How can I check whether a file exists, before opening it for reading in Java (the equivalent of Perl's -e $filename)?
The only similar question on SO deals with writing the file and was thus answered using FileWriter which is obviously not applicable here.
If possible I'd prefer a real API call returning true/false as opposed to some "Call API to open a file and catch when it throws an exception which you check for 'no file' in the text", but I can live with the latter.
Using java.io.File:
File f = new File(filePathString);
if(f.exists() && !f.isDirectory()) {
// do something
}
I would recommend using isFile() instead of exists(). Most of the time you are looking to check if the path points to a file not only that it exists. Remember that exists() will return true if your path points to a directory.
new File("path/to/file.txt").isFile();
new File("C:/").exists() will return true but will not allow you to open and read from it as a file.
By using nio in Java SE 7,
import java.nio.file.*;
Path path = Paths.get(filePathString);
if (Files.exists(path)) {
// file exist
}
if (Files.notExists(path)) {
// file is not exist
}
If both exists and notExists return false, the existence of the file cannot be verified. (maybe no access right to this path)
You can check if path is a directory or regular file.
if (Files.isDirectory(path)) {
// path is directory
}
if (Files.isRegularFile(path)) {
// path is regular file
}
Please check this Java SE 7 tutorial.
Using Java 8:
if(Files.exists(Paths.get(filePathString))) {
// do something
}
File f = new File(filePathString);
This will not create a physical file. Will just create an object of the class File. To physically create a file you have to explicitly create it:
f.createNewFile();
So f.exists() can be used to check whether such a file exists or not.
f.isFile() && f.canRead()
There are multiple ways to achieve this.
In case of just for existence. It could be file or a directory.
new File("/path/to/file").exists();
Check for file
File f = new File("/path/to/file");
if(f.exists() && f.isFile()) {}
Check for Directory.
File f = new File("/path/to/file");
if(f.exists() && f.isDirectory()) {}
Java 7 way.
Path path = Paths.get("/path/to/file");
Files.exists(path) // Existence
Files.isDirectory(path) // is Directory
Files.isRegularFile(path) // Regular file
Files.isSymbolicLink(path) // Symbolic Link
Don't. Just catch the FileNotFoundException. The file system has to test whether the file exists anyway. There is no point in doing all that twice, and several reasons not to, such as:
double the code
the timing window problem whereby the file might exist when you test but not when you open, or vice versa, and
the fact that, as the existence of this question shows, you might make the wrong test and get the wrong answer.
Don't try to second-guess the system. It knows. And don't try to predict the future. In general the best way to test whether any resource is available is just to try to use it.
You can use the following: File.exists()
first hit for "java file exists" on google:
import java.io.*;
public class FileTest {
public static void main(String args[]) {
File f = new File(args[0]);
System.out.println(f + (f.exists()? " is found " : " is missing "));
}
}
For me a combination of the accepted answer by Sean A.O. Harney and the resulting comment by Cort3z seems to be the best solution.
Used the following snippet:
File f = new File(filePathString);
if(f.exists() && f.isFile()) {
//do something ...
}
Hope this could help someone.
I know I'm a bit late in this thread. However, here is my answer, valid since Java 7 and up.
The following snippet
if(Files.isRegularFile(Paths.get(pathToFile))) {
// do something
}
is perfectly satifactory, because method isRegularFile returns false if file does not exist. Therefore, no need to check if Files.exists(...).
Note that other parameters are options indicating how links should be handled. By default, symbolic links are followed.
From Java Oracle documentation
It's also well worth getting familiar with Commons FileUtils https://commons.apache.org/proper/commons-io/javadocs/api-2.5/org/apache/commons/io/FileUtils.html
This has additional methods for managing files and often better than JDK.
Simple example with good coding practices and covering all cases :
private static void fetchIndexSafely(String url) throws FileAlreadyExistsException {
File f = new File(Constants.RFC_INDEX_LOCAL_NAME);
if (f.exists()) {
throw new FileAlreadyExistsException(f.getAbsolutePath());
} else {
try {
URL u = new URL(url);
FileUtils.copyURLToFile(u, f);
} catch (MalformedURLException ex) {
Logger.getLogger(RfcFetcher.class.getName()).log(Level.SEVERE, null, ex);
} catch (IOException ex) {
Logger.getLogger(RfcFetcher.class.getName()).log(Level.SEVERE, null, ex);
}
}
}
Reference and more examples at
https://zgrepcode.com/examples/java/java/nio/file/filealreadyexistsexception-implementations
Don't use File constructor with String.
