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 > > > _______________________________________________ > jQuery mailing list > discuss@jquery.com > http://jquery.com/discuss/ > -- Ⓙⓐⓚⓔ - יעקב ʝǡǩȩ ᎫᎪᏦᎬ _______________________________________________ jQuery mailing list discuss@jquery.com http://jquery.com/discuss/