how to Split name file to count series in java?

Question

i have a list of files in this form:

name_of_file_001.csv
name_of_file_002.csv    
name_of_file_123.csv
or
name_of_file.csv
second_name_of_file.csv

i don't know if the file has 001 or not.
how to take name of file (only name_of_file) in java?

I dont really have the time to write a working answer right now, but have you considered checking with regular expressions? for instance smth like `[a-z]*(?=[0-9]{3})` — posdef, Feb 23 '15 at 08:48

score 2 · Answer 1 · edited May 23 '17 at 11:43

2

Try the following:

int i=0;
while(!fullName.charAt(i).equals('.')&&!fullName.charAt(i).equals('0')){
  i++;
}
String name=fullName.substring(0, i);

Take the string from the beginning of the fullName to the first appearance of . or 0.

EDIT:

Referring to the comments and the case of high numbers greater than 1.. and inspired from this answer:

    int i=0;
    String patternStr = "[0-9\.]";
    Pattern pattern = Pattern.compile(patternStr);
    Matcher matcher = pattern.matcher(fullName);
    if(matcher.find()){
     i=matcher.start();  //this will give you the first index of the regex
    }
    String name=fullName.substring(0, i);

EDIT2:

In the case where there's no Extension and the fullname doesn't match the regex(there's no numbers):

if(matcher.find()){
     i=matcher.start();  //this will give you the first index of the regex
}else {
 i=fullname.length();
}
String name=fullName.substring(0, i);

Or simply we will take all the name.

edited May 23 '17 at 11:43

Community

1
1

answered Feb 23 '15 at 08:15

cнŝdk

31,391
7
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1

there is a possibility that the numbering is high (e.g. name_of_file_142.csv) – Mauro Midolo Feb 23 '15 at 08:31
@mmxx yes good point, well in that case we just add this test `!fullName.charAt(i).equals('1')`. – cнŝdk Feb 23 '15 at 08:52
What about numbers starting with 2, 3, 4, 5, 6, 7, 8, 9 ? – fjf2002 Feb 23 '15 at 08:53
What about filenames that don't have an extension? – Eldad Mor Feb 23 '15 at 09:06
@EldadMor what about them? the name will be taken either it has an extension or not. – cнŝdk Feb 23 '15 at 09:09
Well, you'll fly high on file name like "abc" – Eldad Mor Feb 23 '15 at 09:34
No, the filename in all cases contains the `.` mark. – cнŝdk Feb 23 '15 at 09:48
Yes, I know it is a valid name, in that case if there's no extension and there's no match for the regex we just need to set the `i`var to the length of the **fullname**. – cнŝdk Feb 23 '15 at 16:51

fjf2002 · Answer 2 · 2015-02-23T15:45:16.300

1

I modified chsdk's solution with respect to mmxx's comment:

int i=0;
while(i < fullName.length() && ".0123456789".indexOf(fullName.charAt(i)) == -1) {
  i++;
}
String name=fullName.substring(0, i);

EDIT: Added

i < fullName.length()

edited Feb 23 '15 at 15:45

answered Feb 23 '15 at 08:52

fjf2002

872
5
15

Eldad Mor · Answer 3 · 2015-02-23T09:19:27.690

This little class solves the problem for all the examples shown in main:

public class Example {

   private static boolean isNaturalNumber(String str)
   {
     return str.matches("\\d+(\\.\\d+)?");
   }

   public static String getFileName(String s) {
        String fn = s.split("\\.")[0];
        int idx = fn.lastIndexOf('_');
        if (idx < 0) {
            return fn;
        }
        String lastPart = fn.substring(idx+1);
        System.out.println("last part = " + lastPart);
        if (isNaturalNumber(lastPart)) {
            return fn.substring(0,idx);
        } else {
            return fn;
        }
    }

    public static void main(String []args){
        System.out.println(getFileName("file_name_001.csv"));
        System.out.println(getFileName("file_name_1234.csv"));
        System.out.println(getFileName("file_name.csv"));
        System.out.println(getFileName("file_name"));
        System.out.println(getFileName("file"));
    }
}

EDIT 1: Replaced the exception-based check with a regex check.

EDIT 2: Handling file names without any underscores.

Relying on exceptions in regular execution flow isn't good programming practice. — fjf2002, Feb 23 '15 at 09:00
That depends on the times you need to call this function. See here for alternatives: http://stackoverflow.com/questions/1102891/how-to-check-if-a-string-is-a-numeric-type-in-java — Eldad Mor, Feb 23 '15 at 09:04

score 0 · Accepted Answer · answered Feb 23 '15 at 13:03

i resolved the problem in this mode:

nameOfFile.split("\\.")[0].replaceall("_[0-9]*","");

split("\.")[0] remove ".csv" name_of_file_001.csv => name_of_file_001

.replaceall("_[0-9]*","") "remove, if there is, "_001" name_of_file_001 => name_of_file

the result is the name of file only

how to Split name file to count series in java?

4 Answers4