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

I believe the cleanest way to strip a single character from a string with bash is:

echo ${COMPANY_NAME:: -1}

but I haven't been able to embed the grep piece within the curly braces, so your particular task becomes a two-liner:

COMPANY_NAME=$(grep "company_name" file.txt); COMPANY_NAME=${COMPANY_NAME:: -1} 

This will strip any character, semicolon or not, but can get rid of the semicolon specifically, too. To remove ALL semicolons, wherever they may fall:

echo ${COMPANY_NAME/;/}

To remove only a semicolon at the end:

echo ${COMPANY_NAME%;}

Or, to remove multiple semicolons from the end:

echo ${COMPANY_NAME%%;}

For great detail and more on this approach, The Linux Documentation Project covers a lot of ground at http://tldp.org/LDP/abs/html/string-manipulation.html