This may be a stupid question, but I have to ask because I couldn't find any proper solution.
I am new to Eclipse. I created a Dynamic Web project in Eclipse, In this, I write a simple code to create a text file, Only file name is specified Not the path that where to create, After successful execution, i could not find my text file in my project folder.
If path is specified in the code, I can find the text file in specified directory, My Question is where i can find my text file if i am not specify a path ?
And my code is
try {
FileWriter outFile = new FileWriter("user_details.txt", true);
PrintWriter out1 = new PrintWriter(outFile);
out1.append(request.getParameter("un"));
out1.println();
out1.append(request.getParameter("pw"));
out1.close();
outFile.close();
System.out.println("file created");
} catch(Exception e) {
System.out.println("error in writing a file"+e);
}
I edited my code with following lines,
String path = new File("user_details.txt").getAbsolutePath();
System.out.println(path);
The path that i got is below
D:\Android\eclipse_JE\eclipse\user_details.txt
Why i got it in the eclipse folder ?
Then,
How can i create a text file in my web app, if this is not the right way to create a textfile ?
The file is located in the actual working directory of your application server. Do a
System.out.println(new File("").getAbsolutPath());
and you'll find the location.
However this is not a good idea to write files in web application like this, because first you never know where it is and second you never know whether you write privilege on it.
You need to specify some filesystem root for your application by passing it as init-parameter and use it as parent for everything you need to do on the filesystem. Check this answer to a similar Question.
You could then create your file like this:
String fsroot = getServletContext().getInitParameter("fsroot")
File ud = new File(fsroot, "user_details.txt");
FileWriter outFile = new FileWriter(ud, true);
You may try the getAbsolutePath() method.
String newFile = new File("Demo.txt").getAbsolutePath();
It will show the location where the files will be created.
Related
This question already has answers here:
java.io.FileNotFoundException: the system cannot find the file specified
(8 answers)
Closed 4 years ago.
I have a csv file in the same path as everything else. Now, when I try to create a File object:
public void getMenu() {
File fileMenu = new File("FastFoodMenu.csv");
try {
Scanner inputStream = new Scanner(fileMenu);
while (inputStream.hasNext()) {
String data = inputStream.next();
System.out.println(data);
}
} catch (FileNotFoundException ex) {
Logger.getLogger(FileHandler.class.getName()).log(Level.SEVERE, null, ex);
}
}
it throws a FileNotFoundException.
the absolute path to all files in the project is:
C:\Users\kenyo\Documents\NetBeansProjects\OrderFastFood\src\fastfoodorderingsystem
I also checked the name a couple of times. fileMenu.exists() returns false.
First, in your root/working directory (in your case it's the folder containing your project), create a folder called 'menus', here you can store all your menus (so you can play around with multi-file input).
Second, move your FastFoodMenu.csv file to that menus folder.
The FastFoodMenu.csv relative path should now look like this: OrderFastFood\menus\FastFoodMenu.csv.
Third, get your working directory from the System properties. This is the folder in which your program is working in. Then, get a reference (File object) to the menus folder.
Lastly, get a reference to the file in question inside the menu folder. When you get to multi-file reading (and at some point, multi-folder reading), you're gonna want to get the files inside the menu folder as a list so that's why I say to just get the menus folder as it's own reference (or just get the file without the isolated reference to the parent aka '\menus\').
So your code should really look like this:
public void getMenu() {
final File workingDir = File(System.getProperty("user.dir"));
final File menusDir = File(workingDir, "menus");
final File fastFoodMenu = File(menusDir, "FastFoodMenu.csv");
try {
final FileInputStream fis = new FileInputStream(fastFoodMenu);
final BufferedInputStream bs = new BufferedInputStream(fis);
while((l = bs.readLine()) != null) {
System.out.println(l);
}
} catch(FileNotFoundException e) {
System.out.println(e.getMessage());
e.printStackTrace()
}
}
This is all psuedocode but that should at least get you started. Make sure to use BufferedInputStream for efficiency, and when reading files, always pass them into FileInputStream's. It's much better than using the Scanner class. I should also mention that when creating a File object, you're not actually creating a file. What you're doing is your're creating an object, giving it the data you want it to have (such as whether it's a folder, and if it is, what child files/folders do you want it to have, whether it's protected or not, hidden or not, etc) before actually telling the system to create the file with everything else.
