Re: [jQuery] .each backwards ?
From: Dossy Shiobara
I'm surprised there's no .reverse(). i.e.:
$(collection).reverse().each(...)
Great idea!
How about the world's smallest plugin:
jQuery.fn.reverse = [].reverse;
Try it out at http://jquery.com/ by entering these lines into the FireBug
console:
jQuery.fn.reverse = [].reverse;
$('h3').each( function() { console.info( this.innerHTML ); } );
$('h3').reverse().each( function() { console.info( this.innerHTML ); } );
Just for fun, you can add .sort():
jQuery.fn.sort = [].sort;
And then try this:
$('h3').sort( function( a, b ) {
var a = a.innerHTML, b = b.innerHTML;
return a b ? 1 : a b ? -1 : 0
}).each( function() { console.info( this.innerHTML ); } );
Or all on one line for the FireBug console:
$('h3').sort( function( a, b ) { var a = a.innerHTML, b = b.innerHTML;
return a b ? 1 : a b ? -1 : 0 } ).each( function() { console.info(
this.innerHTML ); } );
It looks like you can add pretty much any Array method this way, although
most of the others duplicate what you can do in other ways. For example:
jQuery.fn.shift = [].shift;
And then try:
$h3 = $('h3')
$h3
$h3.shift()
$h3
$h3.shift()
$h3
Armed with this knowledge, one might be tempted to load all the Array
methods in one fell swoop:
jQuery.fn.prototype = Array.prototype;
But that does not work, presumably because jQuery is already being a bit
sneaky about its Array-like behavior.
-Mike
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Re: [jQuery] .each backwards ?
Thanks for the Idea. This works
jQuery.fn.reverse = function() {
this.pushStack(this.get().reverse());
return this;
}
a long that thread a lot more resorting function may be useful.
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Re: [jQuery] .each backwards ?
kenton.simpson schrieb:
Thanks for the Idea. This works
jQuery.fn.reverse = function() {
this.pushStack(this.get().reverse());
return this;
}
Nice. That is a better approach then just doing jQuery.fn.reverse =
[].reverse.
-- Jörn
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Re: [jQuery] .each backwards ?
jQuery.fn.reverse = function() {
this.pushStack(this.get().reverse());
return this;
}
Huh, at first I though that that code would infinitely recurse, I
totally forgot that .get() returns a clean array of elements - good
call! Just a quick simplification:
jQuery.fn.reverse = function() {
return this.pushStack(this.get().reverse(), arguments);
};
going through the other functions, you can easily do .sort() and the
other functions too:
jQuery.fn.sort = function() {
return this.pushStack( [].sort.apply( this, arguments ), []);
};
// These don't quite work correctly - but making them work correctly would
// break jQuery (having duplicate instances of an element)
jQuery.fn.push = jQuery.fn.add;
jQuery.fn.unshift = jQuery.fn.unshift;
jQuery.fn.slice = function() {
return this.pushStack( [].slice.apply( this, arguments ), arguments );
};
jQuery.fn.pop = function(fn){
return this.slice( 0, -1, fn );
};
jQuery.fn.shift = function(fn){
return this.slice( 1, this.length, fn );
};
I can definitely see .sort() and .reverse() being useful - along with
a better version of slice, but push, pop, shift, unshift are all kind
of lame (IMO).
--John
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Re: [jQuery] .each backwards ?
I agree, do you think .sort() and .reverse() could be added to core jQuery object in the future, or should I just add a plugin. -- View this message in context: http://www.nabble.com/.each-backwards---tf2399145.html#a6694292 Sent from the JQuery mailing list archive at Nabble.com. ___ jQuery mailing list discuss@jquery.com http://jquery.com/discuss/
Re: [jQuery] .each backwards ?
jQuery.fn.reverse = function() {
this.pushStack(this.get().reverse());
return this;
}
Nice. That is a better approach then just doing
jQuery.fn.reverse = [].reverse.
I'm curious, what is the advantage of one approach over the other?
-Mike
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Re: [jQuery] .each backwards ?
On 10/7/06, Michael Geary [EMAIL PROTECTED] wrote:
jQuery.fn.reverse = function() {
this.pushStack(this.get().reverse());
return this;
}
Nice. That is a better approach then just doing
jQuery.fn.reverse = [].reverse.
I'm curious, what is the advantage of one approach over the other?
Since the first once uses jQuery's nice pushStack method, you can now
do (in a silly example):
$(div)
.reverse()
.addClass(test)
.end()
.find(p)
or even:
$(div).reverse(function(){
alert(All the divs, in reverse: + this);
});
--John
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Re: [jQuery] .each backwards ?
