bash regex or sed help

[fris]

never done regex in bash before, what im bascially trying to do is this perl/php regex

/(\d+)_\d+_\d+[_a-z]*\.jpg/i

what is gets is the facebook profile id, from a facebook image.

http://a1.sphotos.ak.fbcdn.net/hphot...56204038_n.jpg

would get 80329313253

in php

preg_match('/(\d+)_\d+_\d+[_a-z]*\.jpg/i',$url,$matches);

match[1] would be the fb id

[fris]

actually figured out doing command substituion

#!/bin/bash

function profile() {
	d=${1#*_}; 
	d=${d#*_}; 
	d=${d%_*}; 
	d=${d%_*}; 
	echo ${d%_*};
}

profile http://a1.sphotos.ak.fbcdn.net/hphotos-ak-ash3/p480x480/558586_10150802565778254_80329313253_9861803_956204038_n.jpg

[Barry-xlovecam]

sed -i 's/x/y/g'

not /i

[fris]

sed -i 's/x/y/g'

not /i

got it with sed

[fris@fris ~]$ echo "http://a1.sphotos.ak.fbcdn.net/hphotos-ak-ash3/p480x480/558586_10150802565778254_80329313253_9861803_95620 4038_n.jpg" | sed 's/.*_\([0-9]*\)_[0-9]*_[0-9]*_n\.jpg$/\1/'
80329313253
[fris@fris ~]$