Используйте низкоуровневый $.ajax()
вызов:
$.ajax({
url: "/yourservlet",
data: { },
complete: function(xmlHttp) {
// xmlHttp is a XMLHttpRquest object
alert(xmlHttp.status);
}
});
Попытка это для перенаправления:
if (xmlHttp.code != 200) {
top.location.href = '/some/other/page';
}
Проще было бы написать что-то вроде этого:
WebClient webClient = new WebClient();
webClient.DownloadFile(remoteFileUrl, localFileName);
You just need to make a basic http request using HttpWebRequest for the URI of the image then grab the resulting byte stream then save that stream to a file.
Here is an example on how to do this...
'As a side note if the image is very large you may want to break up br.ReadBytes(500000) into a loop and grab n bytes at a time writing each batch of bytes as you retrieve them.'
using System;
using System.IO;
using System.Net;
using System.Text;
namespace ImageDownloader
{
class Program
{
static void Main(string[] args)
{
string imageUrl = @"http://www.somedomain.com/image.jpg";
string saveLocation = @"C:\someImage.jpg";
byte[] imageBytes;
HttpWebRequest imageRequest = (HttpWebRequest)WebRequest.Create(imageUrl);
WebResponse imageResponse = imageRequest.GetResponse();
Stream responseStream = imageResponse.GetResponseStream();
using (BinaryReader br = new BinaryReader(responseStream ))
{
imageBytes = br.ReadBytes(500000);
br.Close();
}
responseStream.Close();
imageResponse.Close();
FileStream fs = new FileStream(saveLocation, FileMode.Create);
BinaryWriter bw = new BinaryWriter(fs);
try
{
bw.Write(imageBytes);
}
finally
{
fs.Close();
bw.Close();
}
}
}
}