I am trying to access a text file, output.txt which is inside a project folder. The image below shows the folder structure.
String file=".\\testFiles\\output.txt";
BufferedReader br=new BufferedReader(new FileReader(file));
When I try to access that file using the code above, I get the following exception.
java.io.FileNotFoundException: .\testFiles\output.txt (No such file or directory)
I have tried different file path formats but none have worked. I think the problem is with the file path format.
Thanks in advance.
If I remember correctly you can get a folder/file in the current directory like so:
File folder = new File("testFiles");
Then you can open the file by getting the absolutePath and creating a new file with it, like so:
File file = new File(folder.getAbsoluteFile() + File.separator + "output.txt");
I'm not sure but I think you can also do:
File file = new File("testFiles/output.txt");
I hope this helps :)
P.S. this is all untested so it might not work.
Judging by the fact that you have a webcontent folder i presume this is a web project, possibly packaged as a war? In this case what you will want to do is package the respective files along with the classes and access it with something like this:
Thread.currentThread().getContextClassLoader().getResourceAsStream("output.txt")
The code above will work if you add the testFiles folder as a source folder (this means it will get packaged with the classes and be available at runtime)
The good thing is that this way the path can stay relative, no need to go absolute
I believe that your problem is due to the fact that you rely on a relative path as your path starts with a dot which means that it will by relative to the user directory (value of the system property user.dir) so I believe that your user directory is not what you expect. What you could do to debug is simply this:
System.out.println(new File(file).getAbsolutePath());
Thanks to this approach you will be able to quickly know if the absolute path is correct.
You must declare your file as a new file:
File yourFile = new File("testFiles/output.txt");
Related
I am working on a simple web application in NetBeans where I am getting FileNotFoundException. I have stored files in class path, so I need to use relative paths. When I tried with absolute path, it worked fine for me.
Below image shows my file system hierarchy.
I need to write content data in file DBList.txt.
My code is:
File file = new File("data/application/DBList.txt");
PrintWriter writer = new PrintWriter(new BufferedWriter(new FileWriter(file)));
I have searched lot but not getting solution for reading file using relative path.
The path is relative to the working directory of the server, not your project in NetBeans. Given your FNFE I suspect that the directory structure data/application/ doesn't exist under the working directory.
What server are you running and how are you starting it? You can figure out the server's working directory by logging;
File wd = new File(".");
log.debug("working dir: " + wd.getAbsolutePath());
Edit:
The File class and the classpath are totally unrelated concepts. Don't confuse the two. If you are looking to use classpath resources have a look at the getResource() method in ClassLoader.
try to use "\\" instead of "/"
I am using a method that requires a string path to a file that is in my src directory structure in eclipse. Is the path to this file simply src\fileName.txt or is there a different way i should be getting this file as it doesnt seem to be working currently
Thanks
Run this and you will never forget how to remember.
File file = new File("sample.txt");
System.out.println(file.getAbsolutePath());
If you are following the maven standard directory structure, class files will be located relative to the classpath like so:
If you want SomeClass.class, you would access it by com.mydomain.packageorappname.deeperpackage.SomeClass. Is that what you are asking?
Been looking for this for the past 2 hours and can't find anything (I've found solutions to the same problem but with images, not text files).
Pretty much, I made a program that reads a text file. The file is a list of names and IDs. Using Eclipse, I put the file in my src folder and in the program put the path file to it. Like this:
in = new BufferedReader(new FileReader(curDir+"\\bin\\items.txt"));
Where curDir is the user's current directory (found with System.getProperty("user.dir")).
Now, problem is, the program runs fine when I run it from Eclipse, but when I try to make it a runnable JAR and then run it, the program runs, but the info from the text file does not load. It look like Eclipse is not putting the text file with the JAR.
EDIT: Solved-ish the problem? So the JAR file needs to the in a folder with all the original files? I am so confused, what is a JAR file then?
A more robust way to get a file whether you are running from Eclipse or a JAR is to do
MyClass.getResource("items.txt")
where MyClass is a class in the same package (folder) as the resource you need.
If Eclipse is not putting the file in your JAR you can go to
Run -> Run Configurations -> -> Classpath tab -> Advanced -> Add Folders
Then add the folder containing your file to the classpath. Alternatively, export the Ant script and create a custom build script.
To the point, the FileReader can only read disk file system resources. But a JAR contains classpath resources only. You need to read it as a classpath resource. You need the ClassLoader for this.
Assuming that Foo is your class in the JAR which needs to read the resource and items.txt is put in the classpath root of the JAR, then you should read it as follows (note: leading slash needed!):
InputStream input = Foo.class.getResourceAsStream("/items.txt");
reader = new BufferedReader(new InputStreamReader(input, "UTF-8"));
// ...
