it cannot open the file.) Now FileNotFoundException is a checked exception. That means that a method in which the exception may be thrown must either catch the exception or declare it in the throws clause. The above fix takes the latter approach.

The code itself is working correctly. The problem is, that the program working path is pointing to other place than you think.
Use this line and see where the path is:
System.out.println(new File(".").getAbsoluteFile());

Obviously there are a number of possible causes and the previous answers document them well, but here's how I solved this for in one particular case:
A student of mine had this problem and I nearly tore my hair out trying to figure it out. It turned out that the file didn't exist, even though it looked like it did. The problem was that Windows 7 was configured to "Hide file extensions for known file types." This means that if file appears to have the name "data.txt" its actual filename is "data.txt.txt".
Hope this helps others save themselves some hair.

I recently found interesting case that produces FileNotFoundExeption when file is obviously exists on the disk.
In my program I read file path from another text file and create File object:
//String path was read from file
System.out.println(path); //file with exactly same visible path exists on disk
File file = new File(path);
System.out.println(file.exists()); //false
System.out.println(file.canRead()); //false
FileInputStream fis = new FileInputStream(file); // FileNotFoundExeption
The cause of the problem was that the path contained invisible \r\n characters at the end.
The fix in my case was:
File file = new File(path.trim());
To generalize a bit, the invisible / non-printing characters could have include space or tab characters, and possibly others, and they could have appeared at the beginning of the path, at the end, or embedded in the path. Trim will work in some cases but not all. There are a couple of things that you can help to spot this kind of problem:
Output the pathname with quote characters around it; e.g.
System.out.println("Check me! '" + path + "'");
and carefully check the output for spaces and line breaks where they shouldn't be.
Use a Java debugger to carefully examine the pathname string, character by character, looking for characters that shouldn't be there. (Also check for homoglyph characters!)

An easy fix, which worked for me, is moving my files out of src and into the main folder of the project. It's not the best solution, but depending on the magnitude of the project and your time, it might be just perfect.

Reading and writing from and to a file can be blocked by your OS depending on the file's permission attributes.
If you are trying to read from the file, then I recommend using File's setReadable method to set it to true, or, this code for instance:
String arbitrary_path = "C:/Users/Username/Blah.txt";
byte[] data_of_file;
File f = new File(arbitrary_path);
f.setReadable(true);
data_of_file = Files.readAllBytes(f);
f.setReadable(false); // do this if you want to prevent un-knowledgeable
//programmers from accessing your file.
If you are trying to write to the file, then I recommend using File's setWritable method to set it to true, or, this code for instance:
String arbitrary_path = "C:/Users/Username/Blah.txt";
byte[] data_of_file = { (byte) 0x00, (byte) 0xFF, (byte) 0xEE };
File f = new File(arbitrary_path);
f.setWritable(true);
Files.write(f, byte_array);
f.setWritable(false); // do this if you want to prevent un-knowledgeable
//programmers from changing your file (for security.)

Apart from all the other answers mentioned here, you can do one thing which worked for me.
If you are reading the path through Scanner or through command line args, instead of copy pasting the path directly from Windows Explorer just manually type in the path.
It worked for me, hope it helps someone :)

I had this same error and solved it simply by adding the src directory that is found in Java project structure.
String path = System.getProperty("user.dir") + "\\src\\package_name\\file_name";
File file = new File(path);
Scanner scanner = new Scanner(file);
Notice that System.getProperty("user.dir") and new File(".").getAbsolutePath() return your project root directory path, so you have to add the path to your subdirectories and packages

You'd obviously figure it out after a while but just posting this so that it might help someone. This could also happen when your file path contains any whitespace appended or prepended to it.

Use single forward slash and always type the path manually. For example:
FileInputStream fi= new FileInputStream("D:/excelfiles/myxcel.xlsx");

What worked for me was catching the exception. Without it the compiler complains even if the file exists.
InputStream file = new FileInputStream("filename");
changed to
try{
InputStream file = new FileInputStream("filename");
System.out.println(file.available());
}
catch (Exception e){
System.out.println(e);
}

This works for me. It also can read files such txt, csv and .in
public class NewReader {
public void read() throws FileNotFoundException, URISyntaxException {
File file = new File(Objects.requireNonNull(NewReader.class.getResource("/test.txt")).toURI());
Scanner sc = new Scanner(file);
while (sc.hasNext()) {
String text = sc.next();
System.out.println(text);
}
}
}
the file is located in resource folder generated by maven. If you have other folders nested in, just add it to the file name like "examples/test.txt".

