so generalize the class... ----- Original Message ----- From: Alex McAuley To: prototype-scriptaculous@googlegroups.com Sent: Friday, September 25, 2009 12:25 PM Subject: [Proto-Scripty] Re: Range utility increment
That only gives you even or odd increments to the top number... WHat if you wanted every 5th number ... Better to go with my/TJ's loop Alex Mcauley http://www.thevacancymarket.com ----- Original Message ----- From: "Matt Foster" <mattfoste...@gmail.com> To: "Prototype & script.aculo.us" <prototype-scriptaculous@googlegroups.com> Sent: Friday, September 25, 2009 5:57 PM Subject: [Proto-Scripty] Re: Range utility increment I'd just use an 'iterator' class to do this... var EvenNumber = Class.create( { initialize : function(num){ this.num = (num % 2 == 1) ? num - 1 : num; }, succ : function(){ return new EvenNumber(this.num + 2); }, toString : function(){ return this.num; } }); var bottom = new EvenNumber(1); var top = new EvenNumber(21); var range = $A($R(bottom, top)); console.log(range); I had done this to a much greater extent in my date range selector gadget. Where it became the master range, held a range of calendar objects, which were in themselves ranges of date objects... fun stuff. https://positionabsolute.net/blog/2008/01/google-calendar-date-range-selection.php ps just looking over that, should rename that class to "PositiveEvenNumber", only going to work one way, but you get the idea On Sep 25, 4:15 am, "Alex McAuley" <webmas...@thecarmarketplace.com> wrote: > function everyOther(start, end, increment) { > var n; > var rv = []; > for (n = start; n <= end; n += increment) { > rv[rv.length] = n; > } > return rv; > } > > Just a mod to make it better to be able to change it to 3's, 4's, 5's > > ;) > > Alex Mcauleyhttp://www.thevacancymarket.com > > ----- Original Message ----- > From: "T.J. Crowder" <t...@crowdersoftware.com> > To: "Prototype & script.aculo.us" > <prototype-scriptaculous@googlegroups.com> > Sent: Friday, September 25, 2009 8:29 AM > Subject: [Proto-Scripty] Re: Range utility increment > > Hi, > > > I'm looking for a way to specify that I want to increment by any > > value. Let's say count up by 2's: [2,4,6,8,10]. How do you do this? > > ObjectRange itself is for ranges of consecutive values, but you can > get an array of such values by applying Enumerable#collect to a range: > > var twos; > twos = $R(1, 5).collect(function(x) { return x * 2; }); > > ...although frankly I'd probably go with a straight loop like Alex did > (although a slightly different one): > > function everyOther(start, end) { > var n; > var rv = []; > for (n = start; n <= end; n += 2) { > rv[rv.length] = n; > } > return rv; > } > > FWIW, > -- > T.J. Crowder > tj / crowder software / comwww.crowdersoftware.com > > On Sep 24, 11:15 pm, JoJo <tokyot...@gmail.com> wrote: > >http://www.prototypejs.org/api/utility/dollar-r > > > From the documentation of the Range utility, it seems like it can only > > increment by 1's. For example, $A($R(1,10,true)) gives you: > > [1,2,3,4,5,6,7,8,9,10]. > > > I'm looking for a way to specify that I want to increment by any > > value. Let's say count up by 2's: [2,4,6,8,10]. How do you do this? --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Prototype & script.aculo.us" group. To post to this group, send email to prototype-scriptaculous@googlegroups.com To unsubscribe from this group, send email to prototype-scriptaculous+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/prototype-scriptaculous?hl=en -~----------~----~----~----~------~----~------~--~---
