//Change the link url to be a form and submit it.
//this script need jquery suport..
//url 中必须有?分开地址和参数
//this script need jQuery JavaScript Library v1.4 at least
function jsGetToPost(url, formID, target){
	var orgUrl = url;
	url = url.split("?");
	target = target?"target=\""+target+"\"":"";
	var formObj = $("<form id=\"jsGetToPostForm\" method=\"POST\"  action=\""+url[0]+"\" " + target + "></form>");
	var addPara = formID?$("#"+formID).serialize():"";
	if(addPara.length>0){
		url[1] = url[1]?url[1]+"&"+addPara:addPara;
	}
	var paraArr = url[1].split("&");
	for(var i=0;i<=paraArr.length-1;i++){
		var paraKeyValue = paraArr[i].split("=");
		formObj.append("<input name='"+paraKeyValue[0]+"' type='hidden' value='"+paraKeyValue[1]+"' />");
	}
	$("form[id=jsGetToPostForm]").remove();//Clear the form exist
	formObj.appendTo("body");
	setTimeout('$("#jsGetToPostForm").get(0).submit()',10);
}