GREP: variable in regular expression
bash, grep, regex
Solution
You need to use double quotes so that the shell expands the parameter before passing the resulting argument to `grep`:
if grep "^[0-9a-zA-Z]\{1,$UUID_LEN\}$" <<<$1 ; then ...
`bash` can perform regular expression matching itself, without having to start another process to run `grep`:
if [[ $1 =~ ^[0-9a-zA-Z]{1,$UUID_LEN}$ ]]; then
Problem
If I want to look whether a string is alphanumeric and shorter than a certain value, say 10, I would do like this (in BASH+GREP): `if grep '^[0-9a-zA-Z]\{1,10\}$' <<<$1 ; then ...` (BTW: I'm checking for $1, i.e. the first argument) What if I want the value 10 to be written on a variable, e.g. ``` UUID_LEN=10 if grep '^[0-9a-zA-Z]\{1,$UUID_LEN\}$' <<<$1 ; then ... ``` I tried all sort of escapes, braces and so on, but could not avoid the error message ``` grep: Invalid content of \{\} ``` After googling and reading bash and grep tutorials I'm pretty convinced it can't be done. Am I wrong? Any way to go around this?