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Posted on 2002-03-05
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Last Modified: 2010-03-05
I have several URLs like
http://domain/dir1/dir4/../dir2/../../dir3/file.html
http://domain/dir1/dir2/../../dir3/file.html
http://domain/dir1/../dir3/file.html
etc..
How do I make them look like
http://domain/dir3/file.html (no ../../)
I guess a one liner can do the job!!
Thanks,
0
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Question by:sridhar_dvjs
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7 Comments
 
LVL 5

Expert Comment

by:Peewee
ID: 6841579
sridhar_dvjs,
i'm not fully sure what yo want here as your question looks short of information. Perhaps you could explain it a bit more.

However, it seems you would like to get rid of ../ within your URL's.  These characters tell the operating system to go to the preceeding directory.  If you take these out you will may navigate to the desired directory path.

if you wish to take them out you could use a regular expressions.

peewee



0
 
LVL 5

Expert Comment

by:Peewee
ID: 6841584
sridhar_dvjs,

an example regex is below:

#!/usr/bin/perl
print "start\n";
 
my $input = 'http://domain/dir1/dir4/../dir2/../../dir3/file.html';
 
#http://domain/dir1/dir2/../../dir3/file.html
#http://domain/dir1/../dir3/file.html
 
print "$input\n";
$input =~ s/\/\.\.//ig;
print "$input\n";
 
print "end\n";
0
 
LVL 2

Expert Comment

by:windfall
ID: 6841696
Something like this might work:

#!/usr/bin/perl
##########################################################################
@url = ('http://domain/dir1/dir4/../dir2/../../dir3/file.html',
'http://domain/dir1/dir2/../../dir3/file.html',
'http://domain/dir1/../dir3/file.html');

foreach $l(@url){
     
     $l =~ s/(\.\.\/)//gis;
print "$l\n";
}

Hope this helps..
windfall
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LVL 84

Expert Comment

by:ozo
ID: 6841906
@url = ('http://domain/dir1/dir4/../dir2/../../dir3/file.html',
'http://domain/dir1/dir2/../../dir3/file.html',
'http://domain/dir1/../dir3/file.html');
foreach( @url ){
    1 while s![^/]+/\.\./!!;
    print "$_\n";
}
#assuming no symbolic links
0
 
LVL 16

Expert Comment

by:maneshr
ID: 6842100
sridhar_dvjs,

"..I have several URLs like..."

Where are these URL's in? Are these URL's inside HTML or other types of files?

If yes, then can you post a sample file here?

"..I guess a one liner can do the job!!..."

Do you have command prompt access to run any command?

Please provide as much more detail as you can.

This will help you get a more accurate answer, faster.
0
 
LVL 1

Expert Comment

by:japhyRPI
ID: 6843129
What the poster is asking is "how can I change a directory path with ../'s in it to the path without ../'s?"  In other words, how can I get from

  /this/that/../foo/bar/blat/../it.txt

to

  /this/foo/bar/it.txt

Here is such code:

  $URL = "http://www.pobox.com/../~japhy/regexes/../

  # remove "leading" ../'s
  $URL =~ s{(http://[^/]+/)(?:\.\.(?:/|$))+}{$1};

  # remove other /..'s
  1 while $URL =~ s{[^/]+/\.\.(?:/|$)}{}g;

This code works with URLs such as "http://www.foo.bar/blat/..", even though the ".." doesn't have a trailing slash.
0
 
LVL 1

Accepted Solution

by:
japhyRPI earned 400 total points
ID: 6843137
Sorry, my line assigning to $URL is missing a closing quote and semicolon.

  $URL = "http://www.pobox.com/../~japhy/regexes/../";
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