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Regex to extract the filename and drop the file timestamp from complete path in Java

i have complete file path and i just need to extract the filename and just extension. So my output would be fileName.csv.
For ex: complete path is:
/Dir1/Dir2/Dir3/Dir4/Dir5/Dir6/fileName_20150108_002_20150109013841.csv
My output of Regex should be fileName.csv.
Extension and level of directories are not fixed.
As part of my requirement i need single regex that can extract fileName.csv not fileName_20150108_002_20150109013841.csv.how can i do it in single regular expression ?

Without using regex this can be solved as -public static String getFileName(String args){
args = args.substring(args.lastIndexOf('/')+1);
return args.substring(0,args.indexOf('_')) + args.substring(args.indexOf('.'));
}

Below would work for you might be
[^\\/:*?"<>|\r\n]+$
This regex has been tested on these two examples:
\var\www\www.example.com\index.jsp
\index.jsp
or rather you should use File.getName() for better approach.
String filename = new File("Payload/brownie.app/Info.plist").getName();
System.out.println(filename)
another way is
int index = path.lastIndexOf(File.separatorChar);
String filename = path.substring(index+1);
finally after getting the full filename use below code snippet
String str = filename;// in your case filename will be fileName_20150108_002_20150109013841.csv
str = str.substring(0,str.indexOf('_'))+str.substring(str.lastIndexOf('.'));
System.out.println("filename is ::"+str); // output will be fileName.csv

In the below code, group one will be fileName_timestamp.extension. I've replaced numerics and underscores with empty string. This may look ugly, but still will server your purpose. If the file name contains numerics, we need go for a different approach.
public static void main(String[] args) {
String toBeSplitted = "/Dir1/Dir2/Dir3/Dir4/Dir5/Dir6/fileName_20150108_002_20150109013841.csv";
Pattern r = Pattern.compile("(/[a-zA-Z0-9_.-]+)+/?");
Matcher m = r.matcher(toBeSplitted);
if(m.matches()){
String s = m.group(1).replaceAll("(/|[0-9]|_)", "");
System.out.println(s);
}
}

Java Replacing Help Needed

Hey guy's so am trying to replace all characters and numbers to get the /hello/what/ only without the REMOVEThis4.PNG i don't want to use string.replace("REMOVEThis4.PNG", ""); cause i wanna use it on other strings not only that
Any help is great my code
String sFile = "/hello/what/REMOVEThis4.PNG";
if (sFile.contains("/")){
String Replaced = sFile.replaceAll("(?s)", "");
System.out.println(Replaced);
}
I want the the output to be
/hello/what/
Only thanks alot!

If you are trying to parse a path, I recommend to find the last index of /, and get the substring to this index plus one. So
string = string.substring(0, string.lastIndexOf("/") + 1);

No need to use regular expressions in your case:
String sFile = "/hello/what/REMOVEThis4.PNG";
// TODO check actual last index of "/" against -1
System.out.println(sFile.substring(0, sFile.lastIndexOf("/") + 1));
Output
/hello/what/
Note
In case you are dealing with actual files, you can probably spare yourself the String manipulation and use File.getParent() instead:
File file = new File("/hello/what/REMOVEThis4.PNG");
System.out.println(file.getParent());
Output (may change depending on your system)
\hello\what

Use Java's File API:
String example = "/hello/what/REMOVEThis4.PNG";
File file = new File(example);
System.out.println(example);
String absolutePath = file.getAbsolutePath();
String filePath = absolutePath.substring(0, absolutePath.lastIndexOf(File.separator));
System.out.println(filePath);

How to split the string a android?

I am working on android application. I am getting the image from gallery. Also I am getting the image path from gallery. Now my requirement is I want to get only the image name with the extension . How can I do that? Please help me.
String imgpath = "/mnt/sdcard/joke.png";
The image extension can be anything joke.png or joke.jpeg. I need to get the image name with extension finally.
i.e I want to split the above string and get only joke.png.
How can I achieve that? Please help me in this regard.

String imgpath = "/mnt/sdcard/joke.png";
String result = imgpath.substring(imgpath.lastIndexOf("/") + 1);
System.out.println("Image name " + result);
Output :-
Image name joke.png
You should read How do I get the file name from a String containing the Absolute file path?

