Handling whitespace with jQuery .val() method. (Using Codeigniter)
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I am confused on how the .val() method in jQuery handles whitespace (if it is a jQuery problem at all).

My example is using an AJAX suggestion box feature which I'm slowly developing. It uses Codeigniter as the framework.

I have this jQuery code:

    $(document).ready(function(){
    	$("#suggestbox").hide();
    	$("#searchboxinput").live("keyup", function(){
      		if ($("#searchboxinput").val().length==0){
    			$("#suggestbox").fadeOut(100);
    		}else{
    			$("#suggestbox").fadeIn(100);
    			$("#sugresultcontain").load('http://localhost/ajaxtest/ajaxhandle/suggest/'+$("#searchboxinput").val()+' #loadsuggest');
    		};
    	});
    });

The '#suggestboxinput' is a text input as shown:

    <div id="content" class="clear">
    	<div id="searchwrap" class="clear">
        	<div id="searchbox" class="float-left"><input id="searchboxinput" name="" value="" type="text"></div>
            <div id="searchbutton" class="float-right"><input id="searchbuttoninput" name="" type="button" value="Go"></div>
        </div>
        <div id="suggestbox" class="clear">
        	<div id="suggesthead">Suggestions:</div>
            <div id="sugresultcontain"></div>
        </div>
    </div>

I then have (Using code igniter) a method which handles the .load() as shown:

    public function suggest($q=''){
    		
    		$suggest = $this->ajax_handle->suggest();
    		
    		for($i=0; $i<count($suggest); $i++){
    			$db[$i] = $suggest[$i]['firstname'].' '.$suggest[$i]['lastname'];
    		}
    		
    		if(strlen($q)>0){
    			$sugstr = '';
    			$e = 0;
    			
    			for($f=0; $f<count($db); $f++){
    				if (strtolower($q)==strtolower(substr($db[$f],0,strlen($q)))){
    					$sugresult[$e] = $db[$f];
    					$e++;
    				}
    			}
    			
    			if(isset($sugresult)){
    				$data['sugresult'] = $sugresult;
    				$this->load->view('ajaxref/suggest', $data);
    			}else{
    				$this->load->view('ajaxref/suggest');
    			}
    		}else{
    			$this->load->view('ajaxref/suggest');
    		}
    	}

This works almost as intended, if I load the page then use the text input the names from the database do show (according to text field). It works fine if I type the first name and will continue to show no matter how many spaces I put after the field (which is perfect). 

However, the problem lies with typing the surname even after just a single space (Dominic Sore) or even just the first letter or the last name (Dominic S). No names show in the suggestion box. 

I had read that using a '+' symbol in the input box will work. I tested it, and it did work (Dominic+Sore outputs Dominic Sore in suggestion box). Understandably though, I cannot have users type a '+' for every space they want to use.

So I am just wondering how do I handle this? I've been reading a lot but I cannot find a solution. 

Thanks.