How to remove the last character from a bash grep output

Question

COMPANY_NAME=`cat file.txt | grep \"company_name\" | cut -d \'=\' -f 2` 

outputs something like this

\"Abc Inc\";

Answer 1

In Bash using only one external utility:

IFS='= ' read -r discard COMPANY_NAME <<< $(grep "company_name" file.txt)
COMPANY_NAME=${COMPANY_NAME/%?}

Answer 2

Assuming the quotation marks are actually part of the output, couldn't you just use the -o switch to return everything between the quote marks?

COMPANY_NAME="\"ABC Inc\";" | echo $COMPANY_NAME | grep -o "\"*.*\""

Answer 3

don't have to chain so many tools. Just one awk command does the job

 COMPANY_NAME=$(awk -F"=" '/company_name/{gsub(/;$/,"",$2) ;print $2}' file.txt)

Answer 4

I am not finding that sed 's/;$//' works. It doesn't trim anything, though I'm wondering whether it's because the character I'm trying to trim off happens to be a "$". What does work for me is sed 's/.\{1\}$//'.

Answer 5

Some refinements to answer above. To remove more than one char you add multiple question marks. For example, to remove last two chars from variable $SRC_IP_MSG, you can use:

SRC_IP_MSG=${SRC_IP_MSG%??}

Answer 6

Using sed, if you don't know what the last character actually is:

$ grep company_name file.txt | cut -d '=' -f2 | sed 's/.$//'
"Abc Inc"