You can do that in Android like in any Java program:
String[] parts = imagepath.split("/");
String result = parts[parts.length-1];

String s[] = imgpath.split("/");
String result = s[s.length-1];

String imgName = imgpath.substring((imgpath.lastIndexOf("/") + 1), imgpath.length());

You can get this with Regex too, if that is the hammer you have in your hands:
String fileName = null;
Pattern pattern = Pattern.compile("(^|.*/)([^/]*)$");
Matcher m = pattern.getMatcher(filenameWithPath);
if(matcher.matches()) {
fileName = matcher.group(2);
}
But don't be tempted to do this. This is less readable, and probably even slower than the other methods.

Dynamic String get dynamic image name in java

i have a dynamic String like
age/data/images/four_seasons1.jpg
from above string i need to get the image name alone (i.e.) four_seasons1.jpg
the path and the image will be a dynamic one(any image format will occure)
Please let me know how to do this in java?
thanks in advance

Use the File Object.
new File("/path/to/file").getName()
You could also use String.split().
"/path/to/file/sdf.png".split("/")
This will give you an array in which you pick the last element. But the File Object is better suited.

String text = "age/data/images/four_seasons1.jpg";
String name = text.substring(text.lastIndexOf("/") + 1);
String path = text.substring(0, text.lastIndexOf("/"));
System.out.println(name);
System.out.println(path);
Outputs
four_seasons1.jpg
age/data/images
Take some time and become familiar with the java.lang.String API. You'll be doing this kind of stuff a lot

You can go for regex but if you find the pattern is fixed, A very crude solution can be a straight forward approach
String url = "age/data/images/four_seasons1.jpg";
String imageName = url.substring(url.lastIndexOf( "/" )+1, url.length()) ;

You can parse this path. As a delimiter you must take '/' symbol. After that you can take last parsed element.
String phrase = "age/data/images/four_seasons1.jpg";
String delims = "/";
String[] tokens = phrase.split(delims);
About String.split you can read more here.

String s = "age/data/images/four_seasons1.jpg";
String fileName = new String();
String[] arr = s.split("/");
fileName = arr[arr.length-1];
}

Java - Convert String to valid URI object

I am trying to get a java.net.URI object from a String. The string has some characters which will need to be replaced by their percentage escape sequences. But when I use URLEncoder to encode the String with UTF-8 encoding, even the / are replaced with their escape sequences.
How can I get a valid encoded URL from a String object?
http://www.google.com?q=a b gives http%3A%2F%2www.google.com... whereas I want the output to be http://www.google.com?q=a%20b
Can someone please tell me how to achieve this.
I am trying to do this in an Android app. So I have access to a limited number of libraries.

You might try: org.apache.commons.httpclient.util.URIUtil.encodeQuery in Apache commons-httpclient project
Like this (see URIUtil):
URIUtil.encodeQuery("http://www.google.com?q=a b")
will become:
http://www.google.com?q=a%20b
You can of course do it yourself, but URI parsing can get pretty messy...

Android has always had the Uri class as part of the SDK:
http://developer.android.com/reference/android/net/Uri.html
You can simply do something like:
String requestURL = String.format("http://www.example.com/?a=%s&b=%s", Uri.encode("foo bar"), Uri.encode("100% fubar'd"));

I'm going to add one suggestion here aimed at Android users. You can do this which avoids having to get any external libraries. Also, all the search/replace characters solutions suggested in some of the answers above are perilous and should be avoided.
Give this a try:
String urlStr = "http://abc.dev.domain.com/0007AC/ads/800x480 15sec h.264.mp4";
URL url = new URL(urlStr);
URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
url = uri.toURL();
You can see that in this particular URL, I need to have those spaces encoded so that I can use it for a request.
This takes advantage of a couple features available to you in Android classes. First, the URL class can break a url into its proper components so there is no need for you to do any string search/replace work. Secondly, this approach takes advantage of the URI class feature of properly escaping components when you construct a URI via components rather than from a single string.
The beauty of this approach is that you can take any valid url string and have it work without needing any special knowledge of it yourself.

