how to extract grep and cut into a bash array - linux

Question

I tried:
here is content of file.txt
some other text
#1.something1=kjfk
#2.something2=dfkjdk
#3.something3=3232
some other text
bash script:
ids=( `grep "something" file.txt | cut -d'.' -f1` )
for id in "${ids[#]}"; do
echo $id
done
result:
(nothing newline...)
(nothing newline...)
(nothing newline...)
but all it prints is nothing like newline for every such id found what am i missing?

Answer 1

Your grep and cut should be working but you can use awk and reduce 2 commands into one:
while read -r id;
echo "$id"
done < <(awk -F '\\.' '/something/{print $1}' file.txt)
To populate an array:
ids=()
while read -r id;
ids+=( "$id" )
done < <(awk -F '\\.' '/something/{print $1}' file.txt)

Answer 2

You can use grep's -o option to output only the text matched by a regular expression:
$ ids=($(grep -Eo '^#[0-9]+' file.txt))
$ echo ${ids[#]}
#1 #2 #3
This of course doesn't check for the existence of a period on the line... If that's important, then you could either expand things with another pipe:
$ ids=($(grep -Eo '^#[0-9]+\.something' file.txt | grep -o '^#[0-9]*'))
or you could trim the array values after populating the array:
$ ids=($(grep -Eo '^#[0-9]+\.something' file.txt))
$ echo ${ids[#]}
#1.something #2.something #3.something
$ for key in "${!ids[#]}"; do ids[key]="${ids[key]%.*}"; done
$ echo ${ids[#]}
#1 #2 #3