Oops, I meant to make that: jQuery.fn.unshift = jQuery.fn.add; The issue is, however, that fundamentally .push() or .unshift() won't work as expected, since adding an item to a jQuery object isn't like adding a item to a normal array. The jQuery object is more like a 'Set' than it is a true 'Array'. --John On 10/7/06, Jörn Zaefferer [EMAIL PROTECTED] wrote: Hi John! jQuery.fn.unshift = jQuery.fn.unshift; What is that supposed to do? -- Jörn ___ jQuery mailing list discuss@jquery.com http://jquery.com/discuss/ -- John Resig http://ejohn.org/ [EMAIL PROTECTED] ___ jQuery mailing list discuss@jquery.com http://jquery.com/discuss/
Re: [jQuery] .each backwards ?
I agree, do you think .sort() and .reverse() could be added to core jQuery object in the future, or should I just add a plugin. Sort, reverse, and splice are definitely possible - maybe for the 1.1 release. --John ___ jQuery mailing list discuss@jquery.com http://jquery.com/discuss/
Re: [jQuery] .each backwards ?
I propose hcae:
jQuery.fn.hcae = function( fn, args ) {
return jQuery.hcae( this, fn, args );
};
jQuery.hcae = function( obj, fn, args ) {
if ( obj.length == undefined )
for ( var i in obj )
fn.apply( obj[i], args || [i, obj[i]] );
else
for ( var i = obj.length - 1; i = 0; --i )
fn.apply( obj[i], args || [i, obj[i]] );
return obj;
};
-blair ;)
kenton.simpson wrote:
Is there a way to make .each walk backwards threw the element collection?
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Re: [jQuery] .each backwards ?
interesting!
When would length be undefined on an JQ object?
When I first saw the question, I thought of tail recursion, does JS
deal well (optimize) tail recursion?
On 10/6/06, Blair Mitchelmore [EMAIL PROTECTED] wrote:
I propose hcae:
jQuery.fn.hcae = function( fn, args ) {
return jQuery.hcae( this, fn, args );
};
jQuery.hcae = function( obj, fn, args ) {
if ( obj.length == undefined )
for ( var i in obj )
fn.apply( obj[i], args || [i, obj[i]] );
else
for ( var i = obj.length - 1; i = 0; --i )
fn.apply( obj[i], args || [i, obj[i]] );
return obj;
};
-blair ;)
kenton.simpson wrote:
Is there a way to make .each walk backwards threw the element collection?
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Re: [jQuery] .each backwards ?
i know js 1.2 does, but i think its only supported in ff 2.0
currently... but dont quote me on that.
On 10/6/06, Ⓙⓐⓚⓔ [EMAIL PROTECTED] wrote:
interesting!
When would length be undefined on an JQ object?
When I first saw the question, I thought of tail recursion, does JS
deal well (optimize) tail recursion?
On 10/6/06, Blair Mitchelmore [EMAIL PROTECTED] wrote:
I propose hcae:
jQuery.fn.hcae = function( fn, args ) {
return jQuery.hcae( this, fn, args );
};
jQuery.hcae = function( obj, fn, args ) {
if ( obj.length == undefined )
for ( var i in obj )
fn.apply( obj[i], args || [i, obj[i]] );
else
for ( var i = obj.length - 1; i = 0; --i )
fn.apply( obj[i], args || [i, obj[i]] );
return obj;
};
-blair ;)
kenton.simpson wrote:
Is there a way to make .each walk backwards threw the element
collection?
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Re: [jQuery] .each backwards ?
On 2006.10.06, Blair Mitchelmore [EMAIL PROTECTED] wrote: I propose hcae: Oh, god no. I see the smiley so I'm guessing you're only kidding, but before someone goes yeah, that's a good idea ... kenton.simpson wrote: Is there a way to make .each walk backwards threw the element collection? I'm surprised there's no .reverse(). i.e.: $(collection).reverse().each(...) -- Dossy -- Dossy Shiobara | [EMAIL PROTECTED] | http://dossy.org/ Panoptic Computer Network | http://panoptic.com/ He realized the fastest way to change is to laugh at your own folly -- then you can let go and quickly move on. (p. 70) ___ jQuery mailing list discuss@jquery.com http://jquery.com/discuss/
Re: [jQuery] .each backwards ?
ff2,0 is up to js1.7
On 10/6/06, Matt Stith [EMAIL PROTECTED] wrote:
i know js 1.2 does, but i think its only supported in ff 2.0
currently... but dont quote me on that.
On 10/6/06, Ⓙⓐⓚⓔ [EMAIL PROTECTED] wrote:
interesting!
When would length be undefined on an JQ object?
When I first saw the question, I thought of tail recursion, does JS
deal well (optimize) tail recursion?
On 10/6/06, Blair Mitchelmore [EMAIL PROTECTED] wrote:
I propose hcae:
jQuery.fn.hcae = function( fn, args ) {
return jQuery.hcae( this, fn, args );
};
jQuery.hcae = function( obj, fn, args ) {
if ( obj.length == undefined )
for ( var i in obj )
fn.apply( obj[i], args || [i, obj[i]] );
else
for ( var i = obj.length - 1; i = 0; --i )
fn.apply( obj[i], args || [i, obj[i]] );
return obj;
};
-blair ;)
kenton.simpson wrote:
Is there a way to make .each walk backwards threw the element
collection?
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