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

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 "\"*.*\""