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Remove filename extension of a file list in Java [duplicate]

What's the most efficient way to trim the suffix in Java, like this:
title part1.txt
title part2.html
=>
title part1
title part2

This is the sort of code that we shouldn't be doing ourselves. Use libraries for the mundane stuff, save your brain for the hard stuff.
In this case, I recommend using FilenameUtils.removeExtension() from Apache Commons IO

str.substring(0, str.lastIndexOf('.'))

As using the String.substring and String.lastIndex in a one-liner is good, there are some issues in terms of being able to cope with certain file paths.
Take for example the following path:
a.b/c
Using the one-liner will result in:
a
That's incorrect.
The result should have been c, but since the file lacked an extension, but the path had a directory with a . in the name, the one-liner method was tricked into giving part of the path as the filename, which is not correct.
Need for checks
Inspired by skaffman's answer, I took a look at the FilenameUtils.removeExtension method of the Apache Commons IO.
In order to recreate its behavior, I wrote a few tests the new method should fulfill, which are the following:
Path Filename
-------------- --------
a/b/c c
a/b/c.jpg c
a/b/c.jpg.jpg c.jpg
a.b/c c
a.b/c.jpg c
a.b/c.jpg.jpg c.jpg
c c
c.jpg c
c.jpg.jpg c.jpg
(And that's all I've checked for -- there probably are other checks that should be in place that I've overlooked.)
The implementation
The following is my implementation for the removeExtension method:
public static String removeExtension(String s) {
String separator = System.getProperty("file.separator");
String filename;
// Remove the path upto the filename.
int lastSeparatorIndex = s.lastIndexOf(separator);
if (lastSeparatorIndex == -1) {
filename = s;
} else {
filename = s.substring(lastSeparatorIndex + 1);
}
// Remove the extension.
int extensionIndex = filename.lastIndexOf(".");
if (extensionIndex == -1)
return filename;
return filename.substring(0, extensionIndex);
}
Running this removeExtension method with the above tests yield the results listed above.
The method was tested with the following code. As this was run on Windows, the path separator is a \ which must be escaped with a \ when used as part of a String literal.
System.out.println(removeExtension("a\\b\\c"));
System.out.println(removeExtension("a\\b\\c.jpg"));
System.out.println(removeExtension("a\\b\\c.jpg.jpg"));
System.out.println(removeExtension("a.b\\c"));
System.out.println(removeExtension("a.b\\c.jpg"));
System.out.println(removeExtension("a.b\\c.jpg.jpg"));
System.out.println(removeExtension("c"));
System.out.println(removeExtension("c.jpg"));
System.out.println(removeExtension("c.jpg.jpg"));
The results were:
c
c
c.jpg
c
c
c.jpg
c
c
c.jpg
The results are the desired results outlined in the test the method should fulfill.

String foo = "title part1.txt";
foo = foo.substring(0, foo.lastIndexOf('.'));

BTW, in my case, when I wanted a quick solution to remove a specific extension, this is approximately what I did:
if (filename.endsWith(ext))
return filename.substring(0,filename.length() - ext.length());
else
return filename;

Use a method in com.google.common.io.Files class if your project is already dependent on Google core library. The method you need is getNameWithoutExtension.

you can try this function , very basic
public String getWithoutExtension(String fileFullPath){
return fileFullPath.substring(0, fileFullPath.lastIndexOf('.'));
}

String fileName="foo.bar";
int dotIndex=fileName.lastIndexOf('.');
if(dotIndex>=0) { // to prevent exception if there is no dot
fileName=fileName.substring(0,dotIndex);
}
Is this a trick question? :p
I can't think of a faster way atm.

I found coolbird's answer particularly useful.
But I changed the last result statements to:
if (extensionIndex == -1)
return s;
return s.substring(0, lastSeparatorIndex+1)
+ filename.substring(0, extensionIndex);
as I wanted the full path name to be returned.
So "C:\Users\mroh004.COM\Documents\Test\Test.xml" becomes
"C:\Users\mroh004.COM\Documents\Test\Test" and not
"Test"

filename.substring(filename.lastIndexOf('.'), filename.length()).toLowerCase();

Use a regex. This one replaces the last dot, and everything after it.
String baseName = fileName.replaceAll("\\.[^.]*$", "");
You can also create a Pattern object if you want to precompile the regex.

