Ошибка java io filenotfoundexception

java.io.FileNotFoundException which is a common exception which occurs while we try to access a file. FileNotFoundExcetion is thrown by constructors RandomAccessFile, FileInputStream, and FileOutputStream. FileNotFoundException occurs at runtime so it is a checked exception, we can handle this exception by java code, and we have to take care of the code so that this exception doesn’t occur. 

Declaration : 

public class FileNotFoundException
  extends IOException
    implements ObjectInput, ObjectStreamConstants

Constructors : 

• FileNotFoundException() : It gives FileNotFoundException with null message.
• FileNotFoundException(String s) : It gives FileNotFoundException with detail message.

It doesn’t have any methods. Now let’s understand the hierarchy of this class i.e FileNotFoundException extends IOException which further extends the Exception class which extends the Throwable class and further the Object class. 

Hierarchy Diagram:

Why this Exception occurs? 

There are mainly 2 scenarios when FileNotFoundException occurs. Now let’s see them with examples provided:

1. If the given file is not available in the given location then this error will occur.
2. If the given file is inaccessible, for example, if it is read-only then you can read the file but not modify the file, if we try to modify it, an error will occur or if the file that you are trying to access for the read/write operation is opened by another program then this error will occur.

Scenario 1:

If the given file is not available in the given location then this error will occur.

Example: 

Output

prog.java:14: error: unreported exception FileNotFoundException; must be caught or declared to be thrown
    FileReader reader = new FileReader("file.txt");
                        ^
prog.java:25: error: unreported exception IOException; must be caught or declared to be thrown
    while ((data = br.readLine()) != null) 
                              ^
prog.java:31: error: unreported exception IOException; must be caught or declared to be thrown
    br.close();
            ^
3 errors

Scenario 2:

If the given file is inaccessible, for example, if it is read-only then you can read the file but not modify the file if we try to modify it, an error will occur or if the file that you are trying to access for the read/write operation is opened by another program then this error will occur.

Example:

Output

java.security.AccessControlException: access denied ("java.io.FilePermission" "file.txt" "write")
    at java.base/java.security.AccessControlContext.checkPermission(AccessControlContext.java:472)
    at java.base/java.security.AccessController.checkPermission(AccessController.java:897)
    at java.base/java.lang.SecurityManager.checkPermission(SecurityManager.java:322)
    at java.base/java.lang.SecurityManager.checkWrite(SecurityManager.java:752)
    at java.base/java.io.FileOutputStream.<init>(FileOutputStream.java:225)
    at java.base/java.io.FileOutputStream.<init>(FileOutputStream.java:187)
    at java.base/java.io.FileWriter.<init>(FileWriter.java:96)
    at Example2.main(File.java:19)

Handling Exception:

Firstly we have to use the try-catch block if we know whether the error will occur. Inside try block all the lines should be there if there are chances of errors. There are other remedies to handle the exception:

1. If the message of the exception tells that there is no such file or directory, then you re-verify whether you mentioned the wrong file name in the program or file exists in that directory or not.
2. If the message of the exception tells us that access is denied then we have to check the permissions of the file (read, write, both read and write) and also check whether that file is in use by another program.
3. If the message of the exception tells us that the specified file is a directory then you must either delete the existing directory(if the directory not in use) or change the name of the file.

Last Updated :
16 Nov, 2021

Like Article

Save Article

Put the word.txt directly as a child of the project root folder and a peer of src

Project_Root
    src
    word.txt

Disclaimer: I’d like to explain why this works for this particular case and why it may not work for others.

Why it works:

When you use File or any of the other FileXxx variants, you are looking for a file on the file system relative to the «working directory». The working directory, can be described as this:

When you run from the command line

C:EclipseWorkspaceProjectRootbin > java com.mypackage.Hangman1

the working directory is C:EclipseWorkspaceProjectRootbin. With your IDE (at least all the ones I’ve worked with), the working directory is the ProjectRoot. So when the file is in the ProjectRoot, then using just the file name as the relative path is valid, because it is at the root of the working directory.

Similarly, if this was your project structure ProjectRootsrcword.txt, then the path "src/word.txt" would be valid.

Why it May not Work

For one, the working directory could always change. For instance, running the code from the command line like in the example above, the working directory is the bin. So in this case it will fail, as there is not binword.txt

Secondly, if you were to export this project into a jar, and the file was configured to be included in the jar, it would also fail, as the path will no longer be valid either.

That being said, you need to determine if the file is to be an embedded-resource (or just «resource» — terms which sometimes I’ll use interchangeably). If so, then you will want to build the file into the classpath, and access it via an URL. First thing you would need to do (in this particular) case is make sure that the file get built into the classpath. With the file in the project root, you must configure the build to include the file. But if you put the file in the src or in some directory below, then the default build should put it into the class path.

You can access classpath resource in a number of ways. You can make use of the Class class, which has getResourceXxx method, from which you use to obtain classpath resources.

For example, if you changed your project structure to ProjectRootsrcresourcesword.txt, you could use this:

InputStream is = Hangman1.class.getResourceAsStream("/resources/word.txt");
BufferedReader reader = new BufferedReader(new InputStreamReader(is));

getResourceAsStream returns an InputStream, but obtains an URL under the hood. Alternatively, you could get an URL if that’s what you need. getResource() will return an URL

For Maven users, where the directory structure is like src/main/resources, the contents of the resources folder is put at the root of the classpath. So if you have a file in there, then you would only use getResourceAsStream("/thefile.txt")