Hi, ok I’m none to good at regular expressions but i’ve got to do
something and need a little help.
Basically i’m trying to hunt down a particular word within a url string,
so to test, i define a local variable, var_string,
var_string=“user=22&val=999”
then run a regular expression match on the variable,
user = var_string.match(/=(\w+)&/)
which returns => ‘22’ as it’s looking for the first word occurence which
has = and a & at the start & finish.
so the question is how could i return the second occurence (999) ?
Alle domenica 30 dicembre 2007, John G. ha scritto:
user = var_string.match(/=(\w+)&/)
which returns => ‘22’ as it’s looking for the first word occurence which
has = and a & at the start & finish.
so the question is how could i return the second occurence (999) ?
In ruby 1.8, String#match always returns the first match. If you need to
find
other matches, you can look at StringScanner or String#scan (of course,
you’ll need to remove the & from the regexp, or make it optional, if you
want
to match the string ‘999’. To make the ‘&’ optional, put a ? after
it: /=(\w+)&?).
In ruby 1.9, instead String#match should accept an offset (as far as I
know,
the documentation doesn’t mention this, but it seems to work). You
should
still modify the regexp to make it work.
I hope this helps
Stefano
John G. wrote:
Hi, ok I’m none to good at regular expressions but i’ve got to do
something and need a little help.
Basically i’m trying to hunt down a particular word within a url string,
so to test, i define a local variable, var_string,
var_string=“user=22&val=999”
then run a regular expression match on the variable,
user = var_string.match(/=(\w+)&/)
which returns => ‘22’ as it’s looking for the first word occurence which
has = and a & at the start & finish.
so the question is how could i return the second occurence (999) ?
I think this will be an easy solution…
var_string=“user=22&val=999”
vars = {}
var_string.scan(/(\w+)=(\w+)(?:&|$)/) do |name, val|
vars[name] = val
end
p vars # => {“user”=>“22”, “val”=>“999”}