There are several options for using jquery
							<script src="http://code.jquery.com/jquery-1.8.2.min.js"></script>
						
						
							<script src="//ajax.googleapis.com/ajax/libs/jquery/1.8.3/jquery.min.js"></script>
						
					Pass the $ function CSS style selectors to serch for things. What is returned is an jquery style array thingy filled with jQuery objects.
$('#special')$('.navbutton')$('img')$('div.imgholder p:nth-child(2)')When you have a jQuery object you use.css to set the style of that object. You don't have to relearn css just use the styles as written.
					Assume:  var $par = $('p');
				
				
 obj.css("css property", "setting") $par.css("background-color", "blue");
					
					
					obj.css({ 
	cssProperty : "setting",
	cssProperty2 : "setting2",
	...	
}); obj.css({ 
	"background-color" : "blue",
	"font-family" : "sans-serif",
	"font-weight" : "bold"
}); 
				Simple as pie useing .attr() function.
					Assume:  var $img = $('img');
				
				
$img.attr("src", "snowflake.jpg");$img.attr({
	"src" : "snowflake.jpg",
	"alt" : "snow is falling",
	"title" : "I love snowflakes"
});All about the $ function. Surround the tag name with < and >. 
					
				
				
var img = $("<img>");var img = $('<img>', {
	"src" : "snowflake.jpg",
	"alt" : "snow is falling",
	"title" : "I love snowflakes"
});
					Use the .remove() method. 
$("#bestName").remove();$('p').fadeOut('slow', function() { $(this).remove(); })
						
					
				$.ajax("url", {
  option : value,
  option : value,
  option : value,
  ...
});$.ajax("http://www.example.com/foo.php", {
  data: { username: 'foobar', password: 'abcdef' },   // "password=abcdef"
  success: loginSuccess,
  error: loginFailure,
  complete: dismissDialog // will be called in either case
});
					Default request is get. Make it a post by adding type: "post" to options.
$.ajax("url", {
  option : value,
  ...
});$.ajax("http://www.example.com/foo.php", {
  type: "post",
  success: loginSuccess,
  ...
});
					
					jQuery also has 
					
						$.get
					
					and 
					
						$.post
					
					options if you are too lazy to include this option.
				
				For security reason we are not allowed to perform ajax requests to other domains. This causes a problem: Other domains have all the cool data!
There are 2 ways to work around this
dataType: 'jsonp' as an ajax option.
					JSONP - Basically involves an agreement between developers that we want to be able to share data so lets use a script workaround.
Full explaination on INFO 343 lecture slides link
// get the value from the text box with an id of search
var query = $('#search').val();  
	
$.ajax('https://search.twitter.com/search.json', {
	data: { 'q': query},
	dataType: 'jsonp', 
	success: injectTweets,
	error: ajaxError
});
/* When given json information as data inject 
that information into the page. Assumes there exists
a ul in the page with an id of results. */
function injectTweets(data) {
	var res = data.results; // get all the results of the request.
	var dl = $('#results');
	dl.empty();
	
	// display each result on the page. 
	$.each(res, function() {
		// get the username and tweet from the json data
		var dt = $("<dt>").text(this["from_user"] + " says:");
		var dd = $("<dd>").text(this["text"]);
		dl.append(dt, dd);
	});
}