Count the number of digits in a bash variable

Question

I have a number num=010. I would like to count the number of digits contained in this number. If the number of digits is above a certain number, I would like to do some processing.

In the above example, the number of digits is 3.

Thanks!

Answer 1

Assuming the variable only contains digits then the shell already does what you want here with the length Shell Parameter Expansion.

$ var=012
$ echo "${#var}"
3

Answer 2

In BASH you can do this:

num='a0b1c0d23'
n="${num//[^[:digit:]]/}"
echo ${#n}
5

Using awk you can do:

num='012'
awk -F '[0-9]' '{print NF-1}' <<< "$num"
3

num='00012'
awk -F '[0-9]' '{print NF-1}' <<< "$num"
5

num='a0b1c0d'
awk -F '[0-9]' '{print NF-1}' <<< "$num"
3

Answer 3

Assuming that the variable x is the "certain number" in the question

chars=`echo -n $num | wc -c`
if [ $chars -gt $x ]; then
   ....
fi

Answer 4

this work for arbitrary string mixed with digits and non digits:

ndigits=`echo $str | grep -P -o '\d' | wc -l`

demo:

$ echo sf293gs192 | grep -P -o '\d' | wc -l
       6

Answer 5

Using sed:

s="string934 56 96containing digits98w6"
num=$(echo "$s" |sed  's/[^0-9]//g')
echo ${#num}
10

Using grep:

s="string934 56 96containing digits98w6"
echo "$s" |grep -o "[0-9]" |grep -c ""
10