hi
a=“hi how [are] you”
Can I have the regular expression which returns the string which comes
under ‘[]’. In the above case it’s “are”.
RAJ
On 2013-09-18, at 6:49 AM, Raj pal [email protected] wrote:
Posted via http://www.ruby-forum.com/.
One way to do it might be:
ratdog:~ mike$ pry
[1] pry(main)> a=“hi how [are] you”
=> “hi how [are] you”
[2] pry(main)> /[(.)]/.match a
=> #<MatchData “[are]” 1:“are”>
[3] pry(main)> (/[(.)]/.match a)[1]
=> “are”
Hope this helps,
Mike
–
Mike S. [email protected]
http://www.stok.ca/~mike/
The “`Stok’ disclaimers” apply.
On Wed, Sep 18, 2013 at 12:49 PM, Raj pal [email protected] wrote:
hi
a=“hi how [are] you”
Can I have the regular expression which returns the string which comes
under ‘[]’. In the above case it’s “are”.
What have you tried?
You can use some tools to help you like www.rubular.com or regexpal.com
Jesus.
El miércoles, 18 de septiembre de 2013 12:49:15 UTC+2, Raj pal escribió:
–
Posted via http://www.ruby-forum.com/.
There are several ways. One of them would be:
“hi how [are] you”.scan(/[(.*)]/).flatten.first
=> “are”
Hope it helps.
hi Mike S.,
it’s working. This is what I wanted.Thank you very much.
RAJ
hi Juan P. Avello
Thank you.
hi Mike S. and Juan P. Avello
I used your regular expression,it’s working, Now I need to avoid the
second instance of [], for an example, “hi [are] you,I am [fine]”. Here
I only need “are” and “fine” should be eliminated, How would I do that?
On 2013-09-18, at 7:07 AM, Raj pal [email protected] wrote:
hi Mike S.,
it’s working. Thank you very much.
RAJ
–
Posted via http://www.ruby-forum.com/.
You should take Jesuss advice and use a tool like rubular or regexpal to
make sure you understand what is happening, and test out whether it
works as you want for odd cases.
Consider a string like “this [is] a [test] string”. In a case like that
do you want “is”, “test”, “is] a [test”, [“is”, “test”], or some kind of
error action to be the result?
Hope this helps,
Mike
–
Mike S. [email protected]
http://www.stok.ca/~mike/
The “`Stok’ disclaimers” apply.
hi Jesús Gabriel y Galán
I tried with the tool which you mentioned. Thank you for that, But
unfortunately, i couldn’t find extract the first instance of [] while
this [] appears second time like
hi how [are] [you]
Now it return this
- are] [you
I only need [are], how can i write to extract “are” alone?
hi Jesús Gabriel y Galán
I am sorry I haven’t seen the comment of yours.yes i will check that
out.
hi Mike S.,
Input :hi [how] are you,I am [fine]
My needed output would be : how
But when I use your regular expression it’s returning like
[how] are you,I am [fine]
RAJ
On 2013-09-18, at 7:20 AM, Raj pal [email protected] wrote:
But when I use your regular expression it’s returning like
[how] are you,I am [fine]
RAJ
You can read the Class: Regexp (Ruby 2.0.0), in the
Repetition section there is an example of using a following ? to change
the greedy operator into a lazy operator which matches as little as
possible.
See if you can apply it to the regular expression I gave you to produce
the right results.
A tool like http://www.rubular.com makes it easy to experiment and see
what is going on.
Hope this helps,
Mike
–
Mike S. [email protected]
http://www.stok.ca/~mike/
The “`Stok’ disclaimers” apply.
hi Regis d’Aubarede,
Excellent, This is what I needed.
I changed my code according to my need
puts line[/[([^]])]/].delete(’[]’).delete(’’’) unless
line[/[([^]])]/].nil?
On Sep 18, 2013, at 7:43 AM, Regis d’Aubarede [email protected]
wrote:
Explanation: .* enlarge to maximum length ==> reach last bracket
s.scan /[([^]]*)]/
=> [[“fff”], [“ssss”]]Explanation: [^]]* enlarge to first bracket …
or
s.scan /[(.*?)]/ # don’t be greedy
hi how [are] [you]
Now it return this
- are] [you
s=“ssss[fff]gggg[ssss]ddddddd”
s.scan /[(.*)]/
=> [[“fff]gggg[ssss”]]
Explanation: .* enlarge to maximum length ==> reach last bracket
s.scan /[([^]]*)]/
=> [[“fff”], [“ssss”]]
Explanation: [^]]* enlarge to first bracket …
hi tamouse m
Thank you for your code,it’s working fine too.
hi Robert K.
You have included the 1 in the array like a[/[(.?)]/, 1], But
a[/[(.?)]/] also works in the same way ,It also returns the first
occurrence of the string.
RAJ
On Thu, Sep 19, 2013 at 7:59 AM, Tamara T. [email protected]
wrote:
or
s.scan /[(.*?)]/ # don’t be greedy
I think OP said he needs only the first match. In that case he can do
irb(main):001:0> a=“hi how [are] you”
=> “hi how [are] you”
irb(main):002:0> a[/[([^]])]/, 1]
=> “are”
irb(main):003:0> a[/[(.?)]/, 1]
=> “are”
Kind regards
robert
On Thu, Sep 19, 2013 at 10:56 AM, Raj pal [email protected] wrote:
You have included the 1 in the array like a[/[(.?)]/, 1], But
a[/[(.?)]/] also works in the same way ,It also returns the first
occurrence of the string.
I know. But without the 1 you’ll also get the brackets.
irb(main):001:0> “foo”[/f(o+)/]
=> “foo”
irb(main):002:0> “foo”[/f(o+)/, 1]
=> “oo”
And, btw, it’s not an Array. It’s operator String#[].
Cheers
robert
Hi Robert,
Can you explain me why this 1 removes the Bracket? I refered the
document but it’s not clear enough to understand. why it removes both
opening and closing bracket here?
hi Robert K.
Ohhhhhhhhhhhhhhhhhh, OK, I haven’t noticed that. Thank you.
hi Jesús Gabriel y Galán,
Yes I understood. Thank you.