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:
3
To match both ‘s’ and ‘S’, enter:
x="This is a test. S" grep -o "[s|S]" <<<"$x" | wc -l
Sample outputs:
4
Count Number of Characters Using Bash Only
You can use Bash parameter expansion as follows:
x="This is a test" y="${x//[^s]}" echo "$y" echo "${#y}"
To match both ‘s’ and ‘S’, enter:
x="This is a test. S" y="${x//[^s|S]}" echo "${#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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Hi,
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:
y=”${x//[^s]}”
and
echo “${#y}”
I don’t know what they mean. Thank you.
Find and replace all occurrences of ‘s’ i.e. when pattern begins with /, all matches of pattern are replaced with ‘s’ and stored to $y:
y=”${x//[^s]}”To find string ($y) length:
echo “${#y}”See HowTo: Use Bash Parameter Substitution Like A Pro for more info.
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.
Hi Richard – Thanks a lot for the excellent explanation ……. as more theory more confusion ….some time :) .Thanks again!!
Thank you for your reply with that good link you give.
A=”Ciao Mondo”
echo ${#A}
10
A whitespace is also counted as a character.
thanku
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
[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?
Thanks
#!/usr/bin/env bash
# if $filename more than 32 then
# @param $1 directory
dir=$1
for file in “$(ls $dir)”
do
file=”${#file}”
if [ “$file” -gt “32” ]; then
echo “Filename is more than 32 chars!”
else
echo “Filename is NOT more than 32 chars!”
fi
done
Wrote this on the fly, should work.