This may not work!
Instead of this use URI:
File f = new File(new URI("file:///"+filePathString.replace('\\', '/')));
if(f.exists() && !f.isDirectory()) {
// to do
}
You can make it this way
import java.nio.file.Paths;
String file = "myfile.sss";
if(Paths.get(file).toFile().isFile()){
//...do somethinh
}
There is specific purpose to design these methods. We can't say use anyone to check file exist or not.
isFile(): Tests whether the file denoted by this abstract pathname is a normal file.
exists(): Tests whether the file or directory denoted by this abstract pathname exists.
docs.oracle.com
You must use the file class , create a file instance with the path of the file you want to check if existent . After that you must make sure that it is a file and not a directory . Afterwards you can call exist method on that file object referancing your file . Be aware that , file class in java is not representing a file . It actually represents a directory path or a file path , and the abstract path it represents does not have to exist physically on your computer . It is just a representation , that`s why , you can enter a path of a file as an argument while creating file object , and then check if that folder in that path does really exist , with the exists() method .
If spring framework is used and the file path starts with classpath:
public static boolean fileExists(String sFileName) {
if (sFileName.startsWith("classpath:")) {
String path = sFileName.substring("classpath:".length());
ClassLoader cl = ClassUtils.getDefaultClassLoader();
URL url = cl != null ? cl.getResource(path) : ClassLoader.getSystemResource(path);
return (url != null);
} else {
Path path = Paths.get(sFileName);
return Files.exists(path);
}
}
Related
can someone help me to find Canonical Path like "C:\Agent17-01\_work\50\s\Test Files".
I have created below method to find Canonical path but it gave the path
"C:\Agent17-01\_work\50\Test Files".
\s\ is missing in my method. below is my method code
public static String getRelativePath(String FileName)
{
String path =
null;
File f = new File("..\\..\\Test Files\\" + FileName);
try
{
path = f.getCanonicalPath();
System.out.println(path);
}
catch (Exception e)
{
}
return path;
}
any help is appreciated
Thanks in advance.
Java accepts also "/" chars as directory dividers. Why is it important? - OS independence. My recomendation is to put the needed resource files into the src/main/resources folder, and then:
getClass().getClassLoader().getResource(fileName).getFile();
this code would return you File instance, and you would be sure that it won't relies on absolute path. If you need just File path, and you are not gonna perform IO operations, use the method you've used (getCannonicalPath).
BTW: Please use Java code style while writing in Java: (brackets, parameters naming). I would recomend also to read about Try with Resources construction.
Program should check if the directory is exist?? And if not, tell the user that there is no such folder.
I found many examples in which is explained how to check whether there is a file, but I need to know whether there is a directory? All methods
boolean x = context.getExternalFilesDir("/nicknameOfUser/").exists();
Toast.makeText(context, "ExternalFilesDir : " + x, Toast.LENGTH_SHORT).show();
isAbsolute(), isDirectory(), isFile(), create a new path to the files folder - nicknameOfUser I do not want to they were created, I just need to receive there is a directory or not ... I don't need create new folders...
How to do it? I think it is a question from regular, but I can't understand ...
When i launch app first time - in my filemanager no any file! But after i check .exists(); it create a path to the folder that i need a check... I DON'T NEED IT
To check if there is directory you have to use two conditions
File file = new File(filePath);
boolean isPresent = file.exists() && file.isDirectory();
returns true only if file exists and is directory.
You can ask for isDirectory() as follows :
File f = new File(Environment.getExternalStorageDirectory()+"/nicknameOfUser/");
if(f.isDirectory()) {
}
File file = new File("your path");
file.exists();
But after i check .exists(); it create a path to the folder that i need a check... I DON'T NEED IT
The problem is not the File.exists() call. It is the getExternalFilesDir call that is creating the directory. As the documentation states:
Unlike Environment.getExternalStoragePublicDirectory(), the directory returned here will be automatically created for you.
Eventually i found a solution. I begin to think it is impossible but all is ok
String fileDir = Environment.getExternalStorageDirectory().getAbsoluteFile() + "/nicknameOfUser/";
File file = new File(fileDir);
boolean x = file.exists();
This code work properly
I am trying to understand the way Java resolves relative path in while creating a File object.
