string pattern replacement in bash

Using bash, I have a string:

str='a $s'
echo ${str/\$/\$a}
# a $as
str='a $1'
echo ${str/\$/\$a}
# a $a1

How modify pattern that replacement performs only if word开发者_开发技巧 starts with a letter?

I want

str='a $s'
echo ${str/??/??}
# a $as
str='a $1'
echo ${str/??/??}
# a $1

---

This would have been trivial if the regex support in Bash allowed lookahead. The solution would then simply be ${str/\$(?=[a-zA-Z])/\$a}. Unfortunately, that is not the case.

You can still do it if you don't might using sed.

str='a $s'
echo $str | sed 's/\$\([a-zA-Z]\)/\$a\1/g' # gives you a $as

str='a $1'
echo $str | sed 's/\$\([a-zA-Z]\)/\$a\1/g' # gives you a $1

\$\([a-zA-Z]\) matches $ followed by an alphabet and store the alphabet. If a match is found, it is replaced with $a followed by the stored alphabet (\1). For more details, see this tutorial.

The trailing g means it will match all occurrences in the string. Remove that if you only want to only replace the first occurrence.

---

You can use an IF to check if any any letter follows the dollar sign and in case do the replacement:

if [[ "$str" =~ \$[[:alpha:]] ]]; then 
    echo ${str/$/\$a}; 
else 
    echo $str; 
fi

---

Demonstration:

for str in 'a $1' 'a $s'
do
    [[ $str =~ \$([[:alpha:]]) ]]
    echo ${str/\$/\$${BASH_REMATCH[1]:+a}}
done

Result:

a $1
a $as