Related
What is the best way to check if a URL is valid in Java?
If tried to call new URL(urlString) and catch a MalformedURLException, but it seems to be happy with anything that begins with http://.
I'm not concerned about establishing a connection, just validity. Is there a method for this? An annotation in Hibernate Validator? Should I use a regex?
Edit: Some examples of accepted URLs are http://*** and http://my favorite site!.
Consider using the Apache Commons UrlValidator class
UrlValidator urlValidator = new UrlValidator();
urlValidator.isValid("http://my favorite site!");
There are several properties that you can set to control how this class behaves, by default http, https, and ftp are accepted.
Here is way I tried and found useful,
URL u = new URL(name); // this would check for the protocol
u.toURI(); // does the extra checking required for validation of URI
I'd love to post this as a comment to Tendayi Mawushe's answer, but I'm afraid there is not enough space ;)
This is the relevant part from the Apache Commons UrlValidator source:
/**
* This expression derived/taken from the BNF for URI (RFC2396).
*/
private static final String URL_PATTERN =
"/^(([^:/?#]+):)?(//([^/?#]*))?([^?#]*)(\\?([^#]*))?(#(.*))?/";
// 12 3 4 5 6 7 8 9
/**
* Schema/Protocol (ie. http:, ftp:, file:, etc).
*/
private static final int PARSE_URL_SCHEME = 2;
/**
* Includes hostname/ip and port number.
*/
private static final int PARSE_URL_AUTHORITY = 4;
private static final int PARSE_URL_PATH = 5;
private static final int PARSE_URL_QUERY = 7;
private static final int PARSE_URL_FRAGMENT = 9;
You can easily build your own validator from there.
The most "foolproof" way is to check for the availability of URL:
public boolean isURL(String url) {
try {
(new java.net.URL(url)).openStream().close();
return true;
} catch (Exception ex) { }
return false;
}
My favorite approach, without external libraries:
try {
URI uri = new URI(name);
// perform checks for scheme, authority, host, etc., based on your requirements
if ("mailto".equals(uri.getScheme()) {/*Code*/}
if (uri.getHost() == null) {/*Code*/}
} catch (URISyntaxException e) {
}
I didn't like any of the implementations (because they use a Regex which is an expensive operation, or a library which is an overkill if you only need one method), so I ended up using the java.net.URI class with some extra checks, and limiting the protocols to: http, https, file, ftp, mailto, news, urn.
And yes, catching exceptions can be an expensive operation, but probably not as bad as Regular Expressions:
final static Set<String> protocols, protocolsWithHost;
static {
protocolsWithHost = new HashSet<String>(
Arrays.asList( new String[]{ "file", "ftp", "http", "https" } )
);
protocols = new HashSet<String>(
Arrays.asList( new String[]{ "mailto", "news", "urn" } )
);
protocols.addAll(protocolsWithHost);
}
public static boolean isURI(String str) {
int colon = str.indexOf(':');
if (colon < 3) return false;
String proto = str.substring(0, colon).toLowerCase();
if (!protocols.contains(proto)) return false;
try {
URI uri = new URI(str);
if (protocolsWithHost.contains(proto)) {
if (uri.getHost() == null) return false;
String path = uri.getPath();
if (path != null) {
for (int i=path.length()-1; i >= 0; i--) {
if ("?<>:*|\"".indexOf( path.charAt(i) ) > -1)
return false;
}
}
}
return true;
} catch ( Exception ex ) {}
return false;
}
Judging by the source code for URI, the
public URL(URL context, String spec, URLStreamHandler handler)
constructor does more validation than the other constructors. You might try that one, but YMMV.
validator package:
There seems to be a nice package by Yonatan Matalon called UrlUtil. Quoting its API:
isValidWebPageAddress(java.lang.String address, boolean validateSyntax,
boolean validateExistance)
Checks if the given address is a valid web page address.
