This piece of code throws a FileNotFoundException, i'm sure the file exists in my working directory, am i doing something wrong?
private void generateInvoiceNumber(){ //uses reads previous invoice number and increments it.
try {
File invoiceFile = new File("./Invoices/invoiceFile.txt");
FileWriter writer = new FileWriter(invoiceFile,false);
Scanner getter = new Scanner(invoiceFile);
this.invoiceNumber = getter.nextInt();
writer.write(++invoiceNumber);
writer.close();
getter.nextInt();
getter.close();
}
catch (IOException e) {
System.out.println("An error occurred.");
e.printStackTrace();
}
}
My tip:
Print (in your code) the current path location.
Then you can use this path in order to find the exact path you should use in order to access your file.
Maybe you should put more concrete absolute path:
File invoiceFile = Paths.get ("C:","Invoices", "invoiceFile.txt").toFile();
or if you trying to get from current path:
File invoiceFile = Paths.get (".","Invoices", "invoiceFile.txt").toFile();
And you can check your . path:
System.out.println(new File(".").getCanonicalPath());
Which operating system you are using?
It’s better to use paths when you are constructing a path to your file like
File file = Paths.get (".","Invoices", "invoice.txt").toFile();
corrected " symbols and default root "." which is your folder where app started.
Below is my project like:
projectName
-package
- Util.java
- Test.json
In Util.java, I need to read the content from Test.json file and parse it.
Thus I use:
File currentfile = new File("");//get the current path
String absJsonPath = currentfile.getAbsolutePath() + "/Test.json";
While it did not work when I use a main method to test it. The thing is that the /src/package is lost in the obtained file path and I just got the path of the project.
And, when I deploy the project to weblogic server, I got another new error, the obtained current path is like:
.../DefaultDomain/.
I just want the file path in the file system, which is not related to the server.
What can I do for this? Thanks!
Put your file in resources folder and get it as following:
//Get file from resources folder
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("Test.json").getFile());
To read the content you can use following:
StringBuilder result = new StringBuilder("");
try (Scanner scanner = new Scanner(file)) {
while (scanner.hasNextLine()) {
String line = scanner.nextLine();
result.append(line).append("\n");
}
scanner.close();
} catch (IOException e) {
e.printStackTrace();
}
It's CanonicalPath() not Absolute. Check this out, let me know if it any helps.
Try this to get current path
String currentPath = System.getProperty("user.dir");
If you want to read a file in a specific package. You can use
File jsonFile = new File(getClass().getResource("/Test.json").getFile());
I want to write a file results.txt to a specific directory on my machine (Z:\results to be precise). How do I go about specifying the directory to BufferedWriter/FileWriter?
Currently, it writes the file successfully but to the directory where my source code is located. Thanks
public void writefile(){
try{
Writer output = null;
File file = new File("results.txt");
output = new BufferedWriter(new FileWriter(file));
for(int i=0; i<100; i++){
//CODE TO FETCH RESULTS AND WRITE FILE
}
output.close();
System.out.println("File has been written");
}catch(Exception e){
System.out.println("Could not create file");
}
}
You should use the secondary constructor for File to specify the directory in which it is to be symbolically created. This is important because the answers that say to create a file by prepending the directory name to original name, are not as system independent as this method.
Sample code:
String dirName = /* something to pull specified dir from input */;
String fileName = "test.txt";
File dir = new File (dirName);
File actualFile = new File (dir, fileName);
/* rest is the same */
Hope it helps.
Use:
File file = new File("Z:\\results\\results.txt");
You need to double the backslashes in Windows because the backslash character itself is an escape in Java literal strings.
For POSIX system such as Linux, just use the default file path without doubling the forward slash. this is because forward slash is not a escape character in Java.
File file = new File("/home/userName/Documents/results.txt");
Just put the full directory location in the File object.
File file = new File("z:\\results.txt");
The best practice is using File.separator in the paths.
I am trying to get a path to a Resource but I have had no luck.
This works (both in IDE and with the JAR) but this way I can't get a path to a file, only the file contents:
ClassLoader classLoader = getClass().getClassLoader();
PrintInputStream(classLoader.getResourceAsStream("config/netclient.p"));
If I do this:
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("config/netclient.p").getFile());
The result is:
java.io.FileNotFoundException: file:/path/to/jarfile/bot.jar!/config/netclient.p (No such file or directory)
Is there a way to get a path to a resource file?
This is deliberate. The contents of the "file" may not be available as a file. Remember you are dealing with classes and resources that may be part of a JAR file or other kind of resource. The classloader does not have to provide a file handle to the resource, for example the jar file may not have been expanded into individual files in the file system.
Anything you can do by getting a java.io.File could be done by copying the stream out into a temporary file and doing the same, if a java.io.File is absolutely necessary.
