Hi there!
I'm Alessandro from Italy and I started using ruby some days ago,
so... Hello, Community!
Well, I was trying to match a pattern multiple times. I tried both
with normal match() and scan(), but i can't get the desired result.
The subject string is something like:
"1a2bend" or "beg1a2b3c4dend"
more generally, it should match /^beg(\d\w)*end$/ : always a begin and
ending pattern, and a unspecified number of central pattern.
The problem is that the central pattern must be extracted for every
time it's encountered.
For example, trying with
"x1A2B3C4Dz".scan /^(x)(\d\w)*(z)$/
returns
[["x", "4D", "z"]]
while i need something like
[["x", "1A", "2B", "3C", "4D", "z"]]
Why does ()* match just the last one? How can i get all the ()* that it matches?
Probabily i'm doing something wrong, but can't understand where :\
On 7/31/07, Alessandro Re <akirosspower@gmail.com> wrote:
Hi there!
I'm Alessandro from Italy and I started using ruby some days ago,
so... Hello, Community!
Well, I was trying to match a pattern multiple times. I tried both
with normal match() and scan(), but i can't get the desired result.
The subject string is something like:
"1a2bend" or "beg1a2b3c4dend"
more generally, it should match /^beg(\d\w)*end$/ : always a begin and
ending pattern, and a unspecified number of central pattern.
The problem is that the central pattern must be extracted for every
time it's encountered.
For example, trying with
"x1A2B3C4Dz".scan /^(x)(\d\w)*(z)$/
returns
[["x", "4D", "z"]]
while i need something like
[["x", "1A", "2B", "3C", "4D", "z"]]
Why does ()* match just the last one? How can i get all the ()* that it matches?
Probabily i'm doing something wrong, but can't understand where :\
if "x1A2B3C4Dz" =~ /^(x)((?:\d\w)*)(z)$/
a, b = $1, $3 #
return [a] + $2.scan(/\d\w/).flatten + [b]
end
I don't know if it's possible to do it in one run though, maybe you
could use split as well...
Take care when doing nested searches as they will overwrite $1..9
(that's why I used a and b)
J.
路路路
On 7/31/07, Alessandro Re <akirosspower@gmail.com> wrote:
Hi there!
I'm Alessandro from Italy and I started using ruby some days ago,
so... Hello, Community!
Well, I was trying to match a pattern multiple times. I tried both
with normal match() and scan(), but i can't get the desired result.
The subject string is something like:
"1a2bend" or "beg1a2b3c4dend"
more generally, it should match /^beg(\d\w)*end$/ : always a begin and
ending pattern, and a unspecified number of central pattern.
The problem is that the central pattern must be extracted for every
time it's encountered.
For example, trying with
"x1A2B3C4Dz".scan /^(x)(\d\w)*(z)$/
returns
[["x", "4D", "z"]]
while i need something like
[["x", "1A", "2B", "3C", "4D", "z"]]
Why does ()* match just the last one? How can i get all the ()* that it matches?
Probabily i'm doing something wrong, but can't understand where :\
On 7/31/07, Alessandro Re <akirosspower@gmail.com> wrote:
For example, trying with
"x1A2B3C4Dz".scan /^(x)(\d\w)*(z)$/
returns
[["x", "4D", "z"]]
while i need something like
[["x", "1A", "2B", "3C", "4D", "z"]]
For example, trying with
"x1A2B3C4Dz".scan /^(x)(\d\w)*(z)$/
returns
[["x", "4D", "z"]]
while i need something like
[["x", "1A", "2B", "3C", "4D", "z"]]
Does this goes more into the direction you wanted:
Thanks, but i need to match the pattern OR don't match anything.
"lol1a2vasd".scan(/\d?\w/) => ["l", "o", "l", "1a", "2v", "a", "s", "d"]
while i need to be sure that the pattern begins with a regex "x" and
ends with "z"
(of course, x 1 a 2 b 3 c should be regexes not just chars)
thanks, you help is apreciated
路路路
On 7/31/07, Harry Kakueki <list.push@gmail.com> wrote:
On 7/31/07, Alessandro Re <akirosspower@gmail.com> wrote:
> For example, trying with
> "x1A2B3C4Dz".scan /^(x)(\d\w)*(z)$/
> returns
> [["x", "4D", "z"]]
> while i need something like
> [["x", "1A", "2B", "3C", "4D", "z"]]
>
Hi,
Mh well, to me it seems a normal regex processing (i mean, it *should*
require only one instruction, since this pattern can be read with just
one regex, even if ruby doesn't allow it... but it would be really
bad).
Anyway well, splitting it there are different ways to do it - thanks
for your sudjestion.
But if ruby make it possible with one call, i'd prefer to use it.
Thanks!
路路路
On 7/31/07, Jano Svitok <jan.svitok@gmail.com> wrote:
On 7/31/07, Alessandro Re <akirosspower@gmail.com> wrote:
> Hi there!