Sun's approach - check the network address
Sun's Java site offers connect attempt as a solution for validating URLs.
Other regex code snippets:
There are regex validation attempts at Oracle's site and weberdev.com.
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;
}
Just to be clear, I'm not looking for the MIME type.
Let's say I have the following input: /path/to/file/foo.txt
I'd like a way to break this input up, specifically into .txt for the extension. Is there any built in way to do this in Java? I would like to avoid writing my own parser.
In this case, use FilenameUtils.getExtension from Apache Commons IO
Here is an example of how to use it (you may specify either full path or just file name):
import org.apache.commons.io.FilenameUtils;
// ...
String ext1 = FilenameUtils.getExtension("/path/to/file/foo.txt"); // returns "txt"
String ext2 = FilenameUtils.getExtension("bar.exe"); // returns "exe"
Maven dependency:
<dependency>
<groupId>commons-io</groupId>
<artifactId>commons-io</artifactId>
<version>2.6</version>
</dependency>
Gradle Groovy DSL
implementation 'commons-io:commons-io:2.6'
Gradle Kotlin DSL
implementation("commons-io:commons-io:2.6")
Others https://search.maven.org/artifact/commons-io/commons-io/2.6/jar
Do you really need a "parser" for this?
String extension = "";
int i = fileName.lastIndexOf('.');
if (i > 0) {
extension = fileName.substring(i+1);
}
Assuming that you're dealing with simple Windows-like file names, not something like archive.tar.gz.
Btw, for the case that a directory may have a '.', but the filename itself doesn't (like /path/to.a/file), you can do
String extension = "";
int i = fileName.lastIndexOf('.');
int p = Math.max(fileName.lastIndexOf('/'), fileName.lastIndexOf('\\'));
if (i > p) {
extension = fileName.substring(i+1);
}
private String getFileExtension(File file) {
String name = file.getName();
int lastIndexOf = name.lastIndexOf(".");
if (lastIndexOf == -1) {
return ""; // empty extension
}
return name.substring(lastIndexOf);
}
If you use Guava library, you can resort to Files utility class. It has a specific method, getFileExtension(). For instance:
String path = "c:/path/to/file/foo.txt";
String ext = Files.getFileExtension(path);
System.out.println(ext); //prints txt
In addition you may also obtain the filename with a similar function, getNameWithoutExtension():
String filename = Files.getNameWithoutExtension(path);
System.out.println(filename); //prints foo
If on Android, you can use this:
String ext = android.webkit.MimeTypeMap.getFileExtensionFromUrl(file.getName());
This is a tested method
public static String getExtension(String fileName) {
char ch;
int len;
if(fileName==null ||
(len = fileName.length())==0 ||
(ch = fileName.charAt(len-1))=='/' || ch=='\\' || //in the case of a directory
ch=='.' ) //in the case of . or ..
return "";
int dotInd = fileName.lastIndexOf('.'),
sepInd = Math.max(fileName.lastIndexOf('/'), fileName.lastIndexOf('\\'));
if( dotInd<=sepInd )
return "";
else
return fileName.substring(dotInd+1).toLowerCase();
}
And test case:
#Test
public void testGetExtension() {
assertEquals("", getExtension("C"));
assertEquals("ext", getExtension("C.ext"));
assertEquals("ext", getExtension("A/B/C.ext"));
assertEquals("", getExtension("A/B/C.ext/"));
assertEquals("", getExtension("A/B/C.ext/.."));
assertEquals("bin", getExtension("A/B/C.bin"));
assertEquals("hidden", getExtension(".hidden"));
assertEquals("dsstore", getExtension("/user/home/.dsstore"));
assertEquals("", getExtension(".strange."));
assertEquals("3", getExtension("1.2.3"));
assertEquals("exe", getExtension("C:\\Program Files (x86)\\java\\bin\\javaw.exe"));
}
If you use Spring framework in your project, then you can use StringUtils
import org.springframework.util.StringUtils;
StringUtils.getFilenameExtension("YourFileName")
String path = "/Users/test/test.txt";
String extension = "";
if (path.contains("."))
