怎样获取ASIFormDataRequest的postvalue

iOS客户端


- (IBAction)sendPostValue:(id)sender {
    NSLog(@"send test");
    NSURL *url = [NSURLURLWithString:@"http://localhost:8080/server/register"];
    ASIHTTPRequest *request = [ASIHTTPRequestrequestWithURL:url];
    NSString *string = @"{\"name1\":\"zcq喘着粗气\"}";
    NSMutableData *data = [NSMutableDatadata];
    [data appendData:[string dataUsingEncoding:NSUTF8StringEncoding]];
    [request setRequestMethod:@"POST"];
    [request appendPostData:data];
    [request setDelegate:self];
    [request setDidFinishSelector:@selector(finish:)];
    [request setDidFailSelector:@selector(failFinish:)];
    [request startAsynchronous];

}



Java服务器端


publicvoid doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
response.setContentType("text/html");
response.setCharacterEncoding("utf-8");
PrintWriter out = response.getWriter();
BufferedReader br = new BufferedReader(new InputStreamReader((ServletInputStream)request.getInputStream(),"utf-8"));
         String line = null;
         StringBuilder sb = new StringBuilder();
         while((line = br.readLine())!=null){
         System.out.println(line);
             sb.append(line);
         }
         System.out.println(sb.toString());
         out.print("sb = "+sb.toString());
         ZCQ o = JSON.parseObject(sb.toString(),ZCQ.class);
         System.out.println(o.getName1());
         out.print(o.getName1());
out.flush();
out.close();
}