我正在尝试将文件发送到Servlet。 与此文件一起,我还必须发送一些参数(即名称/ ID,日期和其他一些参数)。我在客户端使用HttpClient,在服务器端使用ServerFileUpload。
这是客户端代码: ...
String url = "http://localhost:8080/RicezioneServlet/RicezioneServlet";
HttpClient httpclient = new DefaultHttpClient();
HttpPost postMethod = new HttpPost(url);
MultipartEntity mpe = new MultipartEntity();
//I'm sending a .zip file
ContentBody cb = new FileBody(fileToSend,"application/zip");
mpe.addPart("file", cb);
postMethod.setEntity(mpe);
HttpResponse resp = httpclient.execute(postMethod);
HttpEntity respEntity = resp.getEntity();
System.out.println(resp.getStatusLine());
...
在服务器端,我们有:
ServletFileUpload sup = new ServletFileUpload();
FileItemIterator it = sup.getItemIterator(request);
FileItemStream item = it.next();
InputStream ios = item.openStream();
//read from ios and write to a fileoutputstream.
现在,我不知道如何将上述参数添加到请求中...我尝试使用StringBody并将其添加到MultiPartEntity,但我在以下位置得到NullPointerException:
String author = request.getParameter("author");
这意味着该参数不被视为参数,也许?
唯一让我工作的是将这些参数设置为Headers(setHeader和getHeader),但这不是一个选项。
有什么建议吗?或者您可以将我重定向到文件+参数上传的完整示例吗?
谢谢,
亚历
答案 0 :(得分:3)
尝试使用粘贴在此处的类似代码:
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(url);
FileBody bin = new FileBody(new File(fileName));
StringBody comment = new StringBody("Filename: " + fileName);
MultipartEntity reqEntity = new MultipartEntity();
reqEntity.addPart("bin", bin);
reqEntity.addPart("comment", comment);
httppost.setEntity(reqEntity);
HttpResponse response = httpclient.execute(httppost);
HttpEntity resEntity = response.getEntity();
你还需要添加外部jar apache-mime4j-0.6.jar(org.apache.james.mime4j)否则
reqEntity.addPart("bin", bin);
不会编译。
答案 1 :(得分:1)
如果使用servlet 3.0,可以尝试将@MultipartConfig添加到servlet中。