Android，在tomcat服务器上传文件答案

【问题标题】：Android, upload a file in the tomcat serverAndroid，在tomcat服务器上传文件
【发布时间】：2012-04-23 00:54:42
【问题描述】：

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

目前我还没有找到要放置在 tomcat 服务器（servlet）中的 java 代码

【问题讨论】：

【解决方案1】：

这里是演示代码。

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() + "<br/>"); 
   out.print("Request content type is " + request.getHeader("Content-Type") + "<br/>");
   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) + "<br/>");}
                else {
                      name = item.getName();
                      if(name != null && !"".equals(name)){
                         String fileName = new File(item.getName()).getName();
                         out.write("Client file: " + item.getName() + " <br/>with file name "
                                                    + fileName + " was uploaded.<br/>");
                         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 <br/>");
                         out.write("File Path is " + file.getPath() + "<br/>");
                      }
                   }
              }
          } catch(FileUploadException fue) {out.write("fue!!!!!!!!!");}
   } 
  }
}

【讨论】：

【解决方案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

【讨论】：

我用的是tomcat 6，不然不知道commons-fileupload-1.2.2怎么用？？