我想发一个ajax请求从数据库中获取一些数据,然后将该数据发送到另一个文件,这将返回另一个结果。似乎没有用,第一个Ajax请求工作正常,但是第一个返回结果时触发的第二个请求没有做任何事情。
var http = getHTTPObject();
var http2 = getHTTPObject();
ids = fetchSelection().toString();
//Make button animate, visual aid that it is working
obj.src = "http://localhost/nightclub_photography/images/buttons/"+dir+"_animated.gif";
http.onreadystatechange = function()
{
if (http.readyState == 4 && http.status == 200) {
http2.onreadystatechange = function()
{
if (http2.readyState == 4 && http2.status == 200)
{
alert("hello");
}
}
http2.open("GET", "http://localhost/nightclub_photography/net/test2.aspx");
http2.send();
}
}
http.open("GET", "http://localhost/nightclub_photography/asp/returnDatabaseData.asp?ids="+ids);
http.send();
答案 0 :(得分:1)
test.aspx返回错误,因为我忘记传递它正在寻找的URL变量。傻傻的,谢谢你们!