How Do I Use a Regex in a Shell Script?
I Am Trying to Match a String with a Regex in a Shell Script. This String Is a Parameter of the Script ( $1 ) and It Is a Date (Mm/Dd/Yyyy) the Regex I'm...
I am trying to match a string with a regex in a shell script. This string is a parameter of the script ( $1 ) and it is a date (MM/DD/YYYY) The regex I'm trying to use is :
^\d{2}[\/\-]\d{2}[\/\-]\d{4}$
It seems to work, I tried it on several regex tests websites.
My shell code is :
REGEX_DATE="^\d{2}[\/\-]\d{2}[\/\-]\d{4}$"
echo "$1" | grep -q $REGEX_DATE
echo $?
The "echo $?" returns 1 no matter is the string I'm putting in parameter.
Do you guys have an idea ?
Thanks !
3 Answers
To complement the existing helpful answers:
Using Bash's own regex-matching operator, =~, is a faster alternative in this case, given that you're only matching a single value already stored in a variable:
set -- '12-34-5678' # set $1 to sample value
kREGEX_DATE='^[0-9]{2}[-/][0-9]{2}[-/][0-9]{4}$' # note use of [0-9] to avoid \d
[[ $1 =~ $kREGEX_DATE ]]
echo $? # 0 with the sample value, i.e., a successful match
Note, however, that the caveat re using flavor-specific regex constructs such as \d equally applies:
While =~ supports EREs (extended regular expressions), it also supports the host platform's specific extension - it's a rare case of Bash's behavior being platform-dependent.
To remain portable (in the context of Bash), stick to the POSIX ERE specification.
Note that =~ even allows you to define capture groups (parenthesized subexpressions) whose matches you can later access through Bash's special ${BASH_REMATCH[@]} array variable.
Further notes:
$kREGEX_DATEis used unquoted, which is necessary for the regex to be recognized as such (quoted parts would be treated as literals).While not always necessary, it is advisable to store the regex in a variable first, because Bash has trouble with regex literals containing
\.- E.g., on Linux, where
\<is supported to match word boundaries,[[ 3 =~ \<3 ]] && echo yesdoesn't work, butre='\<3'; [[ 3 =~ $re ]] && echo yesdoes.
- E.g., on Linux, where
I've changed variable name
REGEX_DATEtokREGEX_DATE(ksignaling a (conceptual) constant), so as to ensure that the name isn't an all-uppercase name, because all-uppercase variable names should be avoided to prevent conflicts with special environment and shell variables.
I think this is what you want:
REGEX_DATE='^\d{2}[/-]\d{2}[/-]\d{4}$'
echo "$1" | grep -P -q $REGEX_DATE
echo $?
I've used the -P switch to get perl regex.
the problem is you're trying to use regex features not supported by grep. namely, your \d won't work. use this instead:
REGEX_DATE="^[[:digit:]]{2}[-/][[:digit:]]{2}[-/][[:digit:]]{4}$"
echo "$1" | grep -qE "${REGEX_DATE}"
echo $?
you need the -E flag to get ERE in order to use {#} style.
Maya is a hardware enthusiast who tests and reviews smart home devices, smartphones, wearables, and audio gear. She focuses on practical consumer value and build quality.
