Count occurrences of a char in a string using Bash

Question

you can for example remove all other chars and count the whats remains, like:

var="text,text,text,text"
res="${var//[^,]}"
echo "$res"
echo "${#res}"

will print

,,,
3

or

tr -dc ',' <<<"$var" | awk '{ print length; }'

or

tr -dc ',' <<<"$var" | wc -c    #works, but i don't like wc.. ;)

or

awk -F, '{print NF-1}' <<<"$var"

or

grep -o ',' <<<"$var" | grep -c .

or

perl -nle 'print s/,//g' <<<"$var"

I would use the following awk command:

string="text,text,text,text"
char=","
awk -F"${char}" '{print NF-1}' <<< "${string}"

I'm splitting the string by $char and print the number of resulting fields minus 1.

If your shell does not support the <<< operator, use echo:

echo "${string}" | awk -F"${char}" '{print NF-1}'

You can do it by combining tr and wc commands. For example, to count e in the string referee

echo "referee" | tr -cd 'e' | wc -c

output

4

Explanations: Command tr -cd 'e' removes all characters other than 'e', and Command wc -c counts the remaining characters.

Multiple lines of input are also good for this solution, like command cat mytext.txt | tr -cd 'e' | wc -c can counts e in the file mytext.txt, even thought the file may contain many lines.

*** Update ***

To solve the multiple spaces in from of the number (@tom10271), simply append a piped tr command:

 tr -d ' '

For example:

echo "referee" | tr -cd 'e' | wc -c | tr -d ' '

Building on everyone's great answers and comments, this is the shortest and sweetest version:

grep -o "$needle" <<< "$haystack" | wc -l

awk is very cool, but why not keep it simple?

num=$(echo $var | grep -o "," | wc -l)