2017-02-28 63 views
0

我需要将jpeg图像发送(上传)到Servlet,而不是将其保存在文件上,我想将其转换为BufferedImage并对其执行一些处理。将jpeg图像发送到Servlet并将其转换为BufferedImage

这是我的客户端代码:

HttpClient httpclient = new DefaultHttpClient(); 
httpclient.getParams().setParameter(CoreProtocolPNames.PROTOCOL_VERSION, HttpVersion.HTTP_1_1); 

HttpPost httppost = new HttpPost("http://localhost:9000/upload"); 
File file = new File("/tmp/lena.jpg"); 

MultipartEntity mpEntity = new MultipartEntity(); 
ContentBody cbFile = new FileBody(file, "image/jpeg"); 
mpEntity.addPart("userfile", cbFile); 


httppost.setEntity(mpEntity); 
System.out.println("executing request " + httppost.getRequestLine()); 
HttpResponse response = httpclient.execute(httppost); 
HttpEntity resEntity = response.getEntity(); 

System.out.println(response.getStatusLine()); 
if (resEntity != null) { 
    System.out.println(EntityUtils.toString(resEntity)); 
} 
if (resEntity != null) { 
    resEntity.consumeContent(); 
} 

httpclient.getConnectionManager().shutdown(); 

我怎么能接受一个Servlet和处理它的形象呢?

我想这对我的接收端的Servlet,但图像是空:

InputStream is = request.getInputStream(); 
BufferedImage bImageFromConvert = ImageIO.read(is); 

最后,我不应该保存在磁盘上的任何东西的过程中。

回答

1

尝试:

Part file = request.getPart("userfile"); 
InputStream is = file.getInputStream(); 
+0

的答案是正确的,但如果你想使用的Servlet 3.0的HttpServletRequest#的getParts()方法,那么你必须注解与@MultipartConfig你的servlet。 之后,事情奏效 – Hossein