Or if you want to be independent from the class or runtime context, then use the context class loader which operates relative to the classpath root (note: no leading slash needed!):
ClassLoader classLoader = Thread.currentThread().getContextClassLoader();
InputStream input = classLoader.getResourceAsStream("items.txt");
reader = new BufferedReader(new InputStreamReader(input, "UTF-8"));
// ...
(UTF-8 is of course the charset the file is encoded with, else you may see Mojibake)
Get the location of your jar file
Firstly create a folder(say myfolder) and put your files inside it
Consider the following function
public String path(String filename)
{
URL url1 = getClass().getResource("");
String ur=url1.toString();
ur=ur.substring(9);
String truepath[]=ur.split("myjar.jar!");
truepath[0]=truepath[0]+"myfolder/";
truepath[0]=truepath[0].replaceAll("%20"," ");
return truepath[0]+filename;
}//This method will work on Windows and Linux as well.
//You can alternatively use the following line to get the path of your jar file
//classname.class.getProtectionDomain().getCodeSource().getLocation().getPath();
Suppose your jar file is in D:\Test\dist
Then path() will return /D:/Test/dist/myfolder/filename
Now you can place 'myfolder' inside the folder where your jar file is residing
OR
If you want to access some read-only file inside your jar you should copy it to one
of your packages and can access it as
yourClassname.getResource("/packagename/filename.txt");
How do you scan a file with java that isn't in the directory the java file is in?
For example: The java file is located at "C:\Files\JavaFiles\test.java" However, the file I want to scan is located at "C:\Data\DataPacket99\data.txt"
Note: I've already tried putting another java file in the "C:\Data" directory and using the test.java file as a class, but it doesn't work. It still tries to scan from the "C:\Files\JavaFiles" Directory.
By using an absolute path instead of a relative.
File file = new File("C:\\Data\\DataPacket99\\data.txt");
Then you can write code that accesses that file object, using a InputStream or similar.
You need to use absolute paths in java.io stuff. Thus not new File("data.txt"), but new File("C:/Data/DataPacket99/data.txt"). Otherwise it will be relative to the current working directory which may not per-se be the same in all environments or the one you'd expect.
You should be using an absolute path instead of a relative path.
You could use File file = new File("C:/Data/DataPacket99/data.txt"); but it might make your life easier in the future to use a file chooser dialog if at any point the user will have to enter a file path.
I would try this:
File file = new File("../../Data/DataPacket99/data.txt");
In my Java app I need to get some files and directories.
This is the program structure:
./main.java
./package1/guiclass.java
./package1/resources/resourcesloader.java
./package1/resources/repository/modules/ -> this is the dir I need to get
./package1/resources/repository/SSL-Key/cert.jks -> this is the file I need to get
guiclass loads the resourcesloader class which will load my resources (directory and file).
As to the file, I tried
resourcesloader.class.getClass().getResource("repository/SSL-Key/cert.jks").toString()
in order to get the real path, but this way does not work.
I have no idea which path to use for the directory.
I had problems with using the getClass().getResource("filename.txt") method.
Upon reading the Java docs instructions, if your resource is not in the same package as the class you are trying to access the resource from, then you have to give it relative path starting with '/'. The recommended strategy is to put your resource files under a "resources" folder in the root directory. So for example if you have the structure:
src/main/com/mycompany/myapp
then you can add a resources folder as recommended by maven in:
src/main/resources
furthermore you can add subfolders in the resources folder
src/main/resources/textfiles
and say that your file is called myfile.txt so you have
src/main/resources/textfiles/myfile.txt
Now here is where the stupid path problem comes in. Say you have a class in your com.mycompany.myapp package, and you want to access the myfile.txt file from your resource folder. Some say you need to give the:
"/main/resources/textfiles/myfile.txt" path
or
"/resources/textfiles/myfile.txt"
both of these are wrong. After I ran mvn clean compile, the files and folders are copied in the:
myapp/target/classes
folder. But the resources folder is not there, just the folders in the resources folder. So you have:
myapp/target/classes/textfiles/myfile.txt
myapp/target/classes/com/mycompany/myapp/*
so the correct path to give to the getClass().getResource("") method is:
"/textfiles/myfile.txt"
here it is:
getClass().getResource("/textfiles/myfile.txt")
This will no longer return null, but will return your class.
