Solved

How to check file's first name?

Posted on 2006-06-23
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Last Modified: 2010-03-31
I have a situation in which I only have a file's first name(NOT EXTENSION) like 'ms0045' that is coming from database.
Now there may or may not be a file in a known directory with that name.  I want to check if 'ms0045' exists in the directory with any extension or not.

how can i do that.
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Question by:jaipur07
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11 Comments
 
LVL 86

Assisted Solution

by:CEHJ
CEHJ earned 250 total points
ID: 16969624
Just iterate the directory

boolean found = false;
String[] files = new File(x).list();
for(int i = 0;i < files.length && !found;i++) {
      found = files[i].getName().startsWith(searchName);
 
}
0
 
LVL 30

Accepted Solution

by:
Mayank S earned 250 total points
ID: 16969634
File f = new File ( "filepath/filename.extension" ) ;
String fullFileName = f.getName () ;
String fileName = fullFileName.substring ( 0, fullFileName.lastIndexOf ( "." ) ) ;

should do it, I guess....
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Author Comment

by:jaipur07
ID: 16969640
startsWith will return all the names like ms00451212', 'ms00455555'

isn't it?
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LVL 30

Expert Comment

by:Mayank S
ID: 16969643
(I mean just getting the name of a file - not searching)
0
 
LVL 30

Expert Comment

by:Mayank S
ID: 16969665
CEHJ's and mine combined should do both.
0
 
LVL 30

Expert Comment

by:Mayank S
ID: 16969671
>> String[] files = new File(x).list();

Use listFiles () if you want to check for only files, not directories.
0
 
LVL 86

Expert Comment

by:CEHJ
ID: 16969674
>>startsWith will return all the names like ms00451212', 'ms00455555'

Yes. If that's a problem you can do


found = files[i].getName().startsWith(searchName) && searchName.equals(files[i].getName().substring(0, files[i].getName().lastIndexOf('.'));
0
 
LVL 86

Expert Comment

by:CEHJ
ID: 16969680
(You can remove the first condition before &&)
0
 

Author Comment

by:jaipur07
ID: 16969737
let me try and will get back to yo usoon
0
 
LVL 30

Expert Comment

by:Mayank S
ID: 16969807
Note that there is small optimization in my approach that you don't have to call getName () multiple times saving a method-call :)
0
 
LVL 86

Expert Comment

by:CEHJ
ID: 16969830
:-)
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