OS used: Windows
For the below snippet, I am getting an IOException as it cannot find the path:
#Test
public void testPathConversion() {
File f = new File("test/test.txt");
try {
f.createNewFile();
System.out.println(f.getPath());
System.out.println(f.getAbsolutePath());
System.out.println(f.getCanonicalPath());
} catch (Exception e) {
e.printStackTrace();
}
}
My understanding here is, Java treats the path provided as absolute and returns an error when the path does not exist. So it makes sense.
When I update the above code to use relative path:
#Test
public void testPathConversion() {
File f = new File("test/../test.txt");
try {
f.createNewFile();
System.out.println(f.getPath());
System.out.println(f.getAbsolutePath());
System.out.println(f.getCanonicalPath());
} catch (Exception e) {
e.printStackTrace();
}
}
It creates a new file and provides the below output:
test\..\test.txt
C:\JavaForTesters\test\..\test.txt
C:\JavaForTesters\test.txt
In this case, my assumption is, even though the path provided doesn't exist, because the path contains "/../", java treats this as a relative path and creates the file in the user.dir. So this also makes sense.
But if I update the relative path as below:
#Test
public void testPathConversion() {
File f = new File("test/../../test.txt");
try {
f.createNewFile();
System.out.println(f.getPath());
System.out.println(f.getAbsolutePath());
System.out.println(f.getCanonicalPath());
} catch (Exception e) {
e.printStackTrace();
}
}
Then I get IOException: Access is denied.
My questions are:
why "test/../test.txt" is treated as a relative path and creates the file in "user.dir" but"test/../../test.txt" returns an error? Where does it attempt to create the file for the path "test/../../test.txt"?
When the specified relative path is not found, the file seems to be created in the user.dir. So, it appears to me that the below two scenarios does the same thing:
//scenario 1
File f = new File("test/../test.txt");
f.createNewFile();
//scenario 2
File f = new File("test.txt");
f.createNewFile();
So is there a real world case where one would use scenario 1 instead of scenario 2?
I suppose I am missing something obvious here or have fundamentally misunderstood relative paths. I went through the Java docs for File and I am not able to find an explanation for this. There are quite a few questions posted in Stack Overflow regarding relative paths, but the ones I looked up were for specific scenarios and not exactly about how relative paths are resolved.
It will be great if someone could please explain me how this works or point to some related links?
There is a concept of a working directory.
This directory is represented by a . (dot).
In relative paths, everything else is relative to it.
Simply put the . (the working directory) is where you run your program.
In some cases the working directory can be changed but in general this is
what the dot represents. I think this is C:\JavaForTesters\ in your case.
So test\..\test.txt means: the sub-directory test
in my working directory, then one level up, then the
file test.txt. This is basically the same as just test.txt.
For more details check here.
http://docs.oracle.com/javase/7/docs/api/java/io/File.html
http://docs.oracle.com/javase/tutorial/essential/io/pathOps.html
When your path starts with a root dir i.e. C:\ in windows or / in Unix or in java resources path, it is considered to be an absolute path. Everything else is relative, so
new File("test.txt") is the same as new File("./test.txt")
new File("test/../test.txt") is the same as new File("./test/../test.txt")
The major difference between getAbsolutePath and getCanonicalPath is that the first one concatenates a parent and a child path, so it may contain dots: .. or .. getCanonicalPath will always return the same path for a particular file.
Note: File.equals uses an abstract form of a path (getAbsolutePath) to compare files, so this means that two File objects for the same might not be equal and Files are unsafe to use in collections like Map or Set.
The working directory is a common concept across virtually all operating systems and program languages etc. It's the directory in which your program is running. This is usually (but not always, there are ways to change it) the directory the application is in.
Relative paths are ones that start without a drive specifier. So in linux they don't start with a /, in windows they don't start with a C:\, etc. These always start from your working directory.
Absolute paths are the ones that start with a drive (or machine for network paths) specifier. They always go from the start of that drive.
Relative paths can be best understood if you know how Java runs the program.
There is a concept of working directory when running programs in Java. Assuming you have a class, say, FileHelper that does the IO under
/User/home/Desktop/projectRoot/src/topLevelPackage/.
