将url和参数分开解决
具体方法如下:
var resource = "result=.......too many data..........";
var url = "aLink?param1="+param1;
。。。。。生成XMLHttpRequest对象程序。。。。
xmlRequest.open("post", url, false);
xmlRequest.setRequestHeader("Content-Type","application/x-www-form-urlencoded");
xmlRequest.send(resource);