Java implementation dynamically upload multiple files and solve the problem of duplicate file name _java

This article is divided into two major areas to explain:

One, Java implementation dynamic upload multiple files

Second, solve file renaming problem Java

For your reference, the specific contents are as follows

1. Upload Multiple files dynamically

 <form name= "xx" action= "<c:url value= '/up3servlet '/>" method= "post" enctype= "Multipart/form-data" > < Table id= "TB" border= "1" > <tr> <td> File: </td> <td> < Input type= "file" name= "file" > <button onclick= "_del (this);" > Delete </button> </td> </tr> </table> <input type= "button" onclick= " _submit (); "value=" Upload "> <input onclick=" _add (); "type=" button "value=" Add "> </form> </body> <
     Script type= "Text/javascript" > Function _add () {var TB = document.getElementById ("TB");
     Writes a line of var tr = Tb.insertrow ();
      Write column var td = Tr.insertcell ();
     Write Data td.innerhtml= "File:";
     Then declare a new TD var TD2 = Tr.insertcell (); Writes an input td2.innerhtml= ' <input type= ' file ' name= ' file '/><button onclick= ' _del (this);
   > Delete </button> '; } function _del (btn) {var tR = Btn.parentNode.parentNode;
     alert (tr.tagname);
     Gets the subscript var index = Tr.rowindex of TR in the table;
     Delete var TB = document.getElementById ("TB");
   Tb.deleterow (index);
     function _submit () {//traversal of the file var files = document.getelementsbyname ("file");
       if (files.length==0) {alert ("There is no file to upload");
     return false;
         for (Var i=0;i<files.length;i++) {if (files[i].value== ") {alert (" First "+ (i+1) +" file cannot be empty ");
       return false;
   } document.forms[' xx '].submit ();
 } </script>  

Traverse all files to be uploaded

2, solve the problem of duplicate file name

Package cn.hx.servlet;
Import Java.io.File;
Import java.io.IOException;
Import Java.io.PrintWriter;
Import java.util.ArrayList;
Import java.util.List;

Import Java.util.UUID;
Import javax.servlet.ServletException;
Import Javax.servlet.http.HttpServlet;
Import Javax.servlet.http.HttpServletRequest;

Import Javax.servlet.http.HttpServletResponse;
Import Org.apache.commons.fileupload.FileItem;
Import Org.apache.commons.fileupload.disk.DiskFileItemFactory;
Import Org.apache.commons.fileupload.servlet.ServletFileUpload;

Import Org.apache.commons.io.FileUtils; public class Upimgservlet extends HttpServlet {public void DoPost (HttpServletRequest request, HttpServletResponse resp
    Onse) throws Servletexception, IOException {request.setcharacterencoding ("UTF-8");
    String path = Getservletcontext (). Getrealpath ("/up");
    Diskfileitemfactory disk = new Diskfileitemfactory (1024*10,new File ("d:/a"));
    Servletfileupload up = new Servletfileupload (disk); try{List<fileitem> list = up.parserequest (request);
      Only receive picture *.jpg-iamge/jpege.,bmp/imge/bmp,png, list<string> IMGs = new arraylist<string> (); for (Fileitem file:list) {if (File.getcontenttype (). Contains ("image/")) {String fileName = File.getname ()
          ;
          
          FileName = filename.substring (Filename.lastindexof ("\") +1); Gets the extension String extname = filename.substring (Filename.lastindexof ("."));
          /.jpg//uuid String UUID = Uuid.randomuuid (). toString (). Replace ("-", "");     New name String newName = uuid+extname;
              Here you use the UUID to generate a new folder name so that it does not cause duplicate names Fileutils.copyinputstreamtofile (File.getinputstream),
          New File (path+ "/" +newname));
        Put it in List Imgs.add (NewName);
      } file.delete ();
      } request.setattribute ("IMGs", IMGs);
    Request.getrequestdispatcher ("/jsps/imgs.jsp"). Forward (request, response);}catch (Exception e) {e.printstacktrace ();
 }
  
  }

}

The above implementation of the Java file upload, to solve the problem of duplicate file name, I hope to help you learn.