我想要什么我想将SD卡中的视频发送到服务器。我也想用它发送一些参数/值。
我试过我尝试过以下代码:
public String SendToServer(String aUrl,File aFile)
{
HttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(aUrl);
try
{
MultipartEntity entity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE);
entity.addPart("file", new FileBody(aFile));
entity.addPart("video[title]", new StringBody("testVideo"));
entity.addPart("video[type]", new StringBody("1"));
httpPost.setEntity(entity);
HttpContext localContext = new BasicHttpContext();
// Bind custom cookie store to the local context
localContext.setAttribute(ClientContext.COOKIE_STORE, Globals.sessionCookie);
HttpResponse response = httpClient.execute(httpPost, localContext);
HttpEntity resEntity = response.getEntity();
String Response = "";
if (response != null)
{
Response = EntityUtils.toString(resEntity);
}
return Response;
}
catch (IOException e)
{
e.printStackTrace();
}
return "Exception";
}
有什么问题当我运行此代码时,我遇到了问题
HttpResponse response = httpClient.execute(httpPost, localContext);
我也不例外,根本没有回应。谁能指导我,这里有什么问题?
答案 0 :(得分:4)
我的问题中的上述代码是完美的,但我遇到了网络问题。我的设备已连接到热点(Connectify Software)。当我连接到原始网络时,此代码完美无缺。
我建议人们永远不要相信热点这种功能。
答案 1 :(得分:1)
如果想要以内容形式发送,请尝试使用此方式,否则我将在今晚上传项目
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(url);
InputStreamEntity reqEntity = new InputStreamEntity(new FileInputStream(filePath), -1);
reqEntity.setContentType("binary/octet-stream");
reqEntity.setChunked(true); // Send in multiple parts if needed
httppost.setEntity(reqEntity);
HttpResponse response = httpclient.execute(httppost);