Mike, I don't like breaking the chains... I just insert a debug in the
middle of the chain...
I use this for my debug:
jQuery.fn.debug = function(message) {
return this.log('debug:' + (message || '')
+"[").each(function(){jQuery.log(this);}).log("]");
}
jQuery.fn.log = jQuery.log = function(message) {
if (!message) message = 'UNDEFINED'
if (typeof message == "object") message = jsO(message)
if(window.console && window.console.log) //safari
window.console.log(message)
else if(window.console && window.console.debug) //firebug
window.console.debug(message)
else
jQuery("body").prepend(message+ "<br/>")
return this
}
On 1/23/07, Michael Geary <[EMAIL PROTECTED]> wrote:
> > Is there a way to just return the jquery object (so I could
> > see it in firebug's watch section) then pass it to another
> > jquery function then join them all up when I know
> > everything works?
> >
> > $('#test :textarea').before('Current length: <span id="'+
> this.id +'_len">'+this.value.length+'</span> characters').keypress(
> function()
> > {
> > $('#'+this.id + '_len').css('color', ((this.value.length >
> parseInt(this.id.split()[1]))?'red':'green')).html(this.value.length)
> > });
>
> Yes, breaking up your chains is one of the first things do do when you're
> having trouble. For example:
>
> var $test = $('#test :textarea');
> $test.before('Current length: <span id="'+ this.id
> +'_len">'+this.value.length+'</span> characters');
> $test.keypress( function() {
> var $len = $('#'+this.id + '_len');
> $len.css('color', ((this.value.length >
> parseInt(this.id.split()[1]))?'red':'green'));
> $len.html(this.value.length);
> });
>
> Now you'll find it easier to debug.
>
> -Mike
>
>
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> http://jquery.com/discuss/
>
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Ⓙⓐⓚⓔ - יעקב ʝǡǩȩ ᎫᎪᏦᎬ
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