Bash Shell Count Number of Characters In a String or Word

by on June 22, 2011 · 10 comments· LAST UPDATED June 22, 2011


How do I count and print particular character (say D or digit 7) in a string or variable under Bash UNIX / Linux shell?

You can use UNIX / Linux command such as sed, grep, wc or bash shell itself to count number of characters in a string or word or shell variable.

grep Command To Count Number of Characters

Use the following syntax to count 's' character in a variable called $x:

x="This is a test"
grep -o "s" <<<"$x" | wc -l

Sample outputs:


To match both 's' and 'S', enter:

x="This is a test. S"
grep -o "[s|S]" <<<"$x" | wc -l

Sample outputs:


Count Number of Characters Using Bash Only

You can use Bash parameter expansion as follows:

x="This is a test"
echo "$y"
echo "${#y}"

To match both 's' and 'S', enter:

x="This is a test. S"
echo "${#y}"

Please note that all instructions were tested using:

  • Debian GNU/Linux, v6.x
  • GNU grep, v2.6.3
  • GNU bash, v4.1.5
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{ 10 comments… read them below or add one }

1 Justin Yang June 23, 2011 at 2:56 pm

Just complete this article and feel it’s helpful to me. However, could you offer us more details about the meaning of the following two lines:
echo “${#y}”
I don’t know what they mean. Thank you.


2 nixCraft June 25, 2011 at 2:01 pm

Find and replace all occurrences of ‘s’ i.e. when pattern begins with /, all matches of pattern are replaced with ‘s’ and stored to $y:


To find string ($y) length:

echo “${#y}”

See HowTo: Use Bash Parameter Substitution Like A Pro for more info.


3 Richard Black March 18, 2015 at 6:02 pm

Here is the breakdown for y=”${x//[^s]}”

This assumes that you already understand y=”${x}” and somewhat understand %, %%, # and ##.

/ indicates we will be doing pattern substitution as described in the bash man page.

The pattern section is “/[^s]”

This pattern has a leading “/” which means bash will replace all matches of pattern.

The pattern itself “[^s]” means to match any character that is not an “s”. Look up leading “^” inside of square brackets “[ ]”.

The string is not included because we want to delete any character that is not an “s”. Because the replace string is empty then the “/” that divides pattern from string is not included.

Then $y is left with only “s”s because everything else has been deleted. So “${#y}” will give you the total count of characters in $y.


4 Justin Yang June 25, 2011 at 3:32 pm

Thank you for your reply with that good link you give.


5 mandela900 June 27, 2011 at 8:22 am

A=”Ciao Mondo”
echo ${#A}


6 nixCraft June 29, 2011 at 4:42 pm

A whitespace is also counted as a character.


7 hitendra April 11, 2013 at 3:44 pm



8 Uchmann July 14, 2011 at 8:12 am

Thanks for the post
is it wrong doing it this way.
1. Use the following syntax to count ‘s’ character in a variable called $x:
echo “$x” | grep -o “s” | wc -l

2. To match both ‘s’ and ‘S’, enter:
echo “$x” | grep -oi “s” | wc -l


9 Adilson April 6, 2013 at 3:20 pm

[Fedora Linux & Debian Linux]

hello, thanks for scripts.
I have a directory with 100 files.
I wonder which of those 100 files has more than 32 characters in the name.
can you help me?


10 Jared June 4, 2014 at 3:10 pm

#!/usr/bin/env bash
# if $filename more than 32 then
# @param $1 directory


for file in “$(ls $dir)”
if [ “$file” -gt “32” ]; then
echo “Filename is more than 32 chars!”
echo “Filename is NOT more than 32 chars!”

Wrote this on the fly, should work.


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