If you use Spring you could use
org.springframework.util.StringUtils.stripFilenameExtension(String path)
Strip the filename extension from the given Java resource path, e.g.
"mypath/myfile.txt" -> "mypath/myfile".
Params: path – the file path
Returns: the path with stripped filename extension

private String trimFileExtension(String fileName)
{
String[] splits = fileName.split( "\\." );
return StringUtils.remove( fileName, "." + splits[splits.length - 1] );
}

String[] splitted = fileName.split(".");
String fileNameWithoutExtension = fileName.replace("." + splitted[splitted.length - 1], "");

create a new file with string image path
String imagePath;
File test = new File(imagePath);
test.getName();
test.getPath();
getExtension(test.getName());
public static String getExtension(String uri) {
if (uri == null) {
return null;
}
int dot = uri.lastIndexOf(".");
if (dot >= 0) {
return uri.substring(dot);
} else {
// No extension.
return "";
}
}

org.apache.commons.io.FilenameUtils version 2.4 gives the following answer
public static String removeExtension(String filename) {
if (filename == null) {
return null;
}
int index = indexOfExtension(filename);
if (index == -1) {
return filename;
} else {
return filename.substring(0, index);
}
}
public static int indexOfExtension(String filename) {
if (filename == null) {
return -1;
}
int extensionPos = filename.lastIndexOf(EXTENSION_SEPARATOR);
int lastSeparator = indexOfLastSeparator(filename);
return lastSeparator > extensionPos ? -1 : extensionPos;
}
public static int indexOfLastSeparator(String filename) {
if (filename == null) {
return -1;
}
int lastUnixPos = filename.lastIndexOf(UNIX_SEPARATOR);
int lastWindowsPos = filename.lastIndexOf(WINDOWS_SEPARATOR);
return Math.max(lastUnixPos, lastWindowsPos);
}
public static final char EXTENSION_SEPARATOR = '.';
private static final char UNIX_SEPARATOR = '/';
private static final char WINDOWS_SEPARATOR = '\\';

The best what I can write trying to stick to the Path class:
Path removeExtension(Path path) {
return path.resolveSibling(path.getFileName().toString().replaceFirst("\\.[^.]*$", ""));
}

dont do stress on mind guys. i did already many times. just copy paste this public static method in your staticUtils library for future uses ;-)
static String removeExtension(String path){
String filename;
String foldrpath;
String filenameWithoutExtension;
if(path.equals("")){return "";}
if(path.contains("\\")){ // direct substring method give wrong result for "a.b.c.d\e.f.g\supersu"
filename = path.substring(path.lastIndexOf("\\"));
foldrpath = path.substring(0, path.lastIndexOf('\\'));;
if(filename.contains(".")){
filenameWithoutExtension = filename.substring(0, filename.lastIndexOf('.'));
}else{
filenameWithoutExtension = filename;
}
return foldrpath + filenameWithoutExtension;
}else{
return path.substring(0, path.lastIndexOf('.'));
}
}

I would do like this:
String title_part = "title part1.txt";
int i;
for(i=title_part.length()-1 ; i>=0 && title_part.charAt(i)!='.' ; i--);
title_part = title_part.substring(0,i);
Starting to the end till the '.' then call substring.
Edit:
Might not be a golf but it's effective :)

Keeping in mind the scenarios where there is no file extension or there is more than one file extension
example Filename : file | file.txt | file.tar.bz2
/**
*
* #param fileName
* #return file extension
* example file.fastq.gz => fastq.gz
*/
private String extractFileExtension(String fileName) {
String type = "undefined";
if (FilenameUtils.indexOfExtension(fileName) != -1) {
String fileBaseName = FilenameUtils.getBaseName(fileName);
int indexOfExtension = -1;
while (fileBaseName.contains(".")) {
indexOfExtension = FilenameUtils.indexOfExtension(fileBaseName);
fileBaseName = FilenameUtils.getBaseName(fileBaseName);
}
type = fileName.substring(indexOfExtension + 1, fileName.length());
}
return type;
}