extension = path.substring(path.lastIndexOf("."));
return ".txt"
if you want only "txt", make path.lastIndexOf(".") + 1
In order to take into account file names without characters before the dot, you have to use that slight variation of the accepted answer:
String extension = "";
int i = fileName.lastIndexOf('.');
if (i >= 0) {
extension = fileName.substring(i+1);
}
"file.doc" => "doc"
"file.doc.gz" => "gz"
".doc" => "doc"
My dirty and may tiniest using String.replaceAll:
.replaceAll("^.*\\.(.*)$", "$1")
Note that first * is greedy so it will grab most possible characters as far as it can and then just last dot and file extension will be left.
As is obvious from all the other answers, there's no adequate "built-in" function. This is a safe and simple method.
String getFileExtension(File file) {
if (file == null) {
return "";
}
String name = file.getName();
int i = name.lastIndexOf('.');
String ext = i > 0 ? name.substring(i + 1) : "";
return ext;
}
Here is another one-liner for Java 8.
String ext = Arrays.stream(fileName.split("\\.")).reduce((a,b) -> b).orElse(null)
It works as follows:
Split the string into an array of strings using "."
Convert the array into a stream
Use reduce to get the last element of the stream, i.e. the file extension
How about (using Java 1.5 RegEx):
String[] split = fullFileName.split("\\.");
String ext = split[split.length - 1];
If you plan to use Apache commons-io,and just want to check the file's extension and then do some operation,you can use this,here is a snippet:
if(FilenameUtils.isExtension(file.getName(),"java")) {
someoperation();
}
How about JFileChooser? It is not straightforward as you will need to parse its final output...
JFileChooser filechooser = new JFileChooser();
File file = new File("your.txt");
System.out.println("the extension type:"+filechooser.getTypeDescription(file));
which is a MIME type...
OK...I forget that you don't want to know its MIME type.
Interesting code in the following link:
http://download.oracle.com/javase/tutorial/uiswing/components/filechooser.html
/*
* Get the extension of a file.
*/
public static String getExtension(File f) {
String ext = null;
String s = f.getName();
int i = s.lastIndexOf('.');
if (i > 0 && i < s.length() - 1) {
ext = s.substring(i+1).toLowerCase();
}
return ext;
}
Related question:
How do I trim a file extension from a String in Java?
Here's a method that handles .tar.gz properly, even in a path with dots in directory names:
private static final String getExtension(final String filename) {
if (filename == null) return null;
final String afterLastSlash = filename.substring(filename.lastIndexOf('/') + 1);
final int afterLastBackslash = afterLastSlash.lastIndexOf('\\') + 1;
final int dotIndex = afterLastSlash.indexOf('.', afterLastBackslash);
return (dotIndex == -1) ? "" : afterLastSlash.substring(dotIndex + 1);
}
afterLastSlash is created to make finding afterLastBackslash quicker since it won't have to search the whole string if there are some slashes in it.
The char[] inside the original String is reused, adding no garbage there, and the JVM will probably notice that afterLastSlash is immediately garbage in order to put it on the stack instead of the heap.
This particular question gave me a lot of trouble then i found a very simple solution for this problem which i'm posting here.
file.getName().toLowerCase().endsWith(".txt");
That's it.
Java 20 EA
As of Java 20 EA (early-access), there is finally a new method Path#getExtension that returns the extension as a String:
Paths.get("/Users/admin/notes.txt").getExtension(); // "txt"
Paths.get("/Users/admin/.gitconfig").getExtension(); // "gitconfig"
Paths.get("/Users/admin/configuration.xml.zip").getExtension(); // "zip"
Paths.get("/Users/admin/file").getExtension(); // null
// Modified from EboMike's answer
String extension = "/path/to/file/foo.txt".substring("/path/to/file/foo.txt".lastIndexOf('.'));
extension should have ".txt" in it when run.
