Android,在tomcat服务器中上传文件

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,我还没有找到要放置在tomcat服务器(servlet)中的Java代码

回答:

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() + "<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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