bash string check

Question

bash string check

w5000 (TechnicalUser)

(OP)

8 Sep 16 13:43

Hello,
why test returns "clean" what is marked in red in belongs strings comparison?

CODE -->

# d=274y82heE
# [[ $d =~ [^[:alnum:]] ]] && echo "not clean $d" || echo "clean $d"
clean 274y82heE
# d=274y82he^E
# [[ $d =~ [^[:alnum:]] ]] && echo "not clean $d" || echo "clean $d"
not clean 274y82he^E

# d=/one/ssfdsf/274y82he^E/end/end
# [[ $d =~ /one/ssfdsf/[^[:alnum:]]/end/end ]] && echo "not clean $d" || echo "clean $d"
clean /one/ssfdsf/274y82he^E/end/end
# [[ $d =~ /one/ssfdsf/[[:alnum:]]/end/end ]] && echo "not clean $d" || echo "clean $d"
clean /one/ssfdsf/274y82he^E/end/end
# [[ $d =~ ^/one/ssfdsf/[[:alnum:]]/end/end ]] && echo "not clean $d" || echo "clean $d"
clean /one/ssfdsf/274y82he^E/end/end
# [[ $d =~ ^/one/ssfdsf/[[:alnum:]]/end/end$ ]] && echo "not clean $d" || echo "clean $d"
clean /one/ssfdsf/274y82he^E/end/end

feherke (Programmer) · Answer 1 · 2016-09-08 13:55:13

Hi

[[:alnum:]] - one alphanumeric character
[^[:alnum:]] - one non-alphanumeric character

In that place you have 8 alphanumeric character, then 1 non-alphanumeric character, then 1 alphanumeric character.

Not sure what is your goal, but try something like this :

CODE --> Bash

[[ $d =~ /one/ssfdsf/.*[^[:alnum:]].*/end/end ]]

Feherke.
feherke.ga

w5000 (TechnicalUser) · Answer 2 · 2016-09-08 14:02:42

hello,

ok. but it doesn't return clean when I expect it clean:

CODE -->

# d=/one/ssfdsf/274y82heE/end/end
# [[ $d =~ ^/one/ssfdsf/[[:alnum:]].*/end/end$ ]] && echo "not clean $d" || echo "clean $d"
not clean /one/ssfdsf/274y82heE/end/end

the goal is to check if there is "alnum" between ^/one/ssfdsf/ and /end/end$

feherke (Programmer) · Answer 3 · 2016-09-08 14:32:00

Hi

Your current regular expression says :

starts with "/one/ssfdsf/"
then 1 alphanumeric character
then any number of any character
ends with "/end/end"

As your current string looks exactly like that, so the 1^st echo gets executed, saying "not clean".

Quote (w5000)
the goal is to check if there is "alnum" between ^/one/ssfdsf/ and /end/end$

Your current regular expression says in that place has to be 1 alnum followed by anything or nothing.

I think you want this, so in that place to have 1 or more alphanumeric characters :

CODE --> Bash

[[ $d =~ ^/one/ssfdsf/[[:alnum:]]+/end/end$ ]] && echo "clean $d" || echo "not clean $d"

Feherke.
feherke.ga

w5000 (TechnicalUser) · Answer 4 · 2016-09-09 06:18:13

thx.
indeed, I was pretty sure I had tested [[:alnum:]]+ in one of my first tries... I must had done some mistake

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