How to Increment filename if file exists

Question

How to increment filename if the file already exists? Here's the code that I am using -

int num = 0;

String save = at.getText().toString() + ".jpg";

File file = new File(myDir, save); 

if (file.exists()) {
    save = at.getText().toString() + num +".jpg";

    file = new File(myDir, save); 
    num++;
}

This code works but only 2 files are saved like file.jpg and file2.jpg

Answer 1

This problem is always initializative num = 0 so if file exists, it save file0.jpg and not check whether file0.jpg is exists ? So, To code work. You should check until available :

int num = 0;
String save = at.getText().toString() + ".jpg";
File file = new File(myDir, save);
while(file.exists()) {
    save = at.getText().toString() + (num++) +".jpg";
    file = new File(myDir, save); 
}

Answer 2

Try this:

File file = new File(myDir, at.getText().toString() + ".jpg"); 

for (int num = 0; file.exists(); num++) {
    file = new File(myDir, at.getText().toString() + num + ".jpg");
}

// Now save/use your file here

Answer 3

In addition to the first answer, I made some more changes

private File getUniqueFileName(String folderName, String searchedFilename) {
        int num = 1;
        String extension = getExtension(searchedFilename);
        String filename = searchedFilename.substring(0, searchedFilename.lastIndexOf("."));
        File file = new File(folderName, searchedFilename);
        while (file.exists()) {
            searchedFilename = filename + "("+(num++)+")"+extension;
            file = new File(folderName, searchedFilename);
        }
        return file;
    }

Answer 4

int i = 0;
String save = at.getText().toString();
String filename = save +".jpg";
File f = new File(filename);
while (f.exists()) {
    i++;
    filename =save+ Integer.toString(i)+".jpg";
    f = new File(filename);
}
f.createNewFile();

Answer 5

This function return the Exactly new file with increment number for all kind of extensions

Helpful for others..

private File getFileName(File file) {
        if (file.exists()){
            String newFileName = file.getName();
            String simpleName = file.getName().substring(0,newFileName.indexOf("."));
            String strDigit="";

            try {
                simpleName = (Integer.parseInt(simpleName)+1+"");
                File newFile = new File(file.getParent()+"/"+simpleName+getExtension(file.getName()));
                return getFileName(newFile);
            }catch (Exception e){}

            for (int i=simpleName.length()-1;i>=0;i--){
                if (!Character.isDigit(simpleName.charAt(i))){
                    strDigit = simpleName.substring(i+1);
                    simpleName = simpleName.substring(0,i+1);
                    break;
                }
            }

            if (strDigit.length()>0){
                simpleName = simpleName+(Integer.parseInt(strDigit)+1);
            }else {
                simpleName+="1";
            }

            File newFile = new File(file.getParent()+"/"+simpleName+getExtension(file.getName()));
            return getFileName(newFile);
        }
        return file;
    }

private String getExtension(String name) {
        return name.substring(name.lastIndexOf("."));
}

Answer 6

You can avoid the code repetition of some of the answers here by using a do while loop

Here's an example using the newer NIO Path API introduced in Java 7

Path candidate = null;
int counter = 0;
do {
    candidate = Paths.get(String.format("%s-%d",
            path.toString(), ++counter));
} while (Files.exists(candidate));
Files.createFile(candidate);

Answer 7

Having needing to solve this problem in my own code I took @Tejas Trivedi 's answer made it work like windows when you happen to download the same file several times

// This function will iteratively to find a unique file name to use when given a file: example (###).txt
//  More or less how windows will save a new file when one already exists example.txt becomes example (1).txt
//  if example.txt already exists
private File getUniqueFileName(File file) {
    File originalFile = file;
    try {
        while (file.exists()) {
            String newFileName = file.getName();
            String baseName = newFileName.substring(0, newFileName.lastIndexOf("."));
            String extension = getExtension(newFileName);

            Pattern pattern = Pattern.compile("( \\(\\d+\\))\\."); // Find ' (###).' in the file name, if it exists
            Matcher matcher = pattern.matcher(newFileName);

            String strDigits = "";
            if (matcher.find()) {
                baseName = baseName.substring(0, matcher.start(0)); // remove the (###)
                strDigits = matcher.group(0); // grab the ### we'll want to increment
                strDigits = strDigits.substring(strDigits.indexOf("(") + 1, strDigits.lastIndexOf(")")); // strip off the ' (' and ').' from the match
                // increment the found digit and convert it back to a string
                int count = Integer.parseInt(strDigits);
                strDigits = Integer.toString(count + 1);
            } else {
                strDigits = "1"; // if there is no (###) match then start with 1
            }
            file = new File(file.getParent() + "/" + baseName + " (" + strDigits + ")" + extension); // put the pieces back together
        }
        return file;
    } catch (Error e) {
        return originalFile; // Just overwrite the original file at this point...
    }

}

private String getExtension(String name) {
    return name.substring(name.lastIndexOf("."));
}

Calling getUniqueFileName(new File('/dir/example.txt') when 'example.txt' already exists while generate a new File targeting '/dir/example (1).txt' if that too exists it'll just keep incrementing number between the parentheses until a unique file is found, if an error happens, it'll just give the original file name.

I hope this helps some one needing to generate a unique file in Java on Android or another platform.

Answer 8

Kotlin version:

private fun checkAndRenameIfExists(name: String): File {
    var filename = name
    val extension = "pdf"
    val root = Environment.getExternalStorageDirectory().absolutePath
    var file = File(root, "$filename.$extension")
    var n = 0
    while (file.exists()) {
        n += 1
        filename = "$name($n)"
        file = File(root, appDirectoryName + File.separator + "$filename.$extension")
    }
    return file
}

Answer 9

This is the solution I use to handle this case. Works for folders as well as for files.

var destination = File(Environment.getExternalStoragePublicDirectory(Environment.DIRECTORY_DOWNLOADS), "MyFolder")
 if (!destination.exists()){
     destination.mkdirs()
 } else {
     val numberOfFileAlreadyExist =
         destination.listFiles().filter { it.name.startsWith("MyFolder") }.size
     destination = File(
         Environment.getExternalStoragePublicDirectory(Environment.DIRECTORY_DOWNLOADS),
         "MyFolder(${numberOfFileAlreadyExist + 1})"
     )
     destination.mkdirs()
 }

Have nice day !

Answer 10

Another simple logic solution to get the unique file name under a directory using apache commons io using WildcardFileFilter to match the file name and get the no of exists with the given name and increment the counter.

public static String getUniqueFileName(String directory, String fileName){ String fName = fileName.substring(0, fileName.lastIndexOf(".")); Collection listFiles = FileUtils.listFiles(new File(directory), new WildcardFileFilter(fName+"*", IOCase.INSENSITIVE), DirectoryFileFilter.DIRECTORY); if(listFiles.isEmpty()){ return fName; } return fName.concat(" ("+listFiles.size()+")"); }