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Question

Answer 1

A:

Try this:

String S =  ([^\>]*)\>([^\;]*)\;
String T = " {:period => $1, :hours => $2}, "
originalString.replaceAll(S,T);

Might have to play with the regexp a little more but that should about do it.

Edit - Well, you asked for the answer in the context of ruby and I gave you the Java answer but the regular expression should work anyway...

Ben 2010-02-02 07:42:40

um, except that I'm guessing the OQ wanted an array of hashes as output, not a string serialisation that needs eval'ing.

p00ya 2010-02-02 08:14:10

@P00ya - Your'e right... Wasn't clear though.

Ben 2010-02-02 08:17:22

Answer 2

+2 A:

this is how I would do it

str="Monday-Friday>10:00-18:00;Saturday>12:00-17:00;Sunday>12:00-15:00"
periods = str.split(';') 
#=> ["Monday-Friday>10:00-18:00", "Saturday>12:00-17:00", "Sunday>12:00-15:00"]
period_array=[]
periods.each do |period| 
  period_with_hours = period.split('>')
  period_array << {:period =>  period_with_hours.first, :hours => period_with_hours.last}
end

period_array

#=> [{:period=>"Monday-Friday", :hours=>"10:00-18:00"}, {:period=>"Saturday", :hours=>"12:00-17:00"}, {:period=>"Sunday", :hours=>"12:00-15:00"}]

nas 2010-02-02 07:55:05

Just for fun... `periods = str.split(';').inject([]) {|n,i| n << Hash[*[:period, :hours].zip(i.split('>')).flatten] })`

kejadlen 2010-02-02 08:43:01

Or... `str.scan(/([\w-]+)>([\d:-]+)/).map {|(p,h)| {:period => p, :hours => h }}`

kejadlen 2010-02-02 08:45:55

Yes, that definitely valid for fun :). Ruby allows us to put a long chain of methods on a single line but I think its not prudent to put everything on a line if the intention of your code gets a bit muddled. I try to put things in temporary variables before calling another method on it if that makes the intention of the code a bit more clear. For instance, if you look at my code, it clearly says what it is doing the moment you look at it, first split string by ';' and create an array and then split on '>' to create each element of the array and create hash.Just following clean code principles

nas 2010-02-02 09:54:27

You don't have to write it in a single line. Functional-style solutions (using `map`, `inject` etc.) are considered more elegant and thus preferred among Rubyists.

Mladen Jablanović 2010-02-02 13:55:46

@Mladen I do agree with that as long as the code intention is clear

nas 2010-02-02 14:15:21

Answer 3

+3 A:

If you prefer one-liners:

s = "Monday-Friday>10:00-18:00;Saturday>12:00-17:00;Sunday>12:00-15:00"
s.split(/;/).map{|i| Hash[[[:period, :hours], i.split(/>/)].transpose]}
# or
s.split(/;/).map{|i| p, h = i.split(/>/); {:period => p, :hours => h}}
#=> [{:period=>"Monday-Friday", :hours=>"10:00-18:00"}, {:period=>"Saturday", :hours=>"12:00-17:00"}, {:period=>"Sunday", :hours=>"12:00-15:00"}]

Edit:

Regarding the reverse, this should do the job:

a.map{|i| "#{i[:period]}>#{i[:opens]}-#{i[:closes]}"}.join(';')
=> "Monday-Friday>10:00-18:00;Saturday>12:00-17:00;Sunday>12:00-15:00"

Mladen Jablanović 2010-02-02 08:27:59

Answer 4

A:

This looks like it works

the_input.split(';').collect{|pair|
    period, hours = pair.split('>')
    {:period => period, :hours => hours}
}
=> [{:hours=>"10:00-18:00", :period=>"Monday-Friday"}, {:hours=>"12:00-17:00", :
period=>"Saturday"}, {:hours=>"12:00-15:00", :period=>"Sunday"}]

Gishu 2010-02-02 09:46:57

Answer 5

A:

str.scan(/([\w-]+)>([\d:-]+)/).map {|(p,h)| {:period => p, :hours => h }}

kejadlen 2010-02-02 19:45:02

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String to Array and Hash with Regexp

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