Depending on the case where you invoke java to run the program, you will have different working directory. If you run your program from within and IDE, it will most probably be projectRoot.
In this case $ projectRoot/src : java topLevelPackage.FileHelper it will be src.
In this case $ projectRoot : java -cp src topLevelPackage.FileHelper it will be projectRoot.
In this case $ /User/home/Desktop : java -cp ./projectRoot/src topLevelPackage.FileHelper it will be Desktop.
(Assuming $ is your command prompt with standard Unix-like FileSystem. Similar correspondence/parallels with Windows system)
So, your relative path root (.) resolves to your working directory. Thus to be better sure of where to write files, it's said to consider below approach.
package topLevelPackage
import java.io.File;
import java.nio.file.Path;
import java.nio.file.Paths;
public class FileHelper {
// Not full implementation, just barebone stub for path
public void createLocalFile() {
// Explicitly get hold of working directory
String workingDir = System.getProperty("user.dir");
Path filePath = Paths.get(workingDir+File.separator+"sampleFile.txt");
// In case we need specific path, traverse that path, rather using . or ..
Path pathToProjectRoot = Paths.get(System.getProperty("user.home"), "Desktop", "projectRoot");
System.out.println(filePath);
System.out.println(pathToProjectRoot);
}
}
Hope this helps.
On windows and Netbeans you can set the relative path as:
new FileReader("src\\PACKAGE_NAME\\FILENAME");
On Linux and Netbeans you can set the relative path as:
new FileReader("src/PACKAGE_NAME/FILENAME");
If you have your code inside Source Packages
I do not know if it is the same for eclipse or other IDE
Only slightly related to the question, but try to wrap your head around this one. So un-intuitive:
import java.nio.file.*;
class Main {
public static void main(String[] args) {
Path p1 = Paths.get("/personal/./photos/./readme.txt");
Path p2 = Paths.get("/personal/index.html");
Path p3 = p1.relativize(p2);
System.out.println(p3); //prints ../../../../index.html !!
}
}
I went off of peter.petrov's answer but let me explain where you make the file edits to change it to a relative path.
Simply edit "AXLAPIService.java" and change
url = new URL("file:C:users..../schema/current/AXLAPI.wsdl");
to
url = new URL("file:./schema/current/AXLAPI.wsdl");
or where ever you want to store it.
You can still work on packaging the wsdl file into the meta-inf folder in the jar but this was the simplest way to get it working for me.
When I pass File file to a method I'm trying to get its full path like file.getAbsolutePath(); I always get the same result no matter which one I use either absolute or canonical path PATH_TO_MY_WORKSPACE/projectName/filename and it is not there, how can I get exact location of the file?
Thank you
DETAILS:
Here is some code and this solutions(its bad but its working):
private static void doSomethingToDirectory(File factDir) throws IOException {
File[] dirContents = factDir.listFiles();
if(factDir.isDirectory() && dirContents.length > 0){
for (int i = 0; i < dirContents.length; i++) {
for (String str : dirContents[i].list()) {
if(str.equals(TEMP_COMPARE_FILE)){
process(new File(dirContents[i].getAbsolutePath() + "\\" + str));
}
}
}
}
}
I'm looping trough directories where factDir is src/main, I'm seeking toBeProcessed.txt files only that is TEMP_COMPARE_FILE value and I'm sending them to process method which reads the file and does processing of it.
If someone could better solution I'd be greatful
This quote from the Javadoc might be helpful:
A pathname, whether abstract or in string form, may be either absolute or relative. An absolute pathname is complete in that no other information is required in order to locate the file that it denotes. A relative pathname, in contrast, must be interpreted in terms of information taken from some other pathname. By default the classes in the java.io package always resolve relative pathnames against the current user directory. This directory is named by the system property user.dir, and is typically the directory in which the Java virtual machine was invoked.
I interpret this so that if you create your File object with new File("filename") where filename is a relative path, that path will not be converted into an absolute path even by a call to file.getAbsolutePath().
Update: now that you posted code, I can think of some ways to improve it:
you could use a FilenameFilter to find the desired files,
note that list and listFiles return null for non-directory objects, so we need an extra check for that,
you could also use listFiles() again in the inner loop, thus avoiding the need to create new File objects with hand-assembled paths. (Btw note that appending \\ manually to the path is not portable; the proper way would be to use File.separator).
