Find just part of string with a regex - ruby

Question

I have a string like so:
"#[30:Larry Middleton]"
I want to return just 30. Where 30 will always be digits, and can be of 1 to infinity in length.
I've tried:
user_id = result.match(/#\[(\d+):.*]/)
But that returns everything. How can I get back just 30?

Answer 1

If that's really all your string, you don't need to match the rest of the pattern; just match the consecutive integers:
irb(main):001:0> result = "#[30:Larry Middleton]"
#=> "#[30:Larry Middleton]"
irb(main):002:0> result[/\d+/]
#=> "30"
However, if you need to match this as part of a larger string that might have digits elsewhere:
irb(main):004:0> result[/#\[(\d+):.*?\]/]
#=> "#[30:Larry Middleton]"
irb(main):005:0> result[/#\[(\d+):.*?\]/,1]
#=> "30"
irb(main):006:0> result[/#\[(\d+):.*?\]/,1].to_i
#=> 30
If you need the name also:
irb(main):002:0> m = result.match /#\[(\d+):(.*?)\]/
#=> #<MatchData "#[30:Larry Middleton]" 1:"30" 2:"Larry Middleton">
irb(main):003:0> m[1]
#=> "30"
irb(main):004:0> m[2]
#=> "Larry Middleton"
In Ruby 1.9 you can even name the matches, instead of using the capture number:
irb(main):005:0> m = result.match /#\[(?<id>\d+):(?<name>.*?)\]/
#=> #<MatchData "#[30:Larry Middleton]" id:"30" name:"Larry Middleton">
irb(main):006:0> m[:id]
#=> "30"
irb(main):007:0> m[:name]
#=> "Larry Middleton"
And if you need to find many of these:
irb(main):008:0> result = "First there was #[30:Larry Middleton], age 17, and then there was #[42:Phrogz], age unknown."
#irb(main):015:0> result.scan /#\[(\d+):.*?\]/
#=> [["30"], ["42"]]
irb(main):016:0> result.scan(/#\[(\d+):.*?\]/).flatten.map(&:to_i)
#=> [30, 42]
irb(main):017:0> result.scan(/#\[(\d+):(.*?)\]/).each{ |id,name| puts "#{name} is #{id}" }
Larry is 30
Phrogz is 42

Answer 2

Try this:
user_id = result.match(/#\[(\d+):.*]/)[1]

Answer 3

You've forgot to escape ']':
user_id = result.match(/#\[(\d+):.*\]/)[1]

Answer 4

I don't know ruby, but if it supports lookbehinds and lookaheads:
user_id = result.match(/(?<#\[)\d+(?=:)/)
If not, you should have some way of retrieving subpattern from the match - again, I wouldn't know how.

Answer 5

I prefer String#scan for most of my regex needs, here's what I would do:
results.scan(/#\[(\d+):/).flatten.map(&:to_i).first
For your second question about getting the name:
results.scan(/(\d+):([A-Za-z ]+)\]$/).flatten[1]
Scan will always return an array of sub string matches:
"#[123:foo bars]".scan(/\d+/) #=> ['123']
If you include a pattern in parens, then each match for those "sub-patterns" will be included in a sub array:
"#[123:foo bars]".scan(/(\d+):(\w+)/) #=> [['123'], ['foo']]
That's why we have to do flatten on results involving sub-patterns:
[['123'], ['foo']].flatten = ['123', 'foo']
Also it always returns strings, that's why conversion to integer is needed in the first example:
['123'].to_i = 123
Hope this is helpful.