Shell Programming and Scripting

Hi,

I have a file which has list of files with some other info too.. say..

1/1/2008./to/path/filename1.:
...
...

something like that.. can anyone tell me how can i just open the file and read the contents and then cut just the filename1 and write it to an @array..

thanks

Is this exactly how the lines of the file are?

1/1/2008./to/path/filename1.:

which part of that is the part you need to open and read the file?

Here is my problem.. I have a file say .. test.lst .. Now i need to open the test.lst file which has contents which looks like ..

1/1/2008 00:00:00 /to/path/filename1:
1/1/2008 00:00:00 /to/path/filename2:
..
..

i need filename (filename1, filename2 ....) part in this line and store it in @array

Thanks

See how well this works:

Code:

my @array;
open (FH, 'yourfile') or die "$!";
while(<FH>){
   chomp;
   tr/://d;
   push @array, (split("/"))[-1];
}
close FH;
print "$_\n" for @array;

If you get too much garbage in the array you will need to use some filtering to get only the lines with a filename in them.

yay!! it worked!.. thanks a lot .. really appreciate your replies on time..

a small problem as i said the contents of the file looked like
1/1/2008 00:00:00 /to/path/filename1:
1/1/2008 00:00:00 /to/path/filename2:
..

but it has one more additional character .. this way

1/1/2008 00:00:00 /to/path/filename1: 0
1/1/2008 00:00:00 /to/path/filename2: 0
..

im getting the filenames as

filename1 0
filename2 0
...

can you tell me how to get rid of this??

Thanks

try this instead:

Code:

my @array;
open (FH, 'yourfile') or die "$!";
while(<FH>){
   chomp;
   tr/://d;
   push @array, (split(/\s+/))[-2];
}
close FH;
print "$_\n" for @array;