Output of Reading the file is not getting as Expected? [duplicate]

I have an assignment for my CS class where it says to read a file with several test scores and asks me to sum and average them. While summing and averaging is easy, I am having problems with the file reading. The instructor said to use this syntax
Scanner scores = new Scanner(new File("scores.dat"));
However, this throws a FileNotFoundException, but I have checked over and over again to see if the file exists in the current folder, and after that, I figured that it had to do something with the permissions. I changed the permissions for read and write for everyone, but it still did not work and it still keeps throwing the error. Does anyone have any idea why this may be occurring?
EDIT: It was actually pointing to a directory up, however, I have fixed that problem. Now file.exists() returns true, but when I try to put it in the Scanner, it throws the FileNotFoundException
Here is all my code
import java.util.Scanner;
import java.io.*;
public class readInt{
public static void main(String args[]){
File file = new File("lines.txt");
System.out.println(file.exists());
Scanner scan = new Scanner(file);
}
}

There are a number situation where a FileNotFoundException may be thrown at runtime.
The named file does not exist. This could be for a number of reasons including:
The pathname is simply wrong
The pathname looks correct but is actually wrong because it contains non-printing characters (or homoglyphs) that you did not notice
The pathname is relative, and it doesn't resolve correctly relative to the actual current directory of the running application. This typically happens because the application's current directory is not what you are expecting or assuming.
The path to the file is is broken; e.g. a directory name of the path is incorrect, a symbolic link on the path is broken, or there is a permission problem with one of the path components.
The named file is actually a directory.
The named file cannot be opened for reading for some reason.
The good news that, the problem will inevitably be one of the above. It is just a matter of working out which. Here are some things that you can try:
Calling file.exists() will tell you if any file system object exists with the given name / pathname.
Calling file.isDirectory() will test if it is a directory.
Calling file.canRead() will test if it is a readable file.
This line will tell you what the current directory is:
System.out.println(new File(".").getAbsolutePath());
This line will print out the pathname in a way that makes it easier to spot things like unexpected leading or trailing whitespace:
System.out.println("The path is '" + path + "'");
Look for unexpected spaces, line breaks, etc in the output.
It turns out that your example code has a compilation error.
I ran your code without taking care of the complaint from Netbeans, only to get the following exception message:
Exception in thread "main" java.lang.RuntimeException: Uncompilable
source code - unreported exception java.io.FileNotFoundException; must
be caught or declared to be thrown
If you change your code to the following, it will fix that problem.
public static void main(String[] args) throws FileNotFoundException {
File file = new File("scores.dat");
System.out.println(file.exists());
Scanner scan = new Scanner(file);
}
Explanation: the Scanner(File) constructor is declared as throwing the FileNotFoundException exception. (It happens the scanner it cannot open the file.) Now FileNotFoundException is a checked exception. That means that a method in which the exception may be thrown must either catch the exception or declare it in the throws clause. The above fix takes the latter approach.

The code itself is working correctly. The problem is, that the program working path is pointing to other place than you think.
Use this line and see where the path is:
System.out.println(new File(".").getAbsoluteFile());

Obviously there are a number of possible causes and the previous answers document them well, but here's how I solved this for in one particular case:
A student of mine had this problem and I nearly tore my hair out trying to figure it out. It turned out that the file didn't exist, even though it looked like it did. The problem was that Windows 7 was configured to "Hide file extensions for known file types." This means that if file appears to have the name "data.txt" its actual filename is "data.txt.txt".
Hope this helps others save themselves some hair.

I recently found interesting case that produces FileNotFoundExeption when file is obviously exists on the disk.
In my program I read file path from another text file and create File object:
//String path was read from file
System.out.println(path); //file with exactly same visible path exists on disk
File file = new File(path);
System.out.println(file.exists()); //false
System.out.println(file.canRead()); //false
FileInputStream fis = new FileInputStream(file); // FileNotFoundExeption
The cause of the problem was that the path contained invisible \r\n characters at the end.
The fix in my case was:
File file = new File(path.trim());
To generalize a bit, the invisible / non-printing characters could have include space or tab characters, and possibly others, and they could have appeared at the beginning of the path, at the end, or embedded in the path. Trim will work in some cases but not all. There are a couple of things that you can help to spot this kind of problem:
Output the pathname with quote characters around it; e.g.
System.out.println("Check me! '" + path + "'");
and carefully check the output for spaces and line breaks where they shouldn't be.
Use a Java debugger to carefully examine the pathname string, character by character, looking for characters that shouldn't be there. (Also check for homoglyph characters!)

An easy fix, which worked for me, is moving my files out of src and into the main folder of the project. It's not the best solution, but depending on the magnitude of the project and your time, it might be just perfect.