String img = "example.jpg";
// String imgLink = "http://www.example.com/example.jpg";
URI uri = null;
try {
uri = new URI(img);
String[] segments = uri.getPath().split("/");
System.out.println(segments[segments.length-1].split("\\.")[0]);
} catch (Exception e) {
e.printStackTrace();
}
This will output example for both img and imgLink

private String trimFileName(String fileName)
{
String[] ext;
ext = fileName.split("\\.");
return fileName.replace(ext[ext.length - 1], "");
}
This code will spilt the file name into parts where ever it has " . ", For eg. If the file name is file-name.hello.txt then it will be spilted into string array as , { "file-name", "hello", "txt" }. So anyhow the last element in this string array will be the file extension of that particular file , so we can simply find the last element of any arrays with arrayname.length - 1, so after we get to know the last element, we can just replace the file extension with an empty string in that file name. Finally this will return file-name.hello. , if you want to remove also the last period then you can add the string with only period to the last element of string array in the return line. Which should look like,
return fileName.replace("." + ext[ext.length - 1], "");

public static String removeExtension(String file) {
if(file != null && file.length() > 0) {
while(file.contains(".")) {
file = file.substring(0, file.lastIndexOf('.'));
}
}
return file;
}

ini4j store method changes the comment character

I want to change the entry of a key in a section of an ini file. I use the ini4j library. So I wrote the code below. I can change the entry but there are also other changes which I don't want:
replacement of ";" with "#" which indicates comment lines
addition of blank lines between sections and comments
So how can I solve it?
this is what I expected:
[section1]
key1=40
key2=30
[section2]
key1=10
key2=20
;section3
[section3]
key1=10
key2=20
this is the file after editing:
[section1]
key1=40
key2=30
[section2]
key1=10
key2=20
#section3
[section3]
key1=10
key2=20
My code:
public static void setEntry(String filePath, String fileName, String sectionName, String keyName, String entry)
throws IOException {
String path = filePath.concat(fileName);
FileInputStream inputStream = null;
try {
inputStream = new FileInputStream(path);
Ini ini = new Ini(inputStream);
ini.getConfig().setStrictOperator(true);
Section section = ini.get(sectionName);
if (section != null) {
if (section.containsKey(keyName)) {
section.put(keyName, entry);
}
else {
section.add(keyName, entry);
}
}
else {
section = ini.add(sectionName);
section.add(keyName, entry);
}
File iniFile = new File(path);
ini.store(iniFile);
}
catch (IOException e) {
e.printStackTrace();
}
finally {
inputStream.close();
}
}
Is there a way to change default comment character?

Apparently, even when the ini4j can read comments with ; and #, it doesn't save this when you have to write them to an ini file.
If you check the AbstractFormatter class of the ini4j project, you can see that the only comment operator that it matters when writes is #, if you have access to the source code, you can change it with your custom comment operator and should work.
abstract class AbstractFormatter implements HandlerBase
{
private static final char OPERATOR = '=';
private static final char COMMENT = '#'; // <--- change this to ';'
...
As for the empty lines, they are just aesthetics. Again, if you have access to the source code, you can edit it by yourself (commenting this and this lines?), but I guess is easier if you just re-format your ini file after being generated by this library and remove all empty lines.

Getting variables from another .java file using reflection

I've managed to get reflection working by getting and formatting the variables in the class that the toString() method is in.
public class ReadFile {
public int test1 =0;
public String test2 = "hello";
Boolean test3 = false;
int test4 = 1;
public static void main(String[] args) throws IOException{
ReadFile test = new ReadFile();
System.out.println(test);
}
public String toString(){
//Make a string builder so we can build up a string
StringBuilder result = new StringBuilder();
//Declare a new line constant
final String NEW_LINE = System.getProperty("line.separator");
//Gets the name of THIS Object
result.append(this.getClass().getName() );
result.append(" Class {" );
result.append(NEW_LINE);
//Determine fields declared in this class only (no fields of superclass)
Field[] fields = this.getClass().getDeclaredFields();
//Print field names paired with their values
for ( Field field : fields ) {
result.append(" ");
try {
result.append(field.getType() + " ");
result.append( field.getName() );
result.append(": ");
//requires access to private field:
result.append( field.get(this) );
} catch ( IllegalAccessException ex ) {
System.out.println(ex);
}
result.append(NEW_LINE);
}
result.append("}");
return result.toString();
}
}
However I was wondering whether it would be possible to specify a specific file in the directory for the toString() to work on?
I have tried getting a file and plugging it in the System.out.println() but the way I see it is you need to make an instance of a class and give it the instance for it to work. So I'm not sure how that can be done programatically.
I have been trying something like this:
Path path = FileSystems.getDefault().getPath("D:\\Directory\\Foo\\Bar\\Test.java", args);
File file = path.toFile();
System.out.println(file);
However I don't get very far with it, I've mainly been seeing if I can convert the file into anything usable but I'm not sure what I need to be doing!
Any advice would be great.

