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
···
--
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
···
On 2013-09-18, at 6:49 AM, Raj pal <lists@ruby-forum.com> 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".RAJ
--
Posted via http://www.ruby-forum.com/\.
--
Mike Stok <mike@stok.ca>
http://www.stok.ca/~mike/
The "`Stok' disclaimers" apply.
What have you tried?
You can use some tools to help you like www.rubular.com or regexpal.com
Jesus.
···
On Wed, Sep 18, 2013 at 12:49 PM, Raj pal <lists@ruby-forum.com> 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".
hi Mike Stok and Juan Pablo 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?
···
--
Posted via http://www.ruby-forum.com/.
hi Jesús Gabriel y Galán
I am sorry I haven't seen the comment of yours.yes i will check that
out.
hi Mike Stok,
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
···
--
Posted via http://www.ruby-forum.com/.
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
1. are] [you
I only need [are], how can i write to extract "are" alone?
···
--
Posted via http://www.ruby-forum.com/.
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?
···
--
Posted via http://www.ruby-forum.com/.
hi tamouse m
Thank you for your code,it's working fine too.
hi Robert Klemme
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
···
--
Posted via http://www.ruby-forum.com/.
hi Robert Klemme
Ohhhhhhhhhhhhhhhhhh, OK, I haven't noticed that. Thank you.
···
--
Posted via http://www.ruby-forum.com/.
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?
···
--
Posted via http://www.ruby-forum.com/.
hi Jesús Gabriel y Galán,
Yes I understood. Thank you.
···
--
Posted via http://www.ruby-forum.com/.
You should take Jesus’s 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
···
On 2013-09-18, at 7:07 AM, Raj pal <lists@ruby-forum.com> wrote:
hi Mike Stok,
it's working. Thank you very much.
RAJ
--
Posted via http://www.ruby-forum.com/\.
--
Mike Stok <mike@stok.ca>
http://www.stok.ca/~mike/
The "`Stok' disclaimers" apply.
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
···
On 2013-09-18, at 7:20 AM, Raj pal <lists@ruby-forum.com> wrote:
hi Jesús Gabriel y Galán
I am sorry I haven't seen the comment of yours.yes i will check that
out.hi Mike Stok,
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
--
Mike Stok <mike@stok.ca>
http://www.stok.ca/~mike/
The "`Stok' disclaimers" apply.
hi how [are] [you]
Now it return this
1. 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 ...
···
--
Posted via http://www.ruby-forum.com/\.
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
···
On Thu, Sep 19, 2013 at 10:56 AM, Raj pal <lists@ruby-forum.com> 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.
--
remember.guy do |as, often| as.you_can - without end
http://blog.rubybestpractices.com/
From the docs at: http://www.ruby-doc.org/core-2.0.0/String.html#method-i-5B-5D
"If a Regexp is supplied, the matching portion of the string is
returned. If a capture follows the regular expression, which may be a
capture group index or name, follows the regular expression that
component of the MatchData is returned instead."
When you surround some part of the regex in parens, that's a capturing
group, and by passing the second argument as a 1, you are telling that
you are interested in the first captured group. You can also have
named groups, instead of by index:
a[/\[(?<word>[^\]]*)\]/, :word]
Hope this helps,
Jesus.
···
On Thu, Sep 19, 2013 at 8:01 PM, Raj pal <lists@ruby-forum.com> wrote:
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?
or
s.scan /\[(.*?)\]/ # don't be greedy
···
On Sep 18, 2013, at 7:43 AM, Regis d'Aubarede <lists@ruby-forum.com> wrote:
hi how [are] [you]
Now it return this
1. are] [yous="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 ...
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 7:59 AM, Tamara Temple <tamouse.lists@gmail.com> wrote:
On Sep 18, 2013, at 7:43 AM, Regis d'Aubarede <lists@ruby-forum.com> wrote:
hi how [are] [you]
Now it return this
1. are] [yous="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 ...
or
s.scan /\[(.*?)\]/ # don't be greedy
--
remember.guy do |as, often| as.you_can - without end
http://blog.rubybestpractices.com/