Hi Pedro.
You have to build an algorithm to make it work like you want. I´ve cut a
piece of code that makes something like you want. I didnt test it, but
probably will work.
Take a look and if you have any questions, send a message again.
// global vars
var NUM_RECORDS=16;
var page=0;
var numberFriends=0;
function loadFriends() {
var req = opensocial.newDataRequest();
var opt_params = { };
opt_params[opensocial.DataRequest.PeopleRequestFields.MAX] = 400;
opt_params[opensocial.DataRequest.PeopleRequestFields.FILTER] =
opensocial.DataRequest.FilterType.HAS_APP;
req.add(req.newFetchPeopleRequest(opensocial.DataRequest.Group.OWNER_FRIENDS,opt_params),
'viewerFriends');
req.send(onLoadFriends);
}
/**
* Callback to receive friends
*/
function onLoadFriends(data)
{
var viewerFriends = data.get('viewerFriends').getData();
i=1;
firstRecord = (page*NUM_RECORDS)+1;
numberPages = 0;
try{
viewerFriends.each(function(person) {
if(i>=firstRecord && i < (firstRecord+NUM_RECORDS)){
// handle your operations here
}
i++;
}
});
// fix pagination
i--;
// total number of friends
numberFriends = i;
// number of pages
numberPages = i/NUM_RECORDS;
// fix number of pages
if(numberPages % NUM_RECORDS > 0 && numberPages > NUM_RECORDS)
numberPages++;
// convert it to integer
numberPages = parseInt(numberPages);
// just to call pages to test.
document.write('<input type="button" value=">"
onclick="javascript:move('+page+'+1)"'+(page+1>=numPag ? '
disabled':'')+'>');
document.write('<input type="button" value="<"
onclick="javascript:move('+page+'-1)"'+(page<=0 ? ' disabled' : '')+'>');
}
/*
* Simple function to move between pages
*/
function move(pageNumber)
{
// assign to global variable
page = pageNumber;
//load friends again
loadFriends();
}
Cheers,
-Robson
You just need to use that and make a loop to iterate
2009/2/27 Pedro Vicente <neteinst...@gmail.com>
>
> Oh and i forgot. The nasty thing is that if i store like the first 50
> friends, and have 10 per page, then every time i change page and try
> to get the data, i will loop through the first 10, ou 20, or.. i get
> the picture.
>
> On Feb 27, 2:33 pm, Pedro Vicente <neteinst...@gmail.com> wrote:
> > I knew that way, but thought that there was a "nicer" or "cleaner"
> > way.
> >
> > Tks
> >
> > On Feb 27, 2:26 pm, Robson Dantas <biu.dan...@gmail.com> wrote:
> >
> > > Hi Pedro.
> >
> > > The simplest way is to use a variable to count first five and then
> exit.
> >
> > > var i=1;
> > > xpto.each(function(person){
> >
> > > if (i==5) return;
> >
> > > do stuff..
> >
> > > i++;
> >
> > > });
> >
> > > Let me know if it is what you want.
> >
> > > Cheers,
> >
> > > --Robson Dantashttp://blogdodantas.dxs.com.br
> >
> > > 2009/2/27 Pedro Vicente <neteinst...@gmail.com>
> >
> > > > Hello,
> >
> > > > I've been looking at some examples of apps in hi5 and OpenSocial
> page.
> >
> > > > All of them make a request of 10 friends for example, and then with
> > > > the xpto.each(function(person) { .... use them all.
> >
> > > > My question is, how do i get only the 5 first friends of the 10 (so
> > > > that i can request a X ammount of friends and do some pages with them
> > > > without having to do a cycle with all of them...
> >
> > > > Thanks in advanced and sorry if the is a more javascript question.
> >
>
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