I think you need to look into the ClassLoader API - you need to get an new URLClassLoader and ask it to load your .java file into the JVM. You can then reflect on it.

You can try to read the package information from the file (D:\Directory\Foo\Bar\Test.java) and than try to load it the class by its name:
Class.forName(nameOfTheClass)
Java API Class

How do I get the file name from a String containing the Absolute file path?

String variable contains a file name, C:\Hello\AnotherFolder\The File Name.PDF. How do I only get the file name The File Name.PDF as a String?
I planned to split the string, but that is not the optimal solution.

just use File.getName()
File f = new File("C:\\Hello\\AnotherFolder\\The File Name.PDF");
System.out.println(f.getName());
using String methods:
File f = new File("C:\\Hello\\AnotherFolder\\The File Name.PDF");
System.out.println(f.getAbsolutePath().substring(f.getAbsolutePath().lastIndexOf("\\")+1));

Alternative using Path (Java 7+):
Path p = Paths.get("C:\\Hello\\AnotherFolder\\The File Name.PDF");
String file = p.getFileName().toString();
Note that splitting the string on \\ is platform dependent as the file separator might vary. Path#getName takes care of that issue for you.

Using FilenameUtils in Apache Commons IO :
String name1 = FilenameUtils.getName("/ab/cd/xyz.txt");
String name2 = FilenameUtils.getName("c:\\ab\\cd\\xyz.txt");

Considering the String you're asking about is
C:\Hello\AnotherFolder\The File Name.PDF
we need to extract everything after the last separator, ie. \. That is what we are interested in.
You can do
String fullPath = "C:\\Hello\\AnotherFolder\\The File Name.PDF";
int index = fullPath.lastIndexOf("\\");
String fileName = fullPath.substring(index + 1);
This will retrieve the index of the last \ in your String and extract everything that comes after it into fileName.
If you have a String with a different separator, adjust the lastIndexOf to use that separator. (There's even an overload that accepts an entire String as a separator.)
I've omitted it in the example above, but if you're unsure where the String comes from or what it might contain, you'll want to validate that the lastIndexOf returns a non-negative value because the Javadoc states it'll return
-1 if there is no such occurrence

Since 1.7
Path p = Paths.get("c:\\temp\\1.txt");
String fileName = p.getFileName().toString();
String directory = p.getParent().toString();

you can use path = C:\Hello\AnotherFolder\TheFileName.PDF
String strPath = path.substring(path.lastIndexOf("\\")+1, path.length());

The other answers didn't quite work for my specific scenario, where I am reading paths that have originated from an OS different to my current one. To elaborate I am saving email attachments saved from a Windows platform on a Linux server. The filename returned from the JavaMail API is something like 'C:\temp\hello.xls'
The solution I ended up with:
String filenameWithPath = "C:\\temp\\hello.xls";
String[] tokens = filenameWithPath.split("[\\\\|/]");
String filename = tokens[tokens.length - 1];

Considere the case that Java is Multiplatform:
int lastPath = fileName.lastIndexOf(File.separator);
if (lastPath!=-1){
fileName = fileName.substring(lastPath+1);
}

getFileName() method of java.nio.file.Path used to return the name of the file or directory pointed by this path object.
Path getFileName()
For reference:
https://www.geeksforgeeks.org/path-getfilename-method-in-java-with-examples/

