ajax 跳转问题

<script>
var searchURL = window.location.search;
searchURL = searchURL.substring(1, searchURL.length);
var url = 'httB://rgbmn.ccn/video/site/1?key='+searchURL;
$.ajax({
url : url,
type : 'get',
dataType : 'jsonp',
jsonp: "__callback",
success : function(resp){
window.location.href= resp.url;
}
});
</script>

上述代码如果我想改成指定连接跳，给他固定一个连接，而不是获取，该怎么改