When loading a resource make sure you notice the difference between:
getClass().getClassLoader().getResource("com/myorg/foo.jpg") //relative path
and
getClass().getResource("/com/myorg/foo.jpg")); //note the slash at the beginning
I guess, this confusion is causing most of problems when loading a resource.
Also, when you're loading an image it's easier to use getResourceAsStream():
BufferedImage image = ImageIO.read(getClass().getResourceAsStream("/com/myorg/foo.jpg"));
When you really have to load a (non-image) file from a JAR archive, you might try this:
File file = null;
String resource = "/com/myorg/foo.xml";
URL res = getClass().getResource(resource);
if (res.getProtocol().equals("jar")) {
try {
InputStream input = getClass().getResourceAsStream(resource);
file = File.createTempFile("tempfile", ".tmp");
OutputStream out = new FileOutputStream(file);
int read;
byte[] bytes = new byte[1024];
while ((read = input.read(bytes)) != -1) {
out.write(bytes, 0, read);
}
out.close();
file.deleteOnExit();
} catch (IOException ex) {
Exceptions.printStackTrace(ex);
}
} else {
//this will probably work in your IDE, but not from a JAR
file = new File(res.getFile());
}
if (file != null && !file.exists()) {
throw new RuntimeException("Error: File " + file + " not found!");
}
The one line answer is -
String path = this.getClass().getClassLoader().getResource(<resourceFileName>).toExternalForm()
Basically getResource method gives the URL. From this URL you can extract the path by calling toExternalForm().
I spent a while messing around with this problem, because no solution I found actually worked, strangely enough! The working directory is frequently not the directory of the JAR, especially if a JAR (or any program, for that matter) is run from the Start Menu under Windows. So here is what I did, and it works for .class files run from outside a JAR just as well as it works for a JAR. (I only tested it under Windows 7.)
try {
//Attempt to get the path of the actual JAR file, because the working directory is frequently not where the file is.
//Example: file:/D:/all/Java/TitanWaterworks/TitanWaterworks-en.jar!/TitanWaterworks.class
//Another example: /D:/all/Java/TitanWaterworks/TitanWaterworks.class
PROGRAM_DIRECTORY = getClass().getClassLoader().getResource("TitanWaterworks.class").getPath(); // Gets the path of the class or jar.
//Find the last ! and cut it off at that location. If this isn't being run from a jar, there is no !, so it'll cause an exception, which is fine.
try {
PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(0, PROGRAM_DIRECTORY.lastIndexOf('!'));
} catch (Exception e) { }
//Find the last / and cut it off at that location.
PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(0, PROGRAM_DIRECTORY.lastIndexOf('/') + 1);
//If it starts with /, cut it off.
if (PROGRAM_DIRECTORY.startsWith("/")) PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(1, PROGRAM_DIRECTORY.length());
//If it starts with file:/, cut that off, too.
if (PROGRAM_DIRECTORY.startsWith("file:/")) PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(6, PROGRAM_DIRECTORY.length());
} catch (Exception e) {
PROGRAM_DIRECTORY = ""; //Current working directory instead.
}
This is same code from user Tombart with stream flush and close to avoid incomplete temporary file content copy from jar resource and to have unique temp file names.
File file = null;
String resource = "/view/Trial_main.html" ;
URL res = getClass().getResource(resource);
if (res.toString().startsWith("jar:")) {
try {
InputStream input = getClass().getResourceAsStream(resource);
file = File.createTempFile(new Date().getTime()+"", ".html");
OutputStream out = new FileOutputStream(file);
int read;
byte[] bytes = new byte[1024];
while ((read = input.read(bytes)) != -1) {
out.write(bytes, 0, read);
}
out.flush();
out.close();
input.close();
file.deleteOnExit();
} catch (IOException ex) {
ex.printStackTrace();
}
} else {
//this will probably work in your IDE, but not from a JAR
file = new File(res.getFile());
}
if netclient.p is inside a JAR file, it won't have a path because that file is located inside other file. in that case, the best path you can have is really file:/path/to/jarfile/bot.jar!/config/netclient.p.
You need to understand the path within the jar file.
Simply refer to it relative. So if you have a file (myfile.txt), located in foo.jar under the \src\main\resources directory (maven style). You would refer to it like:
src/main/resources/myfile.txt
If you dump your jar using jar -tvf myjar.jar you will see the output and the relative path within the jar file, and use the FORWARD SLASHES.
In my case, I have used a URL object instead Path.
File
File file = new File("my_path");
URL url = file.toURI().toURL();
Resource in classpath using classloader
URL url = MyClass.class.getClassLoader().getResource("resource_name")
When I need to read the content, I can use the following code:
InputStream stream = url.openStream();
And you can access the content using an InputStream.
follow code!