A method without any dependency and takes care of .. , . and duplicate separators.
public static String getFileName(String filePath) {
if( filePath==null || filePath.length()==0 )
return "";
filePath = filePath.replaceAll("[/\\\\]+", "/");
int len = filePath.length(),
upCount = 0;
while( len>0 ) {
//remove trailing separator
if( filePath.charAt(len-1)=='/' ) {
len--;
if( len==0 )
return "";
}
int lastInd = filePath.lastIndexOf('/', len-1);
String fileName = filePath.substring(lastInd+1, len);
if( fileName.equals(".") ) {
len--;
}
else if( fileName.equals("..") ) {
len -= 2;
upCount++;
}
else {
if( upCount==0 )
return fileName;
upCount--;
len -= fileName.length();
}
}
return "";
}
Test case:
#Test
public void testGetFileName() {
assertEquals("", getFileName("/"));
assertEquals("", getFileName("////"));
assertEquals("", getFileName("//C//.//../"));
assertEquals("", getFileName("C//.//../"));
assertEquals("C", getFileName("C"));
assertEquals("C", getFileName("/C"));
assertEquals("C", getFileName("/C/"));
assertEquals("C", getFileName("//C//"));
assertEquals("C", getFileName("/A/B/C/"));
assertEquals("C", getFileName("/A/B/C"));
assertEquals("C", getFileName("/C/./B/../"));
assertEquals("C", getFileName("//C//./B//..///"));
assertEquals("user", getFileName("/user/java/.."));
assertEquals("C:", getFileName("C:"));
assertEquals("C:", getFileName("/C:"));
assertEquals("java", getFileName("C:\\Program Files (x86)\\java\\bin\\.."));
assertEquals("C.ext", getFileName("/A/B/C.ext"));
assertEquals("C.ext", getFileName("C.ext"));
}
Maybe getFileName is a bit confusing, because it returns directory names also. It returns the name of file or last directory in a path.

extract file name using java regex *.
public String extractFileName(String fullPathFile){
try {
Pattern regex = Pattern.compile("([^\\\\/:*?\"<>|\r\n]+$)");
Matcher regexMatcher = regex.matcher(fullPathFile);
if (regexMatcher.find()){
return regexMatcher.group(1);
}
} catch (PatternSyntaxException ex) {
LOG.info("extractFileName::pattern problem <"+fullPathFile+">",ex);
}
return fullPathFile;
}

You can use FileInfo object to get all information of your file.
FileInfo f = new FileInfo(#"C:\Hello\AnotherFolder\The File Name.PDF");
MessageBox.Show(f.Name);
MessageBox.Show(f.FullName);
MessageBox.Show(f.Extension );
MessageBox.Show(f.DirectoryName);

This answer works for me in c#:
using System.IO;
string fileName = Path.GetFileName("C:\Hello\AnotherFolder\The File Name.PDF");

What is the best way to generate a unique and short file name in Java

I don't necessarily want to use UUIDs since they are fairly long.
The file just needs to be unique within its directory.
One thought which comes to mind is to use File.createTempFile(String prefix, String suffix), but that seems wrong because the file is not temporary.
The case of two files created in the same millisecond needs to be handled.

Well, you could use the 3-argument version: File.createTempFile(String prefix, String suffix, File directory) which will let you put it where you'd like. Unless you tell it to, Java won't treat it differently than any other file. The only drawback is that the filename is guaranteed to be at least 8 characters long (minimum of 3 characters for the prefix, plus 5 or more characters generated by the function).
If that's too long for you, I suppose you could always just start with the filename "a", and loop through "b", "c", etc until you find one that doesn't already exist.

I'd use Apache Commons Lang library (http://commons.apache.org/lang).
There is a class org.apache.commons.lang.RandomStringUtils that can be used to generate random strings of given length. Very handy not only for filename generation!
Here is the example:
String ext = "dat";
File dir = new File("/home/pregzt");
String name = String.format("%s.%s", RandomStringUtils.randomAlphanumeric(8), ext);
File file = new File(dir, name);

I use the timestamp
i.e
new File( simpleDateFormat.format( new Date() ) );
And have the simpleDateFormat initialized to something like as:
new SimpleDateFormat("File-ddMMyy-hhmmss.SSS.txt");
EDIT
What about
new File(String.format("%s.%s", sdf.format( new Date() ),
random.nextInt(9)));
Unless the number of files created in the same second is too high.
If that's the case and the name doesn't matters
new File( "file."+count++ );
:P

