Servlet 文件上传， HttpServletRequestWrapper

需要使用一些jar包。commons-fileupload-1.2.2.jar commons-io-1.4.jar

<form action="FileUpload" enctype="multipart/form-data" method="post">
<input type="text" name="name"><br>
<input type="file" name="fileName"><br>
<input type="submit">
</form>

这情况下，因为服务器获得是输入流，是得不到name的值的。

System.out.println(request.getParameter("name"));  //null
System.out.println(request.getParameter("fileName")); //null

public void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {

response.setContentType("text/html");
PrintWriter out = response.getWriter();

System.out.println(request.getParameter("name"));
System.out.println(request.getParameter("fileName"));

ServletFileUpload upload = new ServletFileUpload();

try {
FileItemIterator iter = upload.getItemIterator(request);
while(iter.hasNext()){
FileItemStream item = iter.next();
String name = item.getFieldName();

//得到表单域，是一个输入流
InputStream stream = item.openStream();

//如果是普通表单域
if(item.isFormField()){
String value = Streams.asString(stream);
System.out.println(name + " -> " + value);
}else{
String filename = item.getName(); //得到文件上传名
System.out.println("文件:" + filename);
}
}

} catch (FileUploadException e) {
e.printStackTrace();
}

System.out.println(request.getInputStream().available());;
out.flush();
out.close();
}

null

null

name -> fdsf

文件:hello.zip

0

正常上传的话，可以用：

Streams.copy(stream, new FileOutputStream(request.getRealPath("upload")+"\\" + filename), true);

System.out.println("cpy: " + request.getRealPath("upload") + filename);