Something like this?
for i in ; do wget -q -O - http://www.ellinofreneianet.gr/sound.php\?id\=$i | grep ".=podcast/." ; done The wget options are -q quiet, show no progress etc, and -O - write output to stdout.
Not every page has a mp3 link there; Some even ones show a page which could be the 404 error page. The pages starting from 0 also seem empty.
The empty pages have URLs ending in podcast/", so we can exclude them with matching strings which don't have a " there:
... | grep ".=podcast/[^\"]" To get only the .mp3 urls, use
... | grep -o 'bitsnbytesplayer.php.*\.mp3' You found out yourself how to output the page URL before each mp3 URL. Here's an optimiset variant of that, using only one HTTP request per page:
for i in ; do \ wget -q -O - http://www.ellinofreneianet.gr/sound.php\?id\=$i | \ grep -o 'bitsnbytesplayer.php.*\.mp3' && \ echo http://www.ellinofreneianet.gr/sound.php\?id\=$i ; done | sed -n 'h;n;p;g;p' The && echo ... prints the URL if the grep before found an mp3 url. The sed command switches the order of the line pairs.