GREP: variable in regular expression

Question

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?

possible duplicate of Add grep command to bash script

tripleee
– tripleee

2013-09-05 17:15:18 +00:00
Commented Sep 5, 2013 at 17:15 — tripleee
– tripleee, Commented Sep 5, 2013 at 17:15

chepner · Accepted Answer · 2013-07-02 15:58:04Z

3

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

answered Jul 2, 2013 at 15:58

chepner

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1 Comment

KeinReverb Leonardo Gabrielli Over a year ago

Of course! The answer was so simple! The inverted commas " " will do the trick as differently from ' ' they expand what's inside... the answer. Thanks

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