Here's the version with Optional as a return value (cause you can't be sure the file has an extension)... also sanity checks...
import java.io.File;
import java.util.Optional;
public class GetFileExtensionTool {
public static Optional<String> getFileExtension(File file) {
if (file == null) {
throw new NullPointerException("file argument was null");
}
if (!file.isFile()) {
throw new IllegalArgumentException("getFileExtension(File file)"
+ " called on File object that wasn't an actual file"
+ " (perhaps a directory or device?). file had path: "
+ file.getAbsolutePath());
}
String fileName = file.getName();
int i = fileName.lastIndexOf('.');
if (i > 0) {
return Optional.of(fileName.substring(i + 1));
} else {
return Optional.empty();
}
}
}
How about REGEX version:
static final Pattern PATTERN = Pattern.compile("(.*)\\.(.*)");
Matcher m = PATTERN.matcher(path);
if (m.find()) {
System.out.println("File path/name: " + m.group(1));
System.out.println("Extention: " + m.group(2));
}
or with null extension supported:
static final Pattern PATTERN =
Pattern.compile("((.*\\" + File.separator + ")?(.*)(\\.(.*)))|(.*\\" + File.separator + ")?(.*)");
class Separated {
String path, name, ext;
}
Separated parsePath(String path) {
Separated res = new Separated();
Matcher m = PATTERN.matcher(path);
if (m.find()) {
if (m.group(1) != null) {
res.path = m.group(2);
res.name = m.group(3);
res.ext = m.group(5);
} else {
res.path = m.group(6);
res.name = m.group(7);
}
}
return res;
}
Separated sp = parsePath("/root/docs/readme.txt");
System.out.println("path: " + sp.path);
System.out.println("name: " + sp.name);
System.out.println("Extention: " + sp.ext);
result for *nix:
path: /root/docs/
name: readme
Extention: txt
for windows, parsePath("c:\windows\readme.txt"):
path: c:\windows\
name: readme
Extention: txt
String extension = com.google.common.io.Files.getFileExtension("fileName.jpg");
Here I made a small method (however not that secure and doesnt check for many errors), but if it is only you that is programming a general java-program, this is more than enough to find the filetype. This is not working for complex filetypes, but those are normally not used as much.
public static String getFileType(String path){
String fileType = null;
fileType = path.substring(path.indexOf('.',path.lastIndexOf('/'))+1).toUpperCase();
return fileType;
}
Getting File Extension from File Name
/**
* The extension separator character.
*/
private static final char EXTENSION_SEPARATOR = '.';
/**
* The Unix separator character.
*/
private static final char UNIX_SEPARATOR = '/';
/**
* The Windows separator character.
*/
private static final char WINDOWS_SEPARATOR = '\\';
/**
* The system separator character.
*/
private static final char SYSTEM_SEPARATOR = File.separatorChar;
/**
* Gets the extension of a filename.
* <p>
* This method returns the textual part of the filename after the last dot.
* There must be no directory separator after the dot.
* <pre>
* foo.txt --> "txt"
* a/b/c.jpg --> "jpg"
* a/b.txt/c --> ""
* a/b/c --> ""
* </pre>
* <p>
* The output will be the same irrespective of the machine that the code is running on.
*
* #param filename the filename to retrieve the extension of.
* #return the extension of the file or an empty string if none exists.
*/
public static String getExtension(String filename) {
if (filename == null) {
return null;
}
int index = indexOfExtension(filename);
if (index == -1) {
return "";
} else {
return filename.substring(index + 1);
}
}
/**
* Returns the index of the last extension separator character, which is a dot.
* <p>
* This method also checks that there is no directory separator after the last dot.
* To do this it uses {#link #indexOfLastSeparator(String)} which will
* handle a file in either Unix or Windows format.
