[Proto-Scripty] Re: iterator.call is not a function

[ColinFine]

On Jul 1, 6:16 pm, Ian R i...@fairmountfair.com wrote:


 $$('div.fp_YouTube').each(function(el)  {
         var player = el.select('.player');
         var playlist = el.select('.playlist');
         playlist.select('li').each(function(video)      {
                 console.log(video.id);
         });

 });

'select' delivers a list of elements, not an element. I think you mean
 var player = el.down('.player')

etc.
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[Proto-Scripty] Re: iterator.call is not a function

[Alex McAuley]

Is it returning an array ?

Check with

$$('div.fp_YouTube').each(function(el) {
var player = el.select('.player');
var playlist = el.select('.playlist');
alert(playlist); // see if its an object
return;
playlist.select('li').each(function(video) {
console.log(video.id);
});
});

if it is returning an array then try

$$('div.fp_YouTube').each(function(el) {
var player = el.select('.player');
var playlist = el.select('.playlist');
$(playlist).select('li').each(function(video) {
console.log(video.id);
});
});

HTH

Alex




- Original Message - 
From: Ian R i...@fairmountfair.com
To: Prototype  script.aculo.us prototype-scriptaculous@googlegroups.com
Sent: Wednesday, July 01, 2009 6:22 PM
Subject: [Proto-Scripty] iterator.call is not a function




 Ok. So I have a structure like this:

 div class=fp_YouTube

 div class=player
 object width=425 height=349 ... /object
 /div

 ol class=playlist
 li id=YouTubeID_w7naKj-z5So ... /li
 li id=YouTubeID_naiN5V4Q1rI ... /li
 /ol

 /div

 And eventually what I am doing is replacing the contents of player
 with a new video, based on observing a click on one of the list items
 in playlist.  I've done it before but I'm trying to prototypejs it
 up a little nicer.

 SO... I have this JS code going:

 $$('div.fp_YouTube').each(function(el) {
 var player = el.select('.player');
 var playlist = el.select('.playlist');
 playlist.select('li').each(function(video) {
 console.log(video.id);
 });
 });


 But firebug is telling me this:

 iterator.call is not a function -- prototype.js (line 661)

 I've noticed that if I console.log the original el variable, it
 looks like div class=fp_YouTube but if I log the playlist
 variable, it comes out as  [ol.playlist] ... is it returning a 1-item
 array?  Is there something I am fundamentally misunderstanding,
 something I need to do in order to work with playlist the way I've
 been working with el?

 Thanks in advance, and pardon this message for misfiring and sending
 before I was ready.

 Ian

 
 


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[Proto-Scripty] Re: iterator.call is not a function

[Ian R]

I did check on the array situation -- it's so good and bad that
finally turning to this forum and *writing it all down* usually solves
the problem!

In the above case, $(playlist).select('li').each didn't work, but
playlist[0].select('li').each does.

I'm not sure that I exactly LOVE how that works, but it's not
terrible.  If there's a better way, I'd love to know about it.

Thanks!
Ian
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[Proto-Scripty] Re: iterator.call is not a function

[Rick Waldron]

The reason you need to use an index on playlist is because select returns an
array, so... instead of using the index[0] the way you've done, you can send
it back to the el.select():


$$('div.fp_YouTube').each(function(el)  {
   var player   = el.select('.player')[0]; // --- right here!
   var playlist = el.select('.playlist li'); // also... select your li's
here, this will save you another select() call on the next line!
   playlist.each(function(video)  { // -- neater.
   console.log(video.id);
   });
});


Have a look:

http://jsbin.com/avedu


Rick




On Wed, Jul 1, 2009 at 1:37 PM, Ian R i...@fairmountfair.com wrote:



 I did check on the array situation -- it's so good and bad that
 finally turning to this forum and *writing it all down* usually solves
 the problem!

 In the above case, $(playlist).select('li').each didn't work, but
 playlist[0].select('li').each does.

 I'm not sure that I exactly LOVE how that works, but it's not
 terrible.  If there's a better way, I'd love to know about it.

 Thanks!
 Ian
 


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