to find numbers in a string

Shell Programming and Scripting
1

I writing my script and got stuck in this function. Can someone help me?
I need to extract out the numbers inside a string.
Ex:
INPUT -> OUTPUT
abcdef123 -> 123
abc123def -> 123
123abcdef -> 123
a123bc45d -> 123 45
abcdefghi -> -1

Thank you!

2 
sed "s/[^0-9]//g;s/^$/-1/;" filename
3

one way

echo "salkf4243lsfk" | tr -d '[[:alpha:]]' #or [a-z]

or in bash

#s="slfk32432sdf"
#echo ${s//[a-zA-Z]/} # or [!0-9]
4

Thanks all!
It works.
But how can we separate each numbers?
Anyway it's fine to me!

5 
sed "s/[^0-9]*\([0-9][0-9]*\)[^0-9]*/ \1 /g;s/^[^0-9][^0-9]*$/-1/" filename
6 
awk '{ gsub(/[[:alpha:]]|[[:punct:]]/," ")}1' file
7

Hi I have a similar problem, I would like to get a number that falls within a range from a string.

ex. input: "abcd 9034 efg 1234 hi"
output: 9034

I would only want to get the number range from 9000-9999

I've tried to use sed/grep to get what I want for hours but not able to do it, can anyone advice on this? Thanks a lot!!!

8 
grep  '^9[0-9][0-9][0-9]$' <(printf "%s\n" $s)

Or GNU grep:

grep -oE '\b9[0-9]{3}\b' <(printf "$s")
9

I tried

grep '^9[0-9][0-9][0-9]$' <( printf "%s\n", "9012 abc" )

But it returns "Missing name for redirect."

10

and doesn't grep only grep the whole line?

11

I didn't quote the variable in that example.

grep '^9[0-9][0-9][0-9]$' <( printf "%s\n", 9012 abc )

For older shells:

printf "%s\n", 9012 abc|grep '^9[0-9][0-9][0-9]$'
12

Thanks radoulov!! Really appreciate your help! It worked great.

I always use echo, I just wonder why echo doesn't work like printf, I tried

echo "9012 abc"|grep '^9[0-9][0-9][0-9]$'
13

Because you need the newline between the words (\n).

You can do it with GNU grep:

zsh-4.3.4-dev-2% echo "9012 abc 99999"|grep -Eo '\b9[0-9][0-9][0-9]\b'  
9012
zsh-4.3.4-dev-2% echo "9012 abc 99999"|grep -Eo '9[0-9][0-9][0-9]'  
9012
9999