Print multiple-line section between two strings only if this section contains certain keyword using bash?

Question

I have a file named animals which has following entries:

$ cat animals
name: elephant
class: mammal
habitat: India

name: Baltic herring
class: fish
habitat: Baltic sea

name: wolf
class: mammal
habitat: Sweden

name: eagle
class: bird
habitat: USA
$ 

As seen above, this file contains four sections. I would like to print out only those sections which contain string mammal and I would like to do this in bash, ie not use sed or awk . I did it like this:

$ cat animals | while read -r line; do
>   if [[ "$line" =~ ^name ]]; then
>     section="$line"
>     while read -r line2; do
>       if [[ "$line2" =~ ^$ ]]; then
>         break
>       else
>         section+=$'\n'"$line2"
>       fi
>     done
>   echo "$section" | grep -q mamm && echo "$section"$'\n' ; section=
>   fi
> done
name: elephant
class: mammal
habitat: India

name: wolf
class: mammal
habitat: Sweden

$ 

However, is there a more elegant(shorter and not so cumbersome) way to achieve this in bash? Or is this a point where one should look into tools like awk instead?

Answer 1

一个简单的怎么样

grep mammal -C 1 --group-separator="" animals

Answer 2

Yes, you should look into awk. It allows you to do this much more simply:

awk -v RS= -v ORS='\n\n' '/mammal/' file

This unsets the record separator, so that each block is treated as a single record. It then prints each record that matches the pattern "mammal", followed by two newlines, producing the same output as shown in the question.