Even if this is an old post with an already accepted answer, I post my alternative answer because it works well for the present issue and it seems nobody mentioned this method.
With the java.net.URI library:
URI uri = URI.create(URLString);
And if you want a URL-formatted string corresponding to it:
String validURLString = uri.toASCIIString();
Unlike many other methods (e.g. java.net.URLEncoder) this one replaces only unsafe ASCII characters (like ç, é...).
In the above example, if URLString is the following String:
"http://www.domain.com/façon+word"
the resulting validURLString will be:
"http://www.domain.com/fa%C3%A7on+word"
which is a well-formatted URL.

If you don't like libraries, how about this?
Note that you should not use this function on the whole URL, instead you should use this on the components...e.g. just the "a b" component, as you build up the URL - otherwise the computer won't know what characters are supposed to have a special meaning and which ones are supposed to have a literal meaning.
/** Converts a string into something you can safely insert into a URL. */
public static String encodeURIcomponent(String s)
{
StringBuilder o = new StringBuilder();
for (char ch : s.toCharArray()) {
if (isUnsafe(ch)) {
o.append('%');
o.append(toHex(ch / 16));
o.append(toHex(ch % 16));
}
else o.append(ch);
}
return o.toString();
}
private static char toHex(int ch)
{
return (char)(ch < 10 ? '0' + ch : 'A' + ch - 10);
}
private static boolean isUnsafe(char ch)
{
if (ch > 128 || ch < 0)
return true;
return " %$&+,/:;=?#<>#%".indexOf(ch) >= 0;
}

You can use the multi-argument constructors of the URI class. From the URI javadoc:
The multi-argument constructors quote illegal characters as required by the components in which they appear. The percent character ('%') is always quoted by these constructors. Any other characters are preserved.
So if you use
URI uri = new URI("http", "www.google.com?q=a b");
Then you get http:www.google.com?q=a%20b which isn't quite right, but it's a little closer.
If you know that your string will not have URL fragments (e.g. http://example.com/page#anchor), then you can use the following code to get what you want:
String s = "http://www.google.com?q=a b";
String[] parts = s.split(":",2);
URI uri = new URI(parts[0], parts[1], null);
To be safe, you should scan the string for # characters, but this should get you started.

I had similar problems for one of my projects to create a URI object from a string. I couldn't find any clean solution either. Here's what I came up with :
public static URI encodeURL(String url) throws MalformedURLException, URISyntaxException
{
URI uriFormatted = null;
URL urlLink = new URL(url);
uriFormatted = new URI("http", urlLink.getHost(), urlLink.getPath(), urlLink.getQuery(), urlLink.getRef());
return uriFormatted;
}
You can use the following URI constructor instead to specify a port if needed:
URI uri = new URI(scheme, userInfo, host, port, path, query, fragment);

Well I tried using
String converted = URLDecoder.decode("toconvert","UTF-8");
I hope this is what you were actually looking for?

The java.net blog had a class the other day that might have done what you want (but it is down right now so I cannot check).
This code here could probably be modified to do what you want:
http://svn.apache.org/repos/asf/incubator/shindig/trunk/java/common/src/main/java/org/apache/shindig/common/uri/UriBuilder.java
Here is the one I was thinking of from java.net: https://urlencodedquerystring.dev.java.net/

Or perhaps you could use this class:
http://developer.android.com/reference/java/net/URLEncoder.html
Which is present in Android since API level 1.
Annoyingly however, it treats spaces specially (replacing them with + instead of %20). To get round this we simply use this fragment:
URLEncoder.encode(value, "UTF-8").replace("+", "%20");

I ended up using the httpclient-4.3.6:
import org.apache.http.client.utils.URIBuilder;
public static void main (String [] args) {
URIBuilder uri = new URIBuilder();
uri.setScheme("http")
.setHost("www.example.com")
.setPath("/somepage.php")
.setParameter("username", "Hello Günter")
.setParameter("p1", "parameter 1");
System.out.println(uri.toString());
}
Output will be:
http://www.example.com/somepage.php?username=Hello+G%C3%BCnter&p1=paramter+1