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Decode alfresco file name or replace unicode[_x0020_] characters in String/fileName

I am using alfresco download upload services using java.
When I upload the file to alfreco server it gives me the following path :
/app:Home/cm:Company_x0020_Home/cm:Abc/cm:TestFile/cm:V4/cm:BC1X_x0020_0400_x0020_0109-_x0028_1-2_x0029__v2.pdf
When I use the same file path and download using alfresco services I took the file name at the end of the path
i.e ABC1X_x0020_0400_x0020_0109-_x0028_1-2_x0029__v2.pdf
How can I remove or decode the [Unicode] characters in fileName
String decoded = URLDecoder.decode(queryString, "UTF-8");
The above does not work .
These are some Unicode characters which appeared in my file name.
https://en.wikipedia.org/wiki/List_of_Unicode_characters
Please do not mark the question as duplicate as I have searched below links but non of those gave the solution.
Following are the links that I have searched for replacing unicode charectors in String with java.
Java removing unicode characters
Remove non-ASCII characters from String in Java
How can I replace a unicode character in java string
Java Replace Unicode Characters in a String

The solution given by Jeff Potts will be perfect .
But i had a situation where i was using file name in diffrent project where i wont use org.alfresco related jars
I had to take all those dependencies to use for a simple file decoding
So i used java native methods which uses regex to parse the file name and decode it,which gave me the perfect solution which was same from using
ISO9075.decode(test);
This is the code which can be used
public String decode_FileName(String fileName) {
System.out.println("fileName : " + fileName);
String decodedfileName = fileName;
String temp = "";
Matcher m = Pattern.compile("\\_x(.*?)\\_").matcher(decodedfileName); //rejex which matches _x0020_ kind of charectors
List<String> unicodeChars = new ArrayList<String>();
while (m.find()) {
unicodeChars.add(m.group(1));
}
for (int i = 0; i < unicodeChars.size(); i++) {
temp = unicodeChars.get(i);
if (isInteger(temp)) {
String replace_char = String.valueOf(((char) Integer.parseInt(String.valueOf(temp), 16)));//converting
decodedfileName = decodedfileName.replace("_x" + temp + "_", replace_char);
}
}
System.out.println("Decoded FileName :" + decodedfileName);
return decodedfileName;
}
And use this small java util to know Is integer
public static boolean isInteger(String s) {
try {
Integer.parseInt(s);
} catch (NumberFormatException e) {
return false;
} catch (NullPointerException e) {
return false;
}
return true;
}
So the above code works as simple as this :
Example :
0028 Left parenthesis U+0028 You can see in the link
https://en.wikipedia.org/wiki/List_of_Unicode_characters
String replace_char = String.valueOf(((char) Integer.parseInt(String.valueOf("0028"), 16)));
System.out.println(replace_char);
This code gives output : ( which is a Left parenthesis
This is what the logic i have used in my java program.
The above program will give results same as ISO9075.decode(test)
Output :
fileName : ABC1X_x0020_0400_x0020_0109-_x0028_1-2_x0029__v2.pdf
Decoded FileName :ABC1X 0400 0109-(1-2)_v2.pdf

In the org.alfresco.util package you will find a class called ISO9075. You can use it to encode and decode strings according to that spec. For example:
String test = "ABC1X_x0020_0400_x0020_0109-_x0028_1-2_x0029__v2.pdf";
String out = ISO9075.decode(test);
System.out.println(out);
Returns:
ABC1X 0400 0109-(1-2)_v2.pdf
If you want to see what it does behind the scenes, look at the source.

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);
}
}

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.

Java: String manipulation. Fetch last subpath in a URL

Lets say I have a URL http://example.com/files/public_files/test.zip and I want to extract the last subpath so test.zip, How would I be able do this?
I am from Python so I am still new to Java and learning. In Python you could do something like this:
>>> x = "http://example.com/files/public_files/test.zip"
>>> x.split("/")[-1]
'test.zip'

There are many ways. I prefer:
String url = "http://example.com/files/public_files/test.zip";
String fileName = url.substring(url.lastIndexOf("/") + 1);

Using String class method is a way to go. But given that you are having a URL, you can use java.net.URL.getFile():
String url = "http://example.com/files/public_files/test.zip";
String filePart = new URL(url).getFile();
The above code will get you complete path. To get the file name, you can make use of Apache Commons - FilenameUtils.getName():
String url = "http://example.com/files/public_files/test.zip";
String fileName = FilenameUtils.getName(url);
Well, if you don't want to refer to 3rd party library for this task, String class is still an option to go for. I've just given another way.

you can use the following:
String url = "http://example.com/files/public_files/test.zip";
String arr[] = url.split("/");
String name = arr[arr.length - 1];

Most similar to the python syntax is :
String url = "http://example.com/files/public_files/test.zip";
String [] tokens = url.split("/");
String file = tokens[tokens.length-1];
Java lacks the convenient [-n] nth to last selector that Python has. If you wanted to do it all in one line, you'd have to do something gross like this:
String file = url.split("/")[url.split("/").length-1];
I don't recommend the latter

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];
}