Extract number of variable length from string - regex

Question

I want to extract a number of variable length from a string.
The string looks like this:
used_memory:1775220696
I would like to have the 1775220696 part in a variable. There are a lot of questions about this, but I could not find a solution that suits my needs.

Answer 1

You can use cut:
my_val=$(echo "used_memory:1775220696" | cut -d':' -f2)
Or also awk:
my_val=$(echo "used_memory:1775220696" | awk -F':' '{print $2}')

Answer 2

Use parameter expansion:
string=used_memory:1775220696
num=${string#*:} # Delete everything up to the first colon.

Answer 3

I used to use egrep
echo used_memory:1775220696 | egrep -o [0-9]+
Output:
1775220696

Answer 4

use the regex:
s/^[^:]*://g
you use it with sed or perl and get the part you needed.
> echo "used_memory:1775220696" | perl -pe 's/^[^:]*://g'
1775220696

Answer 5

bash supports regular-expression matching, but for a simple case like this it is overkill; use parameter expansion (see choroba's answer).
For the sake of completeness, here's an example using regular expression matching:
[[ $string =~ (.*):([[:digit:]]+) ]] && num=${BASH_REMATCH[2]}

Answer 6

Can be done using awk, like this:
var=`echo "used_memory:1775220696" | awk -F':' '{print $2;}'`
echo $var
output:
1775220696

Answer 7

If your number could be anywhere in the string, but you know that the digits are contiguous, you can use shell parameter expansion to remove everything that is not a digit:
$ str=used_memory:1775220696
$ num=${str//[!0-9]}
$ echo "$num"
1775220696
This works also for used_memory:1775220696andmoretext and 123numberfirst. However, something like abc123def456 would become 123456.