It is strange to me, that the "resources" folder is not copied as well, but only the subfolders and files directly in the "resources" folder. It would seem logical to me that the "resources" folder would also be found under `"myapp/target/classes"
Supply the path relative to the classloader, not the class you're getting the loader from. For instance:
resourcesloader.class.getClassLoader().getResource("package1/resources/repository/SSL-Key/cert.jks").toString();
In the hopes of providing additional information for those who don't pick this up as quickly as others, I'd like to provide my scenario as it has a slightly different setup. My project was setup with the following directory structure (using Eclipse):
Project/
src/ // application source code
org/
myproject/
MyClass.java
test/ // unit tests
res/ // resources
images/ // PNG images for icons
my-image.png
xml/ // XSD files for validating XML files with JAXB
my-schema.xsd
conf/ // default .conf file for Log4j
log4j.conf
lib/ // libraries added to build-path via project settings
I was having issues loading my resources from the res directory. I wanted all my resources separate from my source code (simply for managment/organization purposes). So, what I had to do was add the res directory to the build-path and then access the resource via:
static final ClassLoader loader = MyClass.class.getClassLoader();
// in some function
loader.getResource("images/my-image.png");
loader.getResource("xml/my-schema.xsd");
loader.getResource("conf/log4j.conf");
NOTE: The / is omitted from the beginning of the resource string because I am using ClassLoader.getResource(String) instead of Class.getResource(String).
When you use 'getResource' on a Class, a relative path is resolved based on the package the Class is in. When you use 'getResource' on a ClassLoader, a relative path is resolved based on the root folder.
If you use an absolute path, both 'getResource' methods will start at the root folder.
#GianCarlo:
You can try calling System property user.dir that will give you root of your java project and then do append this path to your relative path for example:
String root = System.getProperty("user.dir");
String filepath = "/path/to/yourfile.txt"; // in case of Windows: "\\path \\to\\yourfile.txt
String abspath = root+filepath;
// using above path read your file into byte []
File file = new File(abspath);
FileInputStream fis = new FileInputStream(file);
byte []filebytes = new byte[(int)file.length()];
fis.read(filebytes);
For those using eclipse + maven. Say you try to access the file images/pic.jpg in src/main/resources. Doing it this way :
ClassLoader loader = MyClass.class.getClassLoader();
File file = new File(loader.getResource("images/pic.jpg").getFile());
is perfectly correct, but may result in a null pointer exception. Seems like eclipse doesn't recognize the folders in the maven directory structure as source folders right away. By removing and the src/main/resources folder from the project's source folders list and putting it back (project>properties>java build path> source>remove/add Folder), I was able to solve this.
resourcesloader.class.getClass()
Can be broken down to:
Class<resourcesloader> clazz = resourceloader.class;
Class<Class> classClass = clazz.getClass();
Which means you're trying to load the resource using a bootstrap class.
Instead you probably want something like:
resourcesloader.class.getResource("repository/SSL-Key/cert.jks").toString()
If only javac warned about calling static methods on non-static contexts...
Doe the following work?
resourcesloader.class.getClass().getResource("/package1/resources/repository/SSL-Key/cert.jks")
Is there a reason you can't specify the full path including the package?
Going with the two answers as mentioned above. The first one
resourcesloader.class.getClassLoader().getResource("package1/resources/repository/SSL-Key/cert.jks").toString();
resourcesloader.class.getResource("repository/SSL-Key/cert.jks").toString()
Should be one and same thing?
In Order to obtain real path to the file you can try this:
URL fileUrl = Resourceloader.class.getResource("resources/repository/SSL-Key/cert.jks");
String pathToClass = fileUrl.getPath;
Resourceloader is classname here.
"resources/repository/SSL-Key/cert.jks" is relative path to the file. If you had your guiclass in ./package1/java with rest of folder structure remaining, you would take "../resources/repository/SSL-Key/cert.jks" as relative path because of rules defining relative path.
This way you can read your file with BufferedReader. DO NOT USE THE STRING to identify the path to the file, because if you have spaces or some characters from not english alphabet in your path, you will get problems and the file will not be found.
BufferedReader bufferedReader = new BufferedReader(
new InputStreamReader(fileUrl.openStream()));
I made a small modification on #jonathan.cone's one liner ( by adding .getFile() ) to avoid null pointer exception, and setting the path to data directory. Here's what worked for me :
String realmID = new java.util.Scanner(new java.io.File(RandomDataGenerator.class.getClassLoader().getResource("data/aa-qa-id.csv").getFile().toString())).next();
Use this:
resourcesloader.class.getClassLoader().getResource("/path/to/file").**getPath();**
One of the stable way to work across all OS would be toget System.getProperty("user.dir")
String filePath = System.getProperty("user.dir") + "/path/to/file.extension";
Path path = Paths.get(filePath);
if (Files.exists(path)) {
return true;
}