Stripping single and double quotes in a string using bash / standard Linux commands only

This should do it:

sed "s/^\([\"']\)\(.*\)\1\$/\2/g" in.txt

Where in.txt is:

"Fo'od'
'Food'
"Food"
"Fo"od'
Food
'Food"
"Food'
'Fo'od'
"Fo'od"
Fo'od
'Fo"od'
"Fo"od"
Fo"od

And expected.txt is:

"Fo'od'
Food
Food
"Fo"od'
Food
'Food"
"Food'
Fo'od
Fo'od
Fo'od
Fo"od
Fo"od
Fo"od

You can check they match with:

diff -s <(sed "s/^\([\"']\)\(.*\)\1\$/\2/g" in.txt) expected.txt

You could use tr:

echo "$string" | tr -d 'chars to delete' 

... also works, however 'tr' is known to be problematic on much older (circa Redhat 9-ish) distributions. tr is an abbreviation for 'translate', commonly used in pipes to transform input. The -d option simply means 'delete'.

Most modern versions also contain predefined macros to transform upper to lower, lower to upper, kill white space, etc. Hence, if you use it, take a second to poke at what else it does (see the help output / man page), comes in handy.

VAR="'FOOD'"

VAR=$(eval echo $VAR)

Explanation: Since quotes are already understood by the shell you can ask the shell to evaluate a command that just echos the quoted string, the same way it does when you type it yourself.

Here, eval echo $VAR expands to eval echo 'FOOD' because the quotes are actually part of the value of VAR. If you were to run echo 'FOOD' into the shell you'd get FOOD (without the quotes). That's what eval does: it takes its input and runs it like a shell command.

⚠CODE INJECTION!

eval expose scripts to code injection.

VAR=';ls -l'

VAR=$(eval echo $VAR)

will cause execution of ls -l.

Much more harmful codes could be injected here.