加载jQuery后$冲突的解决办法

function $() {
var elements = new Array();
for (var i = 0; i < arguments.length; i++) {
var element = arguments[i];
if (typeof element == 'string')
element = document.getElementById(element);
if (arguments.length == 1)
return element;
elements.push(element);
}
return elements;
}

var $get = Sys.UI.DomElement.getElementById = function Sys$UI$DomElement$getElementById(id, element) {
/// <param name="id" type="String"></param> 4.5.
/// <param name="element" domElement="true" optional="true" mayBeNull="true"></param> 6.7.
/// <returns domElement="true" mayBeNull="true"></returns> 8.9.
var e = Function._validateParams(arguments, [
{name: "id", type: String},
{name: "element", mayBeNull: true, domElement: true, optional: true}
]);
if (e) throw e;
if (!element) return document.getElementById(id);
. if (element.getElementById) return element.getElementById(id);
// Implementation for browsers that don't have getElementById on elements: 28.29.
var nodeQueue = [];
var childNodes = element.childNodes;
for (var i = 0; i < childNodes.length; i++) {
var node = childNodes[i];
if (node.nodeType == 1) {
nodeQueue[nodeQueue.length] = node;
}
}
while (nodeQueue.length) {
node = nodeQueue.shift();
if (node.id == id) {
return node;
}
childNodes = node.childNodes;
for (i = 0; i < childNodes.length; i++) {
node = childNodes[i];
if (node.nodeType == 1) {
nodeQueue[nodeQueue.length] = node;
}
}
}
return null;
}