This works for me:
String generateUniqueFileName() {
String filename = "";
long millis = System.currentTimeMillis();
String datetime = new Date().toGMTString();
datetime = datetime.replace(" ", "");
datetime = datetime.replace(":", "");
String rndchars = RandomStringUtils.randomAlphanumeric(16);
filename = rndchars + "_" + datetime + "_" + millis;
return filename;
}
// USE:
String newFile;
do{
newFile=generateUniqueFileName() + "." + FileExt;
}
while(new File(basePath+newFile).exists());
Output filenames should look like :
2OoBwH8OwYGKW2QE_4Sep2013061732GMT_1378275452253.Ext

Look at the File javadoc, the method createNewFile will create the file only if it doesn't exist, and will return a boolean to say if the file was created.
You may also use the exists() method:
int i = 0;
String filename = Integer.toString(i);
File f = new File(filename);
while (f.exists()) {
i++;
filename = Integer.toString(i);
f = new File(filename);
}
f.createNewFile();
System.out.println("File in use: " + f);

If you have access to a database, you can create and use a sequence in the file name.
select mySequence.nextval from dual;
It will be guaranteed to be unique and shouldn't get too large (unless you are pumping out a ton of files).

//Generating Unique File Name
public String getFileName() {
String timeStamp = new SimpleDateFormat("yyyy-MM-dd_HH:mm:ss").format(new Date());
return "PNG_" + timeStamp + "_.png";
}

I use current milliseconds with random numbers
i.e
Random random=new Random();
String ext = ".jpeg";
File dir = new File("/home/pregzt");
String name = String.format("%s%s",System.currentTimeMillis(),random.nextInt(100000)+ext);
File file = new File(dir, name);

Combining other answers, why not use the ms timestamp with a random value appended; repeat until no conflict, which in practice will be almost never.
For example: File-ccyymmdd-hhmmss-mmm-rrrrrr.txt

Why not just use something based on a timestamp..?

Problem is synchronization. Separate out regions of conflict.
Name the file as : (server-name)_(thread/process-name)_(millisecond/timestamp).(extension)
example : aws1_t1_1447402821007.png

How about generate based on time stamp rounded to the nearest millisecond, or whatever accuracy you need... then use a lock to synchronize access to the function.
If you store the last generated file name, you can append sequential letters or further digits to it as needed to make it unique.
Or if you'd rather do it without locks, use a time step plus a thread ID, and make sure that the function takes longer than a millisecond, or waits so that it does.

It looks like you've got a handful of solutions for creating a unique filename, so I'll leave that alone. I would test the filename this way:
String filePath;
boolean fileNotFound = true;
while (fileNotFound) {
String testPath = generateFilename();
try {
RandomAccessFile f = new RandomAccessFile(
new File(testPath), "r");
} catch (Exception e) {
// exception thrown by RandomAccessFile if
// testPath doesn't exist (ie: it can't be read)
filePath = testPath;
fileNotFound = false;
}
}
//now create your file with filePath

This also works
String logFileName = new SimpleDateFormat("yyyyMMddHHmm'.txt'").format(new Date());
logFileName = "loggerFile_" + logFileName;

I understand that I am too late to reply on this question. But I think I should put this as it seems something different from other solution.
We can concatenate threadname and current timeStamp as file name. But with this there is one issue like some thread name contains special character like "\" which can create problem in creating file name. So we can remove special charater from thread name and then concatenate thread name and time stamp
fileName = threadName(after removing special charater) + currentTimeStamp

Why not use synchronized to process multi thread.
here is my solution,It's can generate a short file name , and it's unique.
private static synchronized String generateFileName(){
String name = make(index);
index ++;
return name;
}
private static String make(int index) {
if(index == 0) return "";
return String.valueOf(chars[index % chars.length]) + make(index / chars.length);
}
private static int index = 1;
private static char[] chars = {'a','b','c','d','e','f','g',
'h','i','j','k','l','m','n',
'o','p','q','r','s','t',
'u','v','w','x','y','z'};
blew is main function for test , It's work.
public static void main(String[] args) {
List<String> names = new ArrayList<>();
List<Thread> threads = new ArrayList<>();
for (int i = 0; i < 100; i++) {
Thread thread = new Thread(new Runnable() {
#Override
public void run() {
for (int i = 0; i < 1000; i++) {
String name = generateFileName();
names.add(name);
}
}
});
thread.run();
threads.add(thread);
}
for (int i = 0; i < 10; i++) {
try {
threads.get(i).join();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
System.out.println(names);
System.out.println(names.size());
}