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Home/ Questions/Q 7772643
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Editorial Team
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Editorial Team
Asked: June 1, 20262026-06-01T16:55:56+00:00 2026-06-01T16:55:56+00:00

I am looking for code that can quickly upload an image from my android

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I am looking for code that can quickly upload an image from my android application in tomcat server.

At the moment I have not found the java code to place in the tomcat server (servlet)

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-01T16:55:57+00:00Added an answer on June 1, 2026 at 4:55 pm

    Here is demo code.

    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!!!!!!!!!");}
       } 
      }
    } 
    
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