java实现动态上传多个文件并解决文件重名问题

时间:2022-09-28 15:03:50

本文分为两大方面进行讲解:

一、java实现动态上传多个文件

二、解决文件重命名问题java

供大家参考,具体内容如下

1、动态上传多个文件

 

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<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);">删除</button>
      </td>
    </tr>
  </table>
  <br/>
  <input type="button" onclick="_submit();" value="上传">
  <input onclick="_add();" type="button" value="增加">
  </form>
 </body>
 <script type="text/javascript">
   function _add(){
     var tb = document.getElementById("tb");
     //写入一行
     var tr = tb.insertRow();
     //写入列
     var td = tr.insertCell();
      //写入数据
     td.innerHTML="File:";
     //再声明一个新的td
     var td2 = tr.insertCell();
     //写入一个input
     td2.innerHTML='<input type="file" name="file"/><button onclick="_del(this);">删除</button>';
   }
   function _del(btn){
     var tr = btn.parentNode.parentNode;
     //alert(tr.tagName);
     //获取tr在table中的下标
     var index = tr.rowIndex;
     //删除
     var tb = document.getElementById("tb");
     tb.deleteRow(index);
   }
   function _submit(){
     //遍历所的有文件
     var files = document.getElementsByName("file");
     if(files.length==0){
       alert("没有可以上传的文件");
       return false;
     }
     for(var i=0;i<files.length;i++){
       if(files[i].value==""){
         alert("第"+(i+1)+"个文件不能为空");
         return false;
       }
     }
    document.forms['xx'].submit();
   }
 </script>
</html>

遍历所有要上传的文件

2、解决文件的重名的问题

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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 response)
      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);
      //只接收图片*.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);
          
          //获取扩展
          String extName = fileName.substring(fileName.lastIndexOf("."));//.jpg
          //UUID
          String uuid = UUID.randomUUID().toString().replace("-", "");
          //新名称
          String newName = uuid+extName;     //在这里用UUID来生成新的文件夹名字,这样就不会导致重名
          
          
          FileUtils.copyInputStreamToFile(file.getInputStream(),
              new File(path+"/"+newName));
          //放到list
          imgs.add(newName);
        }
        file.delete();
      }
      request.setAttribute("imgs",imgs);
      request.getRequestDispatcher("/jsