bash Why doesnt my regex work?

Question

I made this regex but it is not working.

I got this example string: ('/test/test/test/test-Test/TEST/test.sql'),

my bash code:

if [[ ${arrayQuery[$i]} =~ ([a-z0-9]+)\/([a-z0-9]+)\/([a-z0-9]+)\/([a-z0-9-]+)\/([a-z0-9]+)\/([a-z0-9]+).([a-z0-9]+) ]]; then
        queryName=$1
        echo "test $queryName"
fi

it is not priting anything can anyone explain my why this is not working?

i tried my regex on regex101.com and the regex did work on this website.

It looks to me like you're only looking for lower case letters and digits, but you have upper case letters in the example string, so it should not match. — Eric Renouf
– Eric Renouf, Commented Oct 15, 2015 at 14:11
You don't need to escape the / in the regex. bash doesn't use them as a delimiter the way sed, for example`, does, so they have no special meaning. — chepner
– chepner, Commented Oct 15, 2015 at 14:27
May I ask you why you're using this regex? are you trying to split your path into its dir components? — gniourf_gniourf
– gniourf_gniourf, Commented Oct 15, 2015 at 14:40
Hi guys, sorry for the late reactions had a busy weekend:S Anyway thanks for the answers! @gniourf_gniourf yes I want to split the path into components. Is regex a good idea todo this? — kaspertje100
– kaspertje100, Commented Oct 19, 2015 at 10:39

marek.jancuska · Accepted Answer · 2015-10-15 14:28:09Z

2

you need to escape the dot, otherwise it matches any character
your example string contains uppercase, but your regex only accepts lowercase letters

(edit: no quoting needed)

edited Oct 15, 2015 at 14:28

answered Oct 15, 2015 at 14:18

marek.jancuska

3101 silver badge7 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

guillaume guerin · Accepted Answer · 2015-10-15 14:57:32Z

1

Your example does not work in regex101 with the string you provided for several reasons:

Your string starts with a '/' but your regex starts with ([a-z0-9]+)
Your string contains Upper case letters yet you don't use [A-Z] in your regex
Your string contains a '-' yet your regex does not parse it, try adding \-
You did not escape the dot replace it with '\.', by default '.' means all character

This regular expression would do the trick (link to regex101) :

\/[a-zA-Z0-9\-]+\/[a-zA-Z0-9\-]+\/[a-zA-Z0-9\-]+\/[a-zA-Z0-9\-]+\/[a-zA-Z0-9\-]+\/[a-zA-Z0-9\-]+\.[a-zA-Z0-9\-]+

I guess this string represent a SQL file in your hard drive, this regex can be shortened to :

(\/[a-zA-Z0-9\-]+)+\.sql

and does not depend on how many folders you have in your directory tree.

edited Oct 15, 2015 at 14:57

answered Oct 15, 2015 at 14:43

guillaume guerin

3772 gold badges6 silver badges17 bronze badges

3 Comments

kaspertje100 Over a year ago

so I changed my code to this:

code echo "${arrayQuery[$i]}" 	if [[ ${arrayQuery[$i]} =~ (\/([a-zA-Z0-9\-]+)\/([a-zA-Z0-9\-]+)\/([a-zA-Z0-9\-]+)\/([a-zA-Z0-9\-]+)\/([a-zA-Z0-9\-]+)\/([a-zA-Z0-9\-]+)\.([a-zA-Z0-9\-]+)) ]]; then 		queryName=$1 		echo "test $queryName" 	fi

kaspertje100 Over a year ago

It does go into the if statement now (in the console I see it prints test) however it doesnt print $queryName because the $1 variable is empty. I dont understand why the $1 is empty could some explain this to me please?

kaspertje100 Over a year ago

Neverm ind fixed it. I Used ${BASH_REMATCH[1]}" to print the result. Apperently $1 doesnt exists in bash??

reynoldsnlp · Accepted Answer · 2015-10-15 14:12:43Z

0

It should work if you add A-Z (capital letters) to all of the character sets.

answered Oct 15, 2015 at 14:12

reynoldsnlp

1,2851 gold badge22 silver badges52 bronze badges

Comments

chepner · Accepted Answer · 2015-10-15 14:54:07Z

0

You can use [[:alnum:]] as a short cut for the (correct) class [a-zA-Z0-9] to simplify your regex while fixing it. Even simpler would be to use a parameter to shorten it:

x='[[:alnum:]]+'
# Or, if you prefer
# x='[a-zA-Z0-9]+'
if [[ ${arrayQuery[$i]} =~ ($x)/($x)/($x)/($x)/($x)/($x)\.($x) ]]; then
    queryName=$1
    echo "test $queryName"
fi

There is also an option to make regular expression matches case-insensitive, which would allow you to use your current regular expression (with a few minor fixes):

shopt -s nocasematch
if [[ ${arrayQuery[$i]} =~ ([a-z0-9]+)/([a-z0-9]+)/([a-z0-9]+)/([a-z0-9-]+)/([a-z0-9]+)/([a-z0-9]+)\.([a-z0-9]+) ]]; then
    queryName=$1
    echo "test $queryName"
fi
shopt -u nocasematch  # Turn it off again

edited Oct 15, 2015 at 14:54

answered Oct 15, 2015 at 14:32

chepner

538k77 gold badges594 silver badges746 bronze badges

4 Comments

marek.jancuska Over a year ago

you missed that one "-" character , needed to match "test-Test" part

chepner Over a year ago

Good point. Not sure if that should be in each component, or just in the component that contains one in the user's input. It's not clear exactly what the regular expression should be.

guillaume guerin Over a year ago

you also missed the first character '/' of the string '/test/test/test/test-Test/TEST/test.sql' because your regex starts with a "([a-z0-9]+)"

chepner Over a year ago

The leading / is only necessary if you want to ensure that the path has exactly six components, rather than at least six components. As is, you will match and capture the last six in either case.

Community · Accepted Answer · 2017-05-23 11:58:39Z

0

From your comment it seems that you want to split the path into its components. The best option, in Bash, is to use this:

mypath=/testa/testb/testc/test-Testd/TESTe/test.sql
IFS=/ read -r -d '' -a components < <(printf '%s\0' "$mypath")

Like so, you'll have an array components that will contain each component of your path:

gniourf$ declare -p components
declare -a components='([0]="" [1]="testa" [2]="testb" [3]="testc" [4]="test-Testd" [5]="TESTe" [6]="test.sql")'

You don't need (and don't want) to use a regex for this.

Also see this question: How do I split a string on a delimiter in Bash?

edited May 23, 2017 at 11:58

CommunityBot

11 silver badge

answered Oct 19, 2015 at 11:11

gniourf_gniourf

47.4k10 gold badges105 silver badges113 bronze badges

Collectives™ on Stack Overflow

bash Why doesnt my regex work?

5 Answers 5

Comments

3 Comments

Comments

4 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

Comments

3 Comments

Comments

4 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related