Reading and writing from and to a file can be blocked by your OS depending on the file's permission attributes.
If you are trying to read from the file, then I recommend using File's setReadable method to set it to true, or, this code for instance:
String arbitrary_path = "C:/Users/Username/Blah.txt";
byte[] data_of_file;
File f = new File(arbitrary_path);
f.setReadable(true);
data_of_file = Files.readAllBytes(f);
f.setReadable(false); // do this if you want to prevent un-knowledgeable
//programmers from accessing your file.
If you are trying to write to the file, then I recommend using File's setWritable method to set it to true, or, this code for instance:
String arbitrary_path = "C:/Users/Username/Blah.txt";
byte[] data_of_file = { (byte) 0x00, (byte) 0xFF, (byte) 0xEE };
File f = new File(arbitrary_path);
f.setWritable(true);
Files.write(f, byte_array);
f.setWritable(false); // do this if you want to prevent un-knowledgeable
//programmers from changing your file (for security.)

Apart from all the other answers mentioned here, you can do one thing which worked for me.
If you are reading the path through Scanner or through command line args, instead of copy pasting the path directly from Windows Explorer just manually type in the path.
It worked for me, hope it helps someone :)

I had this same error and solved it simply by adding the src directory that is found in Java project structure.
String path = System.getProperty("user.dir") + "\\src\\package_name\\file_name";
File file = new File(path);
Scanner scanner = new Scanner(file);
Notice that System.getProperty("user.dir") and new File(".").getAbsolutePath() return your project root directory path, so you have to add the path to your subdirectories and packages

You'd obviously figure it out after a while but just posting this so that it might help someone. This could also happen when your file path contains any whitespace appended or prepended to it.

Use single forward slash and always type the path manually. For example:
FileInputStream fi= new FileInputStream("D:/excelfiles/myxcel.xlsx");

What worked for me was catching the exception. Without it the compiler complains even if the file exists.
InputStream file = new FileInputStream("filename");
changed to
try{
InputStream file = new FileInputStream("filename");
System.out.println(file.available());
}
catch (Exception e){
System.out.println(e);
}

This works for me. It also can read files such txt, csv and .in
public class NewReader {
public void read() throws FileNotFoundException, URISyntaxException {
File file = new File(Objects.requireNonNull(NewReader.class.getResource("/test.txt")).toURI());
Scanner sc = new Scanner(file);
while (sc.hasNext()) {
String text = sc.next();
System.out.println(text);
}
}
}
the file is located in resource folder generated by maven. If you have other folders nested in, just add it to the file name like "examples/test.txt".

Can Files.createFile() have asynchronous behaviour?

I am getting a really strange behavior on a test server using Files.createFile().
With code looking like that :
Files.createFile(myPath);
... code adding that file to a zip file ...
The file created at myPath is SOMETIMES not included in the zip content, most often it is not included, no Exception thrown. I could never reproduce the issue outside of the test server which runs CentOS release 6.6 (Final) and has an ext4 file system. Performing a Files.exists(myPath) after createFile always returns true.
I tried writing a few characters to the file to check if it makes a difference, but it does not.
FileUtils.writeStringToFile(testLog.toFile(), "Test content");
If I add a short sleep in-between, the file is then included in the zip consistently.
Consistently working :
Files.createFile(myPath);
try {
Thread.sleep(1000);
} catch (InterruptedException e) {
//interrupted
}
... code adding that file to a zip file ...
My view on this is that there is some strange asynchronous behavior going on in the file system as a Thread.sleep() should never make any difference in this code.
Does anyone have a definite explanation on how that sleep can make a difference ?
EDIT : My problem here is really not about the zipping method but about the createFile method behaving asynchronously while it should not.
Thanks for your help !

Why don't you just use the zip filesystem provider?
final Path zipPath = Paths.get("pathToZipFileHere");
final URI uri = URI.create("jar:" + zipPath.toUri());
final Path fileInZip;
// Open the zip as a filesystem
try (
final FileSystem fs = FileSystems.newFileSystem(uri,
Collections.emptyMap());
) {
fileInZip = fs.getPath("path/in/zip");
// work with fileInZip
}

What are possible reasons for java.io.IOException: "The filename, directory name, or volume label syntax is incorrect"

I am trying to copy a file using the following code:
File targetFile = new File(targetPath + File.separator + filename);
...
targetFile.createNewFile();
fileInputStream = new FileInputStream(fileToCopy);
fileOutputStream = new FileOutputStream(targetFile);
byte[] buffer = new byte[64*1024];
int i = 0;
while((i = fileInputStream.read(buffer)) != -1) {
fileOutputStream.write(buffer, 0, i);
}
For some users the targetFile.createNewFile results in this exception:
java.io.IOException: The filename, directory name, or volume label syntax is incorrect
at java.io.WinNTFileSystem.createFileExclusively(Native Method)
at java.io.File.createNewFile(File.java:850)
Filename and directory name seem to be correct. The directory targetPath is even checked for existence before the copy code is executed and the filename looks like this: AB_timestamp.xml
The user has write permissions to the targetPath and can copy the file without problems using the OS.
As I don't have access to a machine this happens on yet and can't reproduce the problem on my own machine I turn to you for hints on the reason for this exception.