> I'm Alessandro from Italy and I started using ruby some days ago,
> so... Hello, Community!
>
> Well, I was trying to match a pattern multiple times. I tried both
> with normal match() and scan(), but i can't get the desired result.
>
> The subject string is something like:
> "1a2bend" or "beg1a2b3c4dend"
> more generally, it should match /^beg(\d\w)*end$/ : always a begin and
> ending pattern, and a unspecified number of central pattern.
> The problem is that the central pattern must be extracted for every
> time it's encountered.
> For example, trying with
> "x1A2B3C4Dz".scan /^(x)(\d\w)*(z)$/
> returns
> [["x", "4D", "z"]]
> while i need something like
> [["x", "1A", "2B", "3C", "4D", "z"]]
>
> Why does ()* match just the last one? How can i get all the ()* that it matches?
>
> Probabily i'm doing something wrong, but can't understand where :\
Try:
if "x1A2B3C4Dz" =~ /^(x)((?:\d\w)*)(z)$/
a, b = $1, $3 #
return [a] + $2.scan(/\d\w/).flatten + [b]
end
I don't know if it's possible to do it in one run though, maybe you
could use split as well...
Take care when doing nested searches as they will overwrite $1..9
(that's why I used a and b)
On 7/31/07, Alessandro Re <akirosspower@gmail.com> wrote:
Thanks, but i need to match the pattern OR don't match anything.
"lol1a2vasd".scan(/\d?\w/) => ["l", "o", "l", "1a", "2v", "a", "s", "d"]
while i need to be sure that the pattern begins with a regex "x" and
ends with "z"
2007/7/31, Alessandro Re <akirosspower@gmail.com>:
Mh well, to me it seems a normal regex processing (i mean, it *should*
require only one instruction, since this pattern can be read with just
one regex, even if ruby doesn't allow it... but it would be really
bad).
Anyway well, splitting it there are different ways to do it - thanks
for your sudjestion.
But if ruby make it possible with one call, i'd prefer to use it.
On 7/31/07, Alessandro Re <akirosspower@gmail.com> wrote:
Mh well, to me it seems a normal regex processing (i mean, it *should*
require only one instruction, since this pattern can be read with just
one regex, even if ruby doesn't allow it... but it would be really bad).
On 7/31/07, Robert Klemme <shortcutter@googlemail.com> wrote:
2007/7/31, Alessandro Re <akirosspower@gmail.com>:
> Mh well, to me it seems a normal regex processing (i mean, it *should*
> require only one instruction, since this pattern can be read with just
> one regex, even if ruby doesn't allow it... but it would be really
> bad).
> Anyway well, splitting it there are different ways to do it - thanks
> for your sudjestion.
> But if ruby make it possible with one call, i'd prefer to use it.
Yep, it's like this.
I solved using 2 instructions as you did: first matching extern words,
then the middle ones, but i still think that one regex would have been
nicer
Thanks guys
路路路
On 8/1/07, Harry Kakueki <list.push@gmail.com> wrote:
On 7/31/07, Alessandro Re <akirosspower@gmail.com> wrote:
> while i need to be sure that the pattern begins with a regex "x" and
> ends with "z"
>
> (of course, x 1 a 2 b 3 c should be regexes not just chars)
>
Sorry, I misunderstood what you wanted.
Is this more like it?
str = "lol1a2vasd"
m = /^(\w{3})(.*)(\w{3})$/.match(str).captures
m[1] = m[1].scan(/\d\w/)
p m.flatten #> ["lol","1a","2v","asd"]
Give special attention to my usage of the reluctant qualifier which is mandatory if your input contains multiple begin end pairs.
Kind regards
robert
PS: please do not top post.
路路路
On 31.07.2007 17:18, Alessandro Re wrote:
Thanks, this is an interesting solution!
On 7/31/07, Robert Klemme <shortcutter@googlemail.com> wrote:
2007/7/31, Alessandro Re <akirosspower@gmail.com>:
Mh well, to me it seems a normal regex processing (i mean, it *should*
require only one instruction, since this pattern can be read with just
one regex, even if ruby doesn't allow it... but it would be really
bad).
Anyway well, splitting it there are different ways to do it - thanks
for your sudjestion.
But if ruby make it possible with one call, i'd prefer to use it.
So it's not suited for a one RX approach and still need two levels of
RX. If that's the case then we have seen simpler solutions for that.
(Btw, one reason why it's so awkward is that there is no lookbehind in
Ruby 1.8 - but this will change.)
Kind regards
robert
路路路
2007/8/4, Alessandro Re <akirosspower@gmail.com>:
On 8/2/07, Wolfgang N谩dasi-donner <ed.odanow@wonado.de> wrote:
> irb(main):001:0>
> "x1A2B3C4Dz".scan(/(?:\G|^(?:x))(x|\d\w|z)(?=(?:\d\w)*(?:z|)$)/)
> => [["x"], ["1A"], ["2B"], ["3C"], ["4D"], ["z"]]