* <p>
* The output will be the same irrespective of the machine that the code is running on.
*
* #param filename the filename to find the last path separator in, null returns -1
* #return the index of the last separator character, or -1 if there
* is no such character
*/
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);
}
/**
* Returns the index of the last directory separator character.
* <p>
* This method will handle a file in either Unix or Windows format.
* The position of the last forward or backslash is returned.
* <p>
* The output will be the same irrespective of the machine that the code is running on.
*
* #param filename the filename to find the last path separator in, null returns -1
* #return the index of the last separator character, or -1 if there
* is no such character
*/
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);
}
Credits
Copied from Apache FileNameUtils Class - http://grepcode.com/file/repo1.maven.org/maven2/commons-io/commons-io/1.3.2/org/apache/commons/io/FilenameUtils.java#FilenameUtils.getExtension%28java.lang.String%29
Without use of any library, you can use the String method split as follows :
String[] splits = fileNames.get(i).split("\\.");
String extension = "";
if(splits.length >= 2)
{
extension = splits[splits.length-1];
}
private String getExtension(File file)
{
String fileName = file.getName();
String[] ext = fileName.split("\\.");
return ext[ext.length -1];
}
Just a regular-expression based alternative. Not that fast, not that good.
Pattern pattern = Pattern.compile("\\.([^.]*)$");
Matcher matcher = pattern.matcher(fileName);
if (matcher.find()) {
String ext = matcher.group(1);
}
I like the simplicity of spectre's answer, and linked in one of his comments is a link to another answer that fixes dots in file paths, on another question, made by EboMike.
Without implementing some sort of third party API, I suggest:
private String getFileExtension(File file) {
String name = file.getName().substring(Math.max(file.getName().lastIndexOf('/'),
file.getName().lastIndexOf('\\')) < 0 ? 0 : Math.max(file.getName().lastIndexOf('/'),
file.getName().lastIndexOf('\\')));
int lastIndexOf = name.lastIndexOf(".");
if (lastIndexOf == -1) {
return ""; // empty extension
}
return name.substring(lastIndexOf + 1); // doesn't return "." with extension
}
Something like this may be useful in, say, any of ImageIO's write methods, where the file format has to be passed in.
Why use a whole third party API when you can DIY?
The fluent way:
public static String fileExtension(String fileName) {
return Optional.of(fileName.lastIndexOf(".")).filter(i-> i >= 0)
.filter(i-> i > fileName.lastIndexOf(File.separator))
.map(fileName::substring).orElse("");
}
try this.
String[] extension = "adadad.adad.adnandad.jpg".split("\\.(?=[^\\.]+$)"); // ['adadad.adad.adnandad','jpg']
extension[1] // jpg
I have a string representing an URL containing spaces and want to convert it to an URI object. If I simply try to create it via
String myString = "http://myhost.com/media/File Name that has spaces inside.mp3";
URI myUri = new URI(myString);
it gives me
java.net.URISyntaxException: Illegal character in path at index X
where index X is the position of the first space in the URL string.
How can i parse myString into a URI object?
You should in fact URI-encode the "invalid" characters. Since the string actually contains the complete URL, it's hard to properly URI-encode it. You don't know which slashes / should be taken into account and which not. You cannot predict that on a raw String beforehand. The problem really needs to be solved at a higher level. Where does that String come from? Is it hardcoded? Then just change it yourself accordingly. Does it come in as user input? Validate it and show error, let the user solve itself.
At any way, if you can ensure that it are only the spaces in URLs which makes it invalid, then you can also just do a string-by-string replace with %20:
URI uri = new URI(string.replace(" ", "%20"));
Or if you can ensure that it's only the part after the last slash which needs to be URI-encoded, then you can also just do so with help of android.net.Uri utility class:
int pos = string.lastIndexOf('/') + 1;
URI uri = new URI(string.substring(0, pos) + Uri.encode(string.substring(pos)));
Do note that URLEncoder is insuitable for the task as it's designed to encode query string parameter names/values as per application/x-www-form-urlencoded rules (as used in HTML forms). See also Java URL encoding of query string parameters.
java.net.URLEncoder.encode(finalPartOfString, "utf-8");
This will URL-encode the string.
finalPartOfString is the part after the last slash - in your case, the name of the song, as it seems.
