Hey guys! First off, thanks for any help you can provide me. I am trying to figure out how to write an awk script(or something better if it exists) to read a text file that contains lines like below.

4517-s-1295546289-10:58:09-2011-01-20.wav | 19990 KiB | 2940.5 KiB/s | binary | 100%
4517-s-1303247233-15:07:13-2011-04-19.wav | 167 KiB | 2918.3 KiB/s | binary | 100%
4517-s-1301431748-14:49:08-2011-03-29.wav | 36220 KiB | 2589.3 KiB/s | binary | 100%
4517-s-1295974735-09:58:55-2011-01-25.wav | 20038 KiB | 2646.0 KiB/s | binary | 100%
4517-s-1295648478-15:21:18-2011-01-21.wav | 45561 KiB | 2154.1 KiB/s | binary | 100%
4517-s-1300384178-11:49:38-2011-03-17.wav | 6752 KiB | 2216.0 KiB/s | binary | 100%

I want to find lines that start with 4517-s and end with 100% and delete them from a directory.

I am sorry, I just am too new to know where to even start.

Also, I am not sure if I have posted this in the wrong place or not so if I have, please let me know and I will move it.
Thanks

For example (with a) input file named list.in, b) no spaces in any wav files, c) directory where to delete files located in current directory and named sound) :

First off thanks for your help. After looking at my file, I noticed that I can just go ahead and delete all files ending in 100% so I think I can just write this:

is that correct?

No, you must put a dot before the star :
In a regular expression (regexp), dot means any single character and star means that the previous regexp can be repeated 0 to n times (and dollar means the end of record).
See awk regexp at http://www.gnu.org/software/gawk/man...wk.html#Regexp

Actually, you don't need the dot and star at all. This will work:

closing. Please look at http://www.linuxquestions.org/questi...pt-890051-new/

sorry for duplicate.