extract substring using sed and regular expressions (regexp)
 

I would like to extract a number substring using sed.
echo "ifeelfat398pounds" | sed -n -e '/[0-9]/,/[0-9]/p'

This is a very simple task but I have tried lots of combinations and have failed.

I want to extract this. "398"

isn't a good regexp the "+" requires at least one occurrence and \d is a synonym for [0-9]

see www.regular-expressions.info

EDIT:
I tried your command, which returned the entire line, leading me to believe that sed's behavior is to return any line that matches, when given as you had it. I tried the substitution command, as follows:
with these results:
Using instead the documentation's reference to a special character that represents only what was matched ('&'), I got nothing back:
Hopefully someone with more sed experience than myself can help you out...

Regex (unfortunately) ain't regex. For sed you'll need [:digit:] or [0-9].
And it'll look ugly - grep is a better tool for this (have a look at -o switch)

you can also use perl

I think the first poster was using extended regular expressions with regular sed, so the parenthesis need to be escaped:

Sed will use the longest matching pattern, so I couldn't use '/.*\([0-9][0-9]*\)/' because the first pattern .* would match 'iweigh29', leaving just 7 to match the second pattern.

Jschiwal, thanks it works well.

echo 'iweigh297lbs' | sed 's/.*[^0-9]\([0-9][0-9]*\).*/\1/g'

no need for external tools

ghostdog -what is the difference between these two:
b=${a//[a-zA-Z]/}
b=${a/[a-zA-Z]/}
Is the first example equivalent to using the 'g' with sed? If so I've been looking for that.

yes, it means global replacement. pls check the bash guide in my sig for more details.

This string won't work if there are any punctuation marks or characters outside of the english alphabet.

Try b=${a//[^0-9]/} instead.

But even this would lead to problems if there were more than one set of numbers in the string. Something like "ifeelfat398poundsand15ounces." would give you an output of "39815".

It would be nice if we could use full regex expressions inside of parameter substitution. Does anyone know if it's possible?

ghostdog74 and David the H. that was nice. This is even better. I like bash scripting and am trying to utilize it to it's fullest. A goal is to minimize using other languages or external tools.

Thanks so much for the input so far.

Lex

Thanks ghostdog, I had tried to find that info before but couldn't find or make sense of what I was reading, I guess. The bash man-page is like an epic...

ghostdog74 is "shopt -s -o nounset" the same as this "# set --" from your example?

same problem, sort of
 

I am having the same problem....sort of. I want to extract a combination of character if they exist. If they don't exist, I want nothing. My problem is that if my pattern doesn't exist, I get the whole line returned.

if I have .....AA9999999999999999....., I want AA9999999999999999
if I have ............................, I want nothing.

where AA9999999999999999 is 2 capital alphas followed by 16 numerics.

I use 's/.*\(AA[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]\).*/\1/'

because \{16\} as a repeater doesn't work.