I am working on the web application with Eclipse. I have created one property file for database configuration. (DBProperty.properties)
Please find below screen-shot of the folder structure.
I want to access this property file. I am accessing with below code.
FileInputStream input = new FileInputStream("src/resources/DBProperty.properties");
I have also tried many relative paths but not able to succeed.
I have set build path for this project.
You need to use
MyClass.class.getClassLoader().getResourceAsStream("DBProperty.properties")
FileInputStream input = new FileInputStream("resources/DBProperty.properties");
Please try the above line of code. Hope it will solve your problem.
The src directory isn't there at runtime.
Resources are not files.
You need to look into Class.getResource() and friends.
You have to specify complete file path with File object.
public static void main(String[] args) {
File file = new File("C:\\Path\\workspace\\jbossmqimpl\\Test1\\resources\\NewFile.xml");
try (FileInputStream fis = new FileInputStream(file)) {
System.out.println("Total file size to read (in bytes) : "+ fis.available());
int content;
while ((content = fis.read()) != -1) {
// convert to char and display it
System.out.print((char) content);
}
} catch (IOException e) {
e.printStackTrace();
}
}
I am new to Java , I am currently analyzing a file compare tool with java that compare two files from this link:
http://www.java2s.com/Code/Java/File-Input-Output/Difftextfiledifferenceutility.htm
But no where in the file, the file path is mentioned. Where should I insert the file path? I searched google and checked Java Filestram and buffer input output stream. But did not found any useful information.
I also searched stackoverflow but it seems no such question exists.
Usually, the file path should be updated in main file, right?
But it seems that is missing in main file.
public static void main(String argstrings[])
{
if ( argstrings.length != 2 ) {
System.err.println("Usage: diff oldfile newfile" );
System.exit(1);
}
Diff d = new Diff();
d.doDiff(argstrings[0], argstrings[1]);
return;
}
Your program takes the file names as the parameter. So while giving the command line input you can give the full file paths. Something like this:
java yourClassName volume1:\dir1\filename1 volume2:\dir2\filename2
You can certainly do the way juned told you but if you want to the program to be more user friendly try to manipulate the main method like this
public static void main(String[] args) throws ParseException {
try{
Scanner in = new Scanner(System.in);
System.out.println("Enter the path of old file");
String oldFile = in.nextLine();
System.out.println("Enter the path of new file");
String newFile = in.nextLine();
Diff d = new Diff();
if(!oldFile.equals("") && !newFile.equals("")) {
d.doDiff(oldFile, newFile);
}
}
catch (Exception e){
e.printStackTrace();
}
}
So, this question has been asked a million times i believed and I've been reading them for a couple of hours and trying several options given by some people but none of them work for me.
I want to list all the files inside a directory inside the application's JAR, so in IDE this works:
File f = new File(this.getClass().getResource("/resources/").getPath());
for(String s : f.list){
System.out.println(s);
}
That gives me all the files inside the directory.
Now, i've tried this also:
InputStream in = this.getClass().getClassLoader().getResourceAsStream("resources/");
InputStreamReader inReader = new InputStreamReader(in);
Scanner scan = new Scanner(inReader);
while (scan.hasNext()) {
String s = scan.next();
System.out.println("read: " + s);
}
System.out.println("END OF LINE");
And from IDE it prints ALL the files in the directory. Outside IDE prints: "END OF LINE".
Now, I can find an entry inside a Jar with this too:
String s = new File(this.getClass().getResource("").getPath()).getParent().replaceAll("(!|file:\\\\)", "");
JarFile jar = new JarFile(s);
JarEntry entry = jar.getJarEntry("resources");
if (entry != null){
System.out.println("EXISTS");
System.out.println(entry.getSize());
}
That's some horrible coding i had to do to that String.
Anyway... I can't get the list of resources inside the "resources" directory within the Jar... How can I do this???
There's no way to simply get a filtered list of internal resources without first enumerating over the contents of the Jar file.
Luckily, that's actually not that hard (and luckily for me you've done most of the hardwork).
Basically, once you have a reference to the JarFile, you simple need to ask for its' entries and iterate over that list.
By checking the JarEntry name for the required match (ie resources), you can filter the elements you want...