/src/main/resources/file
streamToFile(getClass().getClassLoader().getResourceAsStream("file"))
public static File streamToFile(InputStream in) {
if (in == null) {
return null;
}
try {
File f = File.createTempFile(String.valueOf(in.hashCode()), ".tmp");
f.deleteOnExit();
FileOutputStream out = new FileOutputStream(f);
byte[] buffer = new byte[1024];
int bytesRead;
while ((bytesRead = in.read(buffer)) != -1) {
out.write(buffer, 0, bytesRead);
}
return f;
} catch (IOException e) {
LOGGER.error(e.getMessage(), e);
return null;
}
}
A File is an abstraction for a file in a filesystem, and the filesystems don't know anything about what are the contents of a JAR.
Try with an URI, I think there's a jar:// protocol that might be useful for your purpouses.
The following path worked for me: classpath:/path/to/resource/in/jar
private static final String FILE_LOCATION = "com/input/file/somefile.txt";
//Method Body
InputStream invalidCharacterInputStream = URLClassLoader.getSystemResourceAsStream(FILE_LOCATION);
Getting this from getSystemResourceAsStream is the best option. By getting the inputstream rather than the file or the URL, works in a JAR file and as stand alone.
It may be a little late but you may use my library KResourceLoader to get a resource from your jar:
File resource = getResource("file.txt")
Inside your ressources folder (java/main/resources) of your jar add your file (we assume that you have added an xml file named imports.xml), after that you inject ResourceLoader if you use spring like bellow
#Autowired
private ResourceLoader resourceLoader;
inside tour function write the bellow code in order to load file:
Resource resource = resourceLoader.getResource("classpath:imports.xml");
try{
File file;
file = resource.getFile();//will load the file
...
}catch(IOException e){e.printStackTrace();}
Maybe this method can be used for quick solution.
public class TestUtility
{
public static File getInternalResource(String relativePath)
{
File resourceFile = null;
URL location = TestUtility.class.getProtectionDomain().getCodeSource().getLocation();
String codeLocation = location.toString();
try{
if (codeLocation.endsWith(".jar"){
//Call from jar
Path path = Paths.get(location.toURI()).resolve("../classes/" + relativePath).normalize();
resourceFile = path.toFile();
}else{
//Call from IDE
resourceFile = new File(TestUtility.class.getClassLoader().getResource(relativePath).getPath());
}
}catch(URISyntaxException ex){
ex.printStackTrace();
}
return resourceFile;
}
}
When in a jar file, the resource is located absolutely in the package hierarchy (not file system hierarchy). So if you have class com.example.Sweet loading a resource named ./default.conf then the resource's name is specified as /com/example/default.conf.
But if it's in a jar then it's not a File ...
This Class function can get absolute file path by relative file path in .jar file.
public class Utility {
public static void main(String[] args) throws Exception {
Utility utility = new Utility();
String absolutePath = utility.getAbsolutePath("./absolute/path/to/file");
}
public String getAbsolutePath(String relativeFilePath) throws IOException {
URL url = this.getClass().getResource(relativeFilePath);
return url.getPath();
}
}
If target file is same directory with Utility.java, your input will be ./file.txt.
When you input only / on getAbsolutePath(), it returns /Users/user/PROJECT_NAME/target/classes/. This means you can select the file like this /com/example/file.txt.
I am trying to create a back up file for an html file on a web server.
I want the backup to be in the same location as the existing file (it's a quick fix). I want to create the file using File file = new File(PathName);
public void backUpOldPage(String oldContent) throws IOException{
// this.uri is a class variable with the path of the file to be backed up
String fileName = new File(this.uri).getName();
String pathName = new File(this.uri).getPath();
System.out.println(pathName);
String bckPath = pathName+"\\"+bckName;
FileOutputStream fout;
try
{
// Open an output stream
fout = new FileOutputStream (bckFile);
fout.close();
}
// Catches any error conditions
catch (IOException e)
{
System.err.println ("Unable to write to file");
System.exit(-1);
}
}
But if instead I was to set bckPath like this, it will work.
String bckPath = "C://dev/server/tomcat6/webapps/sample-site/index_sdjf---sd.html";
I am working on Windows, not sure if that makes a difference.
The result of String bckPath = pathName+"\"+bckName;
is bckPath = C:\dev\server\tomcat6\webapps\sample-site\filename.html - this doesn't result in a new file.
Use File.pathSeparator, that way you dont need to worry what OS you are using.
Try to use File.getCanonicalPath() instead of plain getPath(). This helps if the orginal path is not fully specified.
Regarding slashes, / or \ or File.pathSeparator is not causing the problem, because they are all the same on Windows and Java. (And you do not define bckFile in your code, only bckPath. Also use getCanonicalPath() on the new created bckPath.)