servlet代码
@MultipartConfig()@WebServlet(name = "test", urlPatterns = "*.do", initParams = { @WebInitParam(name = "", value = "") }, loadOnStartup = 3)public class TestServlet3 extends HttpServlet { @Override protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException { // TODO Auto-generated method stub process(req, resp); } @Override protected void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException { // TODO Auto-generated method stub process(req, resp); } private void process(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException { Part p = req.getPart("f"); BufferedReader br = new BufferedReader(new InputStreamReader( p.getInputStream())); String line = null; while ((line = br.readLine()) != null) { System.out.println(line); } }} 非常简单的代码，读取上传文件内容，然后输出到控制台。
页面代码
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<title>Insert title here</title>
</head>
<body>
<form action="a.do" enctype="multipart/form-data">
<input type="file" name="f" >
<input type="submit" value="submit">
</form>
</body>
</html>
问题来了，提交报错
严重: Servlet.service() for servlet [test] in context with path [/tomcat7test] threw exception [org.apache.tomcat.util.http.fileupload.FileUploadBase$InvalidContentTypeException: the request doesn't contain a multipart/form-data or multipart/form-data stream, content type header is null] with root causeorg.apache.tomcat.util.http.fileupload.FileUploadBase$InvalidContentTypeException: the request doesn't contain a multipart/form-data or multipart/form-data stream, content type header is null at org.apache.tomcat.util.http.fileupload.FileUploadBase$FileItemIteratorImpl.(FileUploadBase.java:806) at org.apache.tomcat.util.http.fileupload.FileUploadBase.getItemIterator(FileUploadBase.java:261)
后来查了查资料，form表单必须要设置method=post和enctype=multipart/form-data