How to remove the beginning and end of a string in bash-CodePudding

I know how to extract a string by removing a prefix or suffix, but I don't know how to do both. Concretely, in the example below, how can I display inside my for-loop the names without a_ and _b?

$ touch a_cat_b a_dog_b a_food_b

$ for i in * ; do echo $i without a_ is ${i##a_} ;done;
a_cat_b without a_ is cat_b
a_dog_b without a_ is dog_b
a_food_b without a_ is food_b

$ for i in * ; do echo $i without _b is ${i%_b} ;done;
a_cat_b without _b is a_cat
a_dog_b without _b is a_dog
a_food_b without _b is a_food

CodePudding user response：

You can use the =~ operator:

#!/bin/bash

for f in *; do
    if [[ $f =~ ^a_(.*)_b$ ]]; then
        echo "$f without leading a_ and trailing _b is ${BASH_REMATCH[1]}"
    fi
done

CodePudding user response：

Use parameter expansion to select just the middle of the string?

for i in a_cat_b a_dog_b a_food_b; do
    printf "%s minux prefix and suffix is: %s\n" "$i" "${i:2:-2}"
done

${i:2-2} is the substring starting with the third (0-based indexes) character of $i, stopping at 2 before the end of the string.

This does assume the text you want to strip is fixed-length, of course.

CodePudding user response：

With bash and a regex:

for i in *; do
  if [[ $i =~ ^([^_]*)_(.*)_([^_]*)$ ]]; then
    declare -p BASH_REMATCH
  fi
done

Output:

declare -ar BASH_REMATCH=([0]="a_cat_b" [1]="a" [2]="cat" [3]="b")
declare -ar BASH_REMATCH=([0]="a_dog_b" [1]="a" [2]="dog" [3]="b")
declare -ar BASH_REMATCH=([0]="a_food_b" [1]="a" [2]="food" [3]="b")

CodePudding user response：

If a_ and _b can only possibly match the start and the end of the string respectively then you can also do:

#!/bin/bash
shopt -s extglob

for f in a_cat_b a_dog_b a_food_b
do
    echo "${f//@(a_|_b)}"
done

cat
dog
food