For example...
import java.io.File;
import java.io.IOException;
import java.util.Enumeration;
import java.util.jar.JarEntry;
import java.util.jar.JarFile;
public class ReadMyResources {
public static void main(String[] args) {
new ReadMyResources();
}
public ReadMyResources() {
JarFile jf = null;
try {
String s = new File(this.getClass().getResource("").getPath()).getParent().replaceAll("(!|file:\\\\)", "");
jf = new JarFile(s);
Enumeration<JarEntry> entries = jf.entries();
while (entries.hasMoreElements()) {
JarEntry je = entries.nextElement();
if (je.getName().startsWith("resources")) {
System.out.println(je.getName());
}
}
} catch (IOException ex) {
ex.printStackTrace();
} finally {
try {
jf.close();
} catch (Exception e) {
}
}
}
}
Caveat
This type of question actually gets ask a bit. Rather then trying to read the contents of the Jar at runtime, it would be better to produce some kind of text file which contained a list of the available resources.
This could be produced by your build process dynamically before the Jar file is created. It would be a much simpler solution to then read this file in (via getClass().getResource(), for example) and then look up each resource list in the text file...IMHO
For Spring Framework users, have a look at PathMatchingResourcePatternResolver to do something like the following:
PathMatchingResourcePatternResolver resolver = new PathMatchingResourcePatternResolver();
Resource[] resources = resolver.getResources("classpath:path/to/resource/*.*");
for (Resource resource : resources) {
InputStream inStream = resource.getInputStream();
// Do something with the input stream
}
My case was to read a directory inside resources:
As my requirement was to transform resource directory to io.File, finally it looked like this:
public static File getResourceDirectory(String resource) {
ClassLoader classLoader = Thread.currentThread().getContextClassLoader();
URL res = classLoader.getResource(resource);
File fileDirectory;
if ("jar".equals(res.getProtocol())) {
InputStream input = classLoader.getResourceAsStream(resource);
fileDirectory = Files.createTempDir();
List<String> fileNames = IOUtils.readLines(input, StandardCharsets.UTF_8);
fileNames.forEach(name -> {
String fileResourceName = resource + File.separator + name;
File tempFile = new File(fileDirectory.getPath() + File.pathSeparator + name);
InputStream fileInput = classLoader.getResourceAsStream(resourceFileName);
FileUtils.copyInputStreamToFile(fileInput, tempFile);
});
fileDirectory.deleteOnExit();
} else {
fileDirectory = new File(res.getFile());
}
return fileDirectory;
}
If resources are in jar, we copy it to temp directory that will be deleted on application end.
Then calling getResourceDirectory("migrations") returned me io.File directory for further use.
I have developed a java code that reads files from the folder chosen by the user. It displays how many lines of code are in each file, it reads only .java filesonly and final outcome is shown on console , I was thinking that output to be get displayed on console but along with a text file conataing the same information to be get stored on desktop also, please advise how to that and the name of the file that is generated its name is to be based on timestamp lets assume that name of the output file would be 'output06282012' and that text file should contain the same information that is shown on the console , here is my piece of code...
public static void main(String[] args) throws FileNotFoundException {
JFileChooser chooser = new JFileChooser();
chooser.setCurrentDirectory(new java.io.File("C:" + File.separator));
chooser.setDialogTitle("FILES ALONG WITH LINE NUMBERS");
chooser.setFileSelectionMode(JFileChooser.DIRECTORIES_ONLY);
chooser.setAcceptAllFileFilterUsed(false);
if (chooser.showOpenDialog(null) == JFileChooser.APPROVE_OPTION)
{ Map<String, Integer> result = new HashMap<String, Integer>();
File directory = new File(chooser.getSelectedFile().getAbsolutePath());
int totalLineCount = 0;
File[] files = directory.listFiles(new FilenameFilter(){
#Override
public boolean accept(File directory, String name) {
if(name.endsWith(".java"))
return true;
else
return false;
}
}
);
for (File file : files)
{
if (file.isFile())
{ Scanner scanner = new Scanner(new FileReader(file));
int lineCount = 0;
try
{ for (lineCount = 0; scanner.nextLine() != null; lineCount++) ;
} catch (NoSuchElementException e)
{ result.put(file.getName(), lineCount);
totalLineCount += lineCount;
}
} }
System.out.println("*****************************************");
System.out.println("FILE NAME FOLLOWED BY LOC");
System.out.println("*****************************************");
for (Map.Entry<String, Integer> entry : result.entrySet())
{ System.out.println(entry.getKey() + " ==> " + entry.getValue());
}
System.out.println("*****************************************");
System.out.println("SUM OF FILES SCANNED ==>"+"\t"+result.size());
System.out.println("SUM OF ALL THE LINES ==>"+"\t"+ totalLineCount);
}
}
Now the idea in my mind id for this logic
1) construct the file name you want to use
2) open the file for write
3) each time you call a System.out.println(), make a similar call to write the same message to the file
4) when you are all done, make sure you close the file handle.