Your csv file is probably at the wrong place. You're just specifying the file name, which is a relative path.
Relative paths are always resolved against the working directory of your application, not against the directory where your source file(s) are.
To solve the issue, you can
move the files to the real working directory.
use an absolute path (not advisable!)
specify the folder of your data files as program argument or in a config file (in your working directory)
put the files somewhere into the classpath of your application and load them from there via classloader. Note that files that are in your classpath are usually packed with your application and hence not easily modifiable by the user, so this solution doesn't work if the file must be changed by the user.
I'm making a game that saves information into a binary file so that I can start at the point I left the game on the next use.
My problem is that it works fine on my PC because I chose a path that already existed to save the file, but once I run the game on another PC, I got an error saying the path of the file is invalid (because i doesn't exist yet, obviously).
Basically I'm using the File class to create the file and then the ObjectOutputStream and ObjectInputStream to read/write info.
Sorry for the noob question, I'm still pretty new to using files.
You must first check if the directory exists and if it does not exist then you must create it.
String folderPath = System.getProperty("user.home") + System.getProperty("file.separator") + "MyFolder";
File folder = new File(folderPath);
if(!folder.exists())
{
folder.mkdirs();
}
File saveFile = new File(folderPath, "fileName.ext");
Please note that the mkdirs() method is more useful in this case instead of the mkdir() method as it will create all non existing parent folders.
Hope this helps. Good luck and have fun programming!
Cheers,
Lofty
You are looking for File mkdirs()
Which will create all the directories necessary that are named in your path.
For example:
File dirs= new File("/this/path/does/not/exist/yet");
dirs.mkdir();
File file = new File(dirs, "myFile.txt");
Take in consideration that it may fail, due to proper file permissions.
My solution has been create a subdirectory within the user's home directory (System.getProperty("user.home")), like
File f = new File(System.getProperty("user.home") + "/CtrlAltDelData");
f.mkdir();
File mySaveFile = new File (f, "save1.txt");
I am making a program that opens and reads a file.
This is my code:
import java.io.*;
public class FileRead{
public static void main(String[] args){
try{
File file = new File("hello.txt");
System.out.println(file.getCanonicalPath());
FileInputStream ft = new FileInputStream(file);
DataInputStream in = new DataInputStream(ft);
BufferedReader br = new BufferedReader(new InputStreamReader(in));
String strline;
while((strline = br.readLine()) != null){
System.out.println(strline);
}
in.close();
}catch(Exception e){
System.err.println("Error: " + e.getMessage());
}
}
}
but when I run, I get this error:
C:\Users\User\Documents\Workspace\FileRead\hello.txt
Error: hello.txt (The system cannot find the file specified)
my FileRead.java and hello.txt where in the same directory that can be found in:
C:\Users\User\Documents\Workspace\FileRead
I'm wondering what I am doing wrong?
Try to list all files' names in the directory by calling:
File file = new File(".");
for(String fileNames : file.list()) System.out.println(fileNames);
and see if you will find your files in the list.
I have copied your code and it runs fine.
I suspect you are simply having some problem in the actual file name of hello.txt, or you are running in a wrong directory. Consider verifying by the method suggested by #Eng.Fouad
You need to give the absolute pathname to where the file exists.
File file = new File("C:\\Users\\User\\Documents\\Workspace\\FileRead\\hello.txt");
In your IDE right click on the file you want to read and choose "copy path"
then paste it into your code.
Note that windows hides the file extension so if you create a text file "myfile.txt" it might be actually saved as "myfile.txt.txt"
Generally, just stating the name of file inside the File constructor means that the file is located in the same directory as the java file. However, when using IDEs like NetBeans and Eclipse i.e. not the case you have to save the file in the project folder directory. So I think checking that will solve your problem.
How are you running the program?
It's not the java file that is being ran but rather the .class file that is created by compiling the java code. You will either need to specify the absolute path like user1420750 says or a relative path to your System.getProperty("user.dir") directory. This should be the working directory or the directory you ran the java command from.