To handle spaces, #, and other unsafe characters in arbitrary locations in the url path, Use Uri.Builder in combination with a local instance of URL as I have described here:
private Uri.Builder builder;
public Uri getUriFromUrl(String thisUrl) {
URL url = new URL(thisUrl);
builder = new Uri.Builder()
.scheme(url.getProtocol())
.authority(url.getAuthority())
.appendPath(url.getPath());
return builder.build();
}
URL url = Test.class.getResource(args[0]); // reading demo file path from
// same location where class
File input=null;
try {
input = new File(url.toURI());
} catch (URISyntaxException e1) {
// TODO Auto-generated catch block
e1.printStackTrace();
}
I wrote this function:
public static String encode(#NonNull String uriString) {
if (TextUtils.isEmpty(uriString)) {
Assert.fail("Uri string cannot be empty!");
return uriString;
}
// getQueryParameterNames is not exist then cannot iterate on queries
if (Build.VERSION.SDK_INT < 11) {
return uriString;
}
// Check if uri has valid characters
// See https://tools.ietf.org/html/rfc3986
Pattern allowedUrlCharacters = Pattern.compile("([A-Za-z0-9_.~:/?\\#\\[\\]#!$&'()*+,;" +
"=-]|%[0-9a-fA-F]{2})+");
Matcher matcher = allowedUrlCharacters.matcher(uriString);
String validUri = null;
if (matcher.find()) {
validUri = matcher.group();
}
if (TextUtils.isEmpty(validUri) || uriString.length() == validUri.length()) {
return uriString;
}
// The uriString is not encoded. Then recreate the uri and encode it this time
Uri uri = Uri.parse(uriString);
Uri.Builder uriBuilder = new Uri.Builder()
.scheme(uri.getScheme())
.authority(uri.getAuthority());
for (String path : uri.getPathSegments()) {
uriBuilder.appendPath(path);
}
for (String key : uri.getQueryParameterNames()) {
uriBuilder.appendQueryParameter(key, uri.getQueryParameter(key));
}
String correctUrl = uriBuilder.build().toString();
return correctUrl;
}
I need to scan a particular folder in Java, and be able to return the integer number of files of a particular type (based on not only extension but also naming convention.) For example, I want to know how many JPG files there are in the \src folder that have a simple integer filename (say, 1.JPG through 30.JPG). Can anyone point me in the right direction? Thx
java.io.File.list(FilenameFilter) is the method you're looking for.
I have a method that uses a regex pattern for a rather complicated file structure. Something like that could be used, although I'm sure it could be written more concisely than my example (edited for security).
/**
* Get all non-directory filenames from a given foo/flat directory
*
* #param network
* #param typeRegex
* #param locationRegex
* #return
*/
public List<String> getFilteredFilenames(String network, String typeRegex, String locationRegex) {
String regex = null;
List<String> filenames = new ArrayList<String>();
String directory;
// Look at the something network
if (network.equalsIgnoreCase("foo")) {
// Get the foo files first
directory = this.pathname + "/" + "foo/filtered/flat";
File[] foofiles = getFilenames(directory);
// run the regex if need be.
if (locationRegex != null && typeRegex != null ) {
regex = typeRegex + "." + locationRegex;
//System.out.println(regex);
}
for (int i = 0; i < foofiles.length; i++) {
if (foofiles[i].isFile()) {
String file = foofiles[i].getName();
if (regex == null) {
filenames.add(file);
}
else {
if (file.matches(regex)) {
filenames.add(file);
}
}
}
}
}
return filenames;
}