我一直在寻找这个,没有什么对我有用.
我正在尝试将图像从Android应用程序上传到java servlet并将其保存在服务器中.
我找到的每个解决方案都不适用于我.
我的代码目前做的是：android应用程序将图像发送到servlet,
当我试图保存它时,文件被创建,但它是空的:(
谢谢你的帮助！
我在android客户端的代码(i_file是设备上的文件位置)：
public static void uploadPictureToServer(String i_file) throws ClientProtocolException,IOException {
// TODO Auto-generated method stub
HttpClient httpclient = new DefaultHttpClient();
httpclient.getParams().setParameter(CoreProtocolPNames.PROTOCOL_VERSION,HttpVersion.HTTP_1_1);
HttpPost httppost = new HttpPost("http://192.168.1.106:8084/Android_Server/GetPictureFromClient");
File file = new File(i_file);
MultipartEntity mpEntity = new MultipartEntity();
ContentBody cbFile = new FileBody(file,"image/jpeg");
mpEntity.addPart("userfile",cbFile);
httppost.setEntity(mpEntity);
System.out.println("executing request " + httppost.getRequestLine());
HttpResponse response = httpclient.execute(httppost);
HttpEntity resEntity = response.getEntity();
System.out.println(response.getStatusLine());
if (resEntity != null) {
System.out.println(EntityUtils.toString(resEntity));
}
if (resEntity != null) {
resEntity.consumeContent();
}
httpclient.getConnectionManager().shutdown();
}
我在服务器端的代码：
protected void doPost(HttpServletRequest request,HttpServletResponse response)
throws ServletException,IOException {
processRequest(request,response);
InputStream in = request.getInputStream();
OutputStream out = new FileOutputStream("C:\\myfile.jpg");
IoUtils.copy(in,out); //The function is below
out.flush();
out.close();
}
public static long copy(InputStream input,OutputStream output) throws IOException {
byte[] buffer = new byte[4096];
long count = 0L;
int n = 0;
while (-1 != (n = input.read(buffer))) {
output.write(buffer,n);
count += n;
}
return count;
}
解决方法
你误解了这个问题.图像文件不为空,但图像文件已损坏,因为您将整个HTTP多部分请求主体存储为图像文件,而不是从HTTP多部分请求主体中提取包含该图像的部分.
您需要HttpServletRequest#getPart()才能获取多部分请求正文的各个部分.如果您已经使用Servlet 3.0(Tomcat 7,Glassfish 3等),请先使用@MultipartConfig注释您的servlet
@WebServlet("/GetPictureFromClient")
@MultipartConfig
public class GetPictureFromClient extends HttpServlet {
// ...
}
然后按如下方式修复你的doPost()以按名称抓取零件,然后将其身体作为输入流：
InputStream in = request.getPart("userfile").getInputStream();
// ...
如果您还没有使用Servlet 3.0,请抓住Apache Commons FileUpload.有关详细示例,请参阅此答案：How to upload files to server using JSP/Servlet?
哦,请摆脱Netbeans生成的processRequest()方法.将doGet()和doPost()委托给单个processRequest()方法绝对不是正确的方法,它只会混淆不使用Netbeans的其他开发人员和维护人员.