First Create folder same as path which you Specified. after then create File
File dir = new File("C:\\USER\\Semple_file\\");
File file = new File("C:\\USER\\Semple_file\\abc.txt");
if(!file.exists())
{
dir.mkdir();
file.createNewFile();
System.out.println("File,Folder Created.);
}
When you run a jar, your Main class itself becomes args[0] and your filename comes immediately after.
I had the same issue: I could locate my file when provided the absolute path from eclipse (because I was referring to the file as args[0]). Yet when I run the same from jar, it was trying to locate my main class - which is when I got the idea that I should be reading my file from args[1].
FileOutputStream fos = openFileOutput(FILENAME, Context.MODE_WORLD_READABLE);
fos.write(string.getBytes());
fos.close();
When trying to delete one of those files, this is what I use, but it's returning false.
String tag = v.getTag().toString();
File file = new File(System.getProperty("user.dir")+"/"+tag);
String s = new Boolean (file.exists()).toString();
Toast.makeText(getApplicationContext(), s, 1500).show();
file.delete();
How can I overcome this problem?
Use getFileStreamPath(FILENAME) to find your file. From the docs:
Returns the absolute path on the filesystem where a file created with openFileOutput(String, int) is stored.
Your current working directory.
To help diagnose the problem, use file.getAbsolutePath() to see the full path.
It could also be a permissions problem, if you're trying to delete from another application. If so, you may need to change to MODE_WORLD_WRITEABLE (insecure), or restructure your code so the create and delete are called by the same app.
EDIT: That was mostly incorrect. I didn't realize that openFileOutput didn't use the current working directory.
Use same contents as 'FILENAME' variable in your first snippet in the second snippet while trying to delete.
String RootDir = Environment.getExternalStorageDirectory()
+ File.separator + "Video";
File RootFile = new File(RootDir);
RootFile.mkdir();
FileOutputStream f = new FileOutputStream(new File(RootFile, "Sample.mp4"));
i used this code to save the video files to non-default location. Hope this will be useful to you.By default it is storing in sd card
For each application the Android system creates a "data/data/package of the application" directory.
Files are saved in the "files" folder under this directory
to change the default directory the above code will be used
the default working directory can be displayed using fileobject.getAbsolutePath()
I think I am really close, but I am unable to open a file I have called LocalNews.txt. Error says can't find file specified.
String y = "LocalNews.txt";
FileInputStream fstream = new FileInputStream(y);
DataInputStream in = new DataInputStream(fstream);
BufferedReader br = new BufferedReader(new InputStreamReader(in));
Name of file is LocalNews.txt in library called News....anyone know why the file will not open?
The file is in the same Java Project that I am working on.
Error: LocalNews.txt (The system cannot find the file specified)
Project is named Bst, package is src in subPackage newsFinder, and library that the text files are stored in is called News.
Found out it was looking in
C:\EclipseIndigoWorkspace1\Bst\bin\LocalNews.txt
But I want it to look in (I believe)
C:\EclipseIndigoWorkspace1\Bst\News\LocalNews.txt
But if I make the above url a string, I get an error.
String y = "LocalNews.txt";
instead use
String y = "path from root/LocalNews.txt"; //I mean the complete path of the file
Your program can probably not find the file because it is looking in another folder.
Try using a absolute path like
String y = "c:\\temp\\LocalNews.txt";
By 'library called News' I assume you mean a jar file like News.jar which is on the classpath and contains the LocalNews.txt file you need. If this is the case, then you can get an InputStream for it by calling:
InputStream is = Thread.currentThread().getContextClassLoader()
.getResourceAsStream("LocalNews.txt");
Use
System.out.println(System.getProperty("user.dir") );
to find out what your current directory is. Then you'll know for sure whether your file is in the current directory or not. If it is not, then you have to specify the path so that it looks in the right directory.
Also, try this -
File file = new File (y);
System.out.println(file.getCanonicalPath());
This will tell you the exact path of your file on the system, provided your file is in the current directory. If it does not, then you know your file is not in the current directory.