Graham Hi!

how can i open each file in a directory,  and add that file name and line number on each line (there is data on each line).


Thanks in advance


Code:

opendir DH,"."; foreach $thing (readdir DH) {        next unless (-f $thing and -r $thing);        open (FH,$thing);        $ln = 0;        while (<FH>) {                $ln++;                print "$thing $ln: ";                print;                }        }

Graham thanks for the reply. however, when i run the program the output has all the files merged together and i am getting a single file. i would like the output files to be separate (same as before) but want the file name and line number added to each of them. sorry if i was not clear in my first posting.


The output of this code is


file1 1: some data
file1 2: some data
file1 3: some data
file1 4: some data
file1 5: some data
file1 6: some data
file2 1: some data
file2 2: some data
file2 3: some data
file2 4: some data
file2 5: some data
and so on......

thanks



You can't output directly to a file of the same name, as you'll overwrite the original before you have read it. SO ...

1. Open a file for output (with an extra extension such as .new) at the same time as you open the input file

2. Print to the new file handle rather than to stdout

3. Use the ren function to rename the new file once you've completed the writing to it.

There are exceptions, but it's generally very bad practise to include the name of a file within the file data - it makes it very hard to rename the file and still have it function as you wish, for example.  What are you doing?  

i want  to  merge the files in the next step and be able to track the origin of the data as there is duplicate data in these files and also wanted to keep them individually in the modified form as  back-up . After what you advised i have dropped that  idea.

thanks for the help

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