The end result is
private static void doSomethingToDirectory(File factDir) throws IOException {
if (factDir.isDirectory()) {
for (File file : factDir.listFiles()) {
if (file.isDirectory()) {
for (File child : file.listFiles(new MyFilter())) {
process(child);
}
}
}
}
}
class MyFilter implements FilenameFilter {
public boolean accept(File dir, String name) {
return name.equals(TEMP_COMPARE_FILE);
}
}
Note that this code mimics the behaviour of your original piece of code as much as I understood it; most notably, it finds the files with the proper name only in the direct subdirectories of factDir, nonrecursively.
I think there is a way it may help you if and only if the file is in the program directory.
first you get the program directory by :
new File(".").getCanonicalPath()
then :
if file is inside a specific directory like folder\\filename
the full path will be
(new File(".").getCanonicalPath() + "\\folder\\filename")
or if file is directly inside the program directory:
the full path will be
(new File(".").getCanonicalPath() + "\\filename")
i wish this answer help you :)
I have a method that writes to a log file. If the file exists it should append to it, if not then I want it to create a new file.
if (!file.exists() && !file.createNewFile()) {
System.err.println("Error with output file: " + outFile
+ "\nCannot create new file.");
continue;
}
I have that to check that a file can be created. file is a java.io.File object. createNewFile is throwing an IOException: No such file or directory. This method has been working perfectly since I wrote it a few weeks ago and has only recently starting doing this although I don't know what I could have changed. I have checked, the directory exists and I have write permissions for it, but then I thought it should just return false if it can't make the file for any reason.
Is there anything that I am missing to get this working?
try to ensure the parent directory exists with:
file.getParentFile().mkdirs()
Perhaps the directory the file is being created in doesn't exist?
normally this is something you changed recently, first off your sample code is if not file exists and not create new file - you are trying to code away something - what is it?
Then, look at a directory listing to see if it actually exists and do a println / toString() on the file object and getMessage() on the exception, as well as print stack trace.
Then, start from zero knowledge again and re factor from the get-go each step you are using to get here. It's probably a duh you stuck in there somewhere while conceptualizing in code ( because it was working ) - you just retrace each step in detail, you will find it.
I think the exception you get is likely the result from the file check of the atomic method file.createNewFile(). The method can't check if the file does exist because some of the parent directories do not exist or you have no permissions to access them. I would suggest this:
if (file.getParentFile() != null && !file.getParentFile().mkDirs()) {
// handle permission problems here
}
// either no parent directories there or we have created missing directories
if (file.createNewFile() || file.isFile()) {
// ready to write your content
} else {
// handle directory here
}
If you take concurrency into account, all these checks are useless because in every case some other thread is able to create, delete or do anything else with your file. In this case you have to use file locks which I would not suggest doing ;)
According to the [java docs](http://java.sun.com/j2se/1.5.0/docs/api/java/io/File.html#createNewFile() ) createNewFile will create a new file atomically for you.
Atomically creates a new, empty file named by this abstract pathname if and only if a file with this name does not yet exist.
Given that createNewFile is atomic and won't over-write an existing file you can re-write your code as
try {
if(!file.createNewFile()) {
System.out.println("File already exists");
}
} catch (IOException ex) {
System.out.println(ex);
}
This may make any potential threading issues, race-conditions, etc, easier to spot.
You are certainly getting this Exception
'The system cannot find the path specified'
Just print 'file.getAbsoluteFile()' , this will let you know what is the file you wanted to create.
This exception will occur if the Directory where you are creating the file doesn't exist.
//Create New File if not present
if (!file.exists()) {
file.getParentFile().mkdirs();
file.createNewFile();
Log.e(TAG, "File Created");
}
This could be a threading issue (checking and creating together are not atomic: !file.exists() && !file.createNewFile()) or the "file" is already a directory.
Try (file.isFile()) :
if (file.exists() && !file.isFile()){
//handle directory is there
}else if(!file.createNewFile()) {
//as before
}
In my case was just a lack of permission:
<uses-permission android:name="android.permission.WRITE_EXTERNAL_STORAGE" />
Use
yourAppsMainActivityContext.getExternalCacheDir()
instead of
Environment.getExternalStorageDriectory()
to get the file storage path.
Alternatively, you can also try getExternalFilesDir(String type), getExternalCacheDir(), getExternalMediaDirs().