I have an rough idea something like this
try{
java.util.Date date= new java.util.Date();
System.out.println(new Timestamp(date.getTime()));
BufferedWriter out = new BufferedWriter(new FileWriter("C://Desktop//output"+new Timestamp(date.getTime())+".txt"));
out.write("some information");
out.close;
}catch(IOException e){
e.printStackTrace();
}
please advise how to that and the name of the file that is generated its name is to be based on timestamp lets assume that name of the output file would be 'output06282012' and that text file should contain the same information that is shown on the console
As far as I can tell, this is the only thing you are actually asking:
Please advise how to that and the name of the file that is generated its name is to be based on timestamp. Lets assume that name of the output file would be 'output06282012' ...
The simple answer is:
String fileName = "output" + new Date().getTime();
You then go on to say:
.... and that text file should contain the same information that is shown on the console
You've got two choices:
You can change where System.out goes to by calling System.setOut(...). (Check the javadoc for details.)
You can create a PrintWriter or PrintStream wrapper for your file stream and write to that instead of writing to System.out.
In my opinion, it is a bad idea to use System.setOut(...) unless you've got no choice. It is a "global action" that affects the entire application. It is better to pass the writer that you want to use as a parameter ...
could you please sow in code as I have done that will clear the understanding
Sorry, I don't write people's programs for them (unless it is an interesting problem!). You need to write and debug the code yourself, using the information provided in the relevant javadocs. You can find the Java documentation online on the Oracle website: http://docs.oracle.com/javase/7/docs/
Can we rename a file say test.txt to test1.txt ?
If test1.txt exists will it rename ?
How do I rename it to the already existing test1.txt file so the new contents of test.txt are added to it for later use?
Copied from http://exampledepot.8waytrips.com/egs/java.io/RenameFile.html
// File (or directory) with old name
File file = new File("oldname");
// File (or directory) with new name
File file2 = new File("newname");
if (file2.exists())
throw new java.io.IOException("file exists");
// Rename file (or directory)
boolean success = file.renameTo(file2);
if (!success) {
// File was not successfully renamed
}
To append to the new file:
java.io.FileWriter out= new java.io.FileWriter(file2, true /*append=yes*/);
In short:
Files.move(source, source.resolveSibling("newname"));
More detail:
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;
import java.nio.file.StandardCopyOption;
The following is copied directly from http://docs.oracle.com/javase/7/docs/api/index.html:
Suppose we want to rename a file to "newname", keeping the file in the same directory:
Path source = Paths.get("path/here");
Files.move(source, source.resolveSibling("newname"));
Alternatively, suppose we want to move a file to new directory, keeping the same file name, and replacing any existing file of that name in the directory:
Path source = Paths.get("from/path");
Path newdir = Paths.get("to/path");
Files.move(source, newdir.resolve(source.getFileName()), StandardCopyOption.REPLACE_EXISTING);
You want to utilize the renameTo method on a File object.
First, create a File object to represent the destination. Check to see if that file exists. If it doesn't exist, create a new File object for the file to be moved. call the renameTo method on the file to be moved, and check the returned value from renameTo to see if the call was successful.
If you want to append the contents of one file to another, there are a number of writers available. Based on the extension, it sounds like it's plain text, so I would look at the FileWriter.
For Java 1.6 and lower, I believe the safest and cleanest API for this is Guava's Files.move.
Example:
File newFile = new File(oldFile.getParent(), "new-file-name.txt");
Files.move(oldFile.toPath(), newFile.toPath());
The first line makes sure that the location of the new file is the same directory, i.e. the parent directory of the old file.
EDIT:
I wrote this before I started using Java 7, which introduced a very similar approach. So if you're using Java 7+, you should see and upvote kr37's answer.