This can occur when filename has timestamp with colons, eg. myfile_HH:mm:ss.csv Removing colons fixed the issue.

Try this, as it takes more care of adjusting directory separator characters in the path between targetPath and filename:
File targetFile = new File(targetPath, filename);

I just encountered the same problem. I think it has to something do with write access permission. I got the error while trying to write to c:\ but on changing to D:\ everything worked fine.
Apparently Java did not have permission to write to my System Drive (Running Windows 7 installed on C:)

Here is the test program I use
import java.io.File;
public class TestWrite {
public static void main(String[] args) {
if (args.length!=1) {
throw new IllegalArgumentException("Expected 1 argument: dir for tmp file");
}
try {
File.createTempFile("bla",".tmp",new File(args[0]));
} catch (Exception e) {
System.out.println("exception:"+e);
e.printStackTrace();
}
}
}

Try to create the file in a different directory - e.g. "C:\" after you made sure you have write access to that directory. If that works, the path name of the file is wrong.
Take a look at the comment in the Exception and try to vary all the elements in the path name of the file. Experiment. Draw conclusions.

Remove any special characters in the file/folder name in the complete path.

Do you check that the targetPath is a directory, or just that something exists with that name? (I know you say the user can copy it from the operating system, but maybe they're typing something else).
Does targetPath end with a File.separator already?
(It would help if you could log and tell us what the value of targetPath and filename are on a failing case)

Maybe the problem is that it is copying the file over the network, to a shared drive? I think java can have problems when writing files using NFS when the path is something like \mypc\myshared folder.
What is the path where this problem happens?

Try adding some logging to see exactly what is the name and path the file is trying to create, to ensure that the parent is well a directory.
In addition, you can also take a look at Channels instead of using a loop. ;-)

You say "for some users" - so it works for others? What is the difference here, are the users running different instances on different machines, or is this a server that services concurrent users?
If the latter, I'd say it is a concurrency bug somehow - two threads check try to create the file with WinNTFileSystem.createFileExclusively(Native Method) simultaniously.
Neither createNewFile or createFileExclusively are synchronized when I look at the OpenJDK source, so you may have to synchronize this block yourself.

Maybe the file already exists. It could be the case if your timestamp resolution is not good enough. As it is an IOException that you are getting, it might not be a permission issue (in which case you would get a SecurityException).
I would first check for file existence before trying to create the file and try to log what's happening.
Look at public boolean createNewFile() for more information on the method you are using.

As I was not able to reproduce the error on my own machine or get hands on the machine of the user where the code failed I waited until now to declare an accepted answer.
I changed the code to the following:
File parentFolder = new File(targetPath);
... do some checks on parentFolder here ...
File targetFile = new File(parentFolder, filename);
targetFile.createNewFile();
fileInputStream = new FileInputStream(fileToCopy);
fileOutputStream = new FileOutputStream(targetFile);
byte[] buffer = new byte[64*1024];
int i = 0;
while((i = fileInputStream.read(buffer)) != -1) {
fileOutputStream.write(buffer, 0, i);
}
After that it worked for the user reporting the problem.
So it seems Alexanders answer did the trick - although I actually use a slightly different constructor than he gave, but along the same lines.
I yet have to talk that user into helping me verifying that the code change fixed the error (instead of him doing something differently) by running the old version again and checking if it still fails.
btw. logging was in place and the logged path seemed ok - sorry for not mentioning that. I took that for granted and found it unnecessarily complicated the code in the question.
Thanks for the helpful answers.

A very similar error:-
" ... java.io.IOException: The filename, directory name, or volume label syntax is incorrect"
was generated in Eclipse for me when the TOMCAT home setting had a training backslash.
The minor edit suggested at:-
http://www.coderanch.com/t/556633/Tomcat/java-io-IOException-filename-directory
fixed it for me.

FileUtils.copyFile(src,new File("C:\\Users\\daiva\\eclipse-workspace\\PracticeProgram\\Screenshot\\adi.png"));
Try to copy file like this.