Android，在tomcat服务器上传文件

我正在寻找可以从tomcat服务器中的android应用程序快速上传图像的代码。

目前我还没有找到要放在tomcat服务器（servlet）中的java代码

Answer 1

这是演示代码。

import javax.servlet.*; import javax.servlet.http.*; import java.io.*; import org.apache.commons.fileupload.*; import org.apache.commons.fileupload.util.*; import org.apache.commons.fileupload.servlet.ServletFileUpload; public class UploadServlet extends HttpServlet{ protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { PrintWriter out = response.getWriter(); out.print("Request content length is " + request.getContentLength() + "
"); out.print("Request content type is " + request.getHeader("Content-Type") + "
"); boolean isMultipart = ServletFileUpload.isMultipartContent(request); if(isMultipart){ ServletFileUpload upload = new ServletFileUpload(); try{ FileItemIterator iter = upload.getItemIterator(request); FileItemStream item = null; String name = ""; InputStream stream = null; while (iter.hasNext()){ item = iter.next(); name = item.getFieldName(); stream = item.openStream(); if(item.isFormField()){out.write("Form field " + name + ": " + Streams.asString(stream) + "
");} else { name = item.getName(); if(name != null && !"".equals(name)){ String fileName = new File(item.getName()).getName(); out.write("Client file: " + item.getName() + " 
with file name " + fileName + " was uploaded.
"); File file = new File(getServletContext().getRealPath("/" + fileName)); FileOutputStream fos = new FileOutputStream(file); long fileSize = Streams.copy(stream, fos, true); out.write("Size was " + fileSize + " bytes 
"); out.write("File Path is " + file.getPath() + "
"); } } } } catch(FileUploadException fue) {out.write("fue!!!!!!!!!");} } } } 

Answer 2

使用
http://commons.apache.org/fileupload/
或者如果您使用的是Tomcat 7，则使用Servlet 3.0 API