Springmvc multipartfile file upload and parameter receive

Reprinted from: http://blog.csdn.net/u013771277/article/details/47384817

Springmvc file Upload, first two base, 1. Add Enctype= "Multipart/form-data" to form form properties

Emphasis: Form form <form method= "post" ..., method must have, I am here to use is post, as to get line not have tried, no method= "POST" will also report not multipart request error.

2. Configuration file in configuration Multipartresolver

File out-of-bounds will throw an exception before entering the controller, without affecting the configuration within the allowed range

3. Simple Receive method, idea: Multipartfile accept the file and enter it into the fileoutstream by IO binary stream (Multipartfile.getinputstream ()) to save the file, then why do you do it?

The parameters receive the same as Multipartfile receive.

Accept files and parameters named file and ID in the form screenshot of the form. As follows

@RequestMapping (value = "Attendee_uploadexcel.do")

@ResponseBody

public void Uploadexcel (@RequestParam ("file") Multipartfile file, @RequestParam ("id") String ID) throws Exception {

parameter of form form submission test to String type

if (file = = null) return;

String fileName = File.getoriginalfilename ();

String path = Getrequest (). Getservletcontext (). Getrealpath ("/upload/excel");

Gets the actual path of the specified file or folder in the project, Getrequest () This method is to return a httpservletrequest that encapsulates this method in order to handle the encoding problem

FileOutputStream fos = fileutils.openoutputstream (new File (path+ "/" +filename));//Open Fileoutstrean stream

Ioutils.copy (File.getinputstream (), FOS);//convert multipartfile file into binary stream and input to Fileoutstrean

Fos.close ();//

......

}

Other methods, will httpservletrequest req strong turn into Multiparthttpservletrequest req after req.getparameter ("id");

HttpServletRequest request;

Multiparthttpservletrequest multipartrequest = (multiparthttpservletrequest) request;

Multipartfile file = multipartrequest.getfile ("file");

String id = multipartrequest.getparameter ("id"); Receive parameters that are carried by the client incoming file

String fileName = File.getoriginalfilename (); .........