Renaming the file by moving it to a new name. (FileUtils is from Apache Commons IO lib)
String newFilePath = oldFile.getAbsolutePath().replace(oldFile.getName(), "") + newName;
File newFile = new File(newFilePath);
try {
FileUtils.moveFile(oldFile, newFile);
} catch (IOException e) {
e.printStackTrace();
}
This is an easy way to rename a file:
File oldfile =new File("test.txt");
File newfile =new File("test1.txt");
if(oldfile.renameTo(newfile)){
System.out.println("File renamed");
}else{
System.out.println("Sorry! the file can't be renamed");
}
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;
import static java.nio.file.StandardCopyOption.*;
Path yourFile = Paths.get("path_to_your_file\text.txt");
Files.move(yourFile, yourFile.resolveSibling("text1.txt"));
To replace an existing file with the name "text1.txt":
Files.move(yourFile, yourFile.resolveSibling("text1.txt"),REPLACE_EXISTING);
Try This
File file=new File("Your File");
boolean renameResult = file.renameTo(new File("New Name"));
// todo: check renameResult
Note :
We should always check the renameTo return value to make sure rename file is successful because it’s platform dependent(different Operating system, different file system) and it doesn’t throw IO exception if rename fails.
Yes, you can use File.renameTo(). But remember to have the correct path while renaming it to a new file.
import java.util.Arrays;
import java.util.List;
public class FileRenameUtility {
public static void main(String[] a) {
System.out.println("FileRenameUtility");
FileRenameUtility renameUtility = new FileRenameUtility();
renameUtility.fileRename("c:/Temp");
}
private void fileRename(String folder){
File file = new File(folder);
System.out.println("Reading this "+file.toString());
if(file.isDirectory()){
File[] files = file.listFiles();
List<File> filelist = Arrays.asList(files);
filelist.forEach(f->{
if(!f.isDirectory() && f.getName().startsWith("Old")){
System.out.println(f.getAbsolutePath());
String newName = f.getAbsolutePath().replace("Old","New");
boolean isRenamed = f.renameTo(new File(newName));
if(isRenamed)
System.out.println(String.format("Renamed this file %s to %s",f.getName(),newName));
else
System.out.println(String.format("%s file is not renamed to %s",f.getName(),newName));
}
});
}
}
}
If it's just renaming the file, you can use File.renameTo().
In the case where you want to append the contents of the second file to the first, take a look at FileOutputStream with the append constructor option or The same thing for FileWriter. You'll need to read the contents of the file to append and write them out using the output stream/writer.
As far as I know, renaming a file will not append its contents to that of an existing file with the target name.
About renaming a file in Java, see the documentation for the renameTo() method in class File.
Files.move(file.toPath(), fileNew.toPath());
works, but only when you close (or autoclose) ALL used resources (InputStream, FileOutputStream etc.) I think the same situation with file.renameTo or FileUtils.moveFile.
Here is my code to rename multiple files in a folder successfully:
public static void renameAllFilesInFolder(String folderPath, String newName, String extension) {
if(newName == null || newName.equals("")) {
System.out.println("New name cannot be null or empty");
return;
}
if(extension == null || extension.equals("")) {
System.out.println("Extension cannot be null or empty");
return;
}
File dir = new File(folderPath);
int i = 1;
if (dir.isDirectory()) { // make sure it's a directory
for (final File f : dir.listFiles()) {
try {
File newfile = new File(folderPath + "\\" + newName + "_" + i + "." + extension);
if(f.renameTo(newfile)){
System.out.println("Rename succesful: " + newName + "_" + i + "." + extension);
} else {
System.out.println("Rename failed");
}
i++;
} catch (Exception e) {
e.printStackTrace();
}
}
}
}
and run it for an example:
renameAllFilesInFolder("E:\\Downloads\\Foldername", "my_avatar", "gif");
I do not like java.io.File.renameTo(...) because sometimes it does not renames the file and you do not know why! It just returns true of false. It does not thrown an exception if it fails.
On the other hand, java.nio.file.Files.move(...) is more useful as it throws an exception when it fails.
Running code is here.
private static void renameFile(File fileName) {
FileOutputStream fileOutputStream =null;
BufferedReader br = null;
FileReader fr = null;
String newFileName = "yourNewFileName"
try {
fileOutputStream = new FileOutputStream(newFileName);
fr = new FileReader(fileName);
br = new BufferedReader(fr);
String sCurrentLine;
while ((sCurrentLine = br.readLine()) != null) {
fileOutputStream.write(("\n"+sCurrentLine).getBytes());
}
fileOutputStream.flush();
} catch (IOException e) {
e.printStackTrace();
} finally {
try {
fileOutputStream.close();
if (br != null)
br.close();
if (fr != null)
fr.close();
} catch (IOException ex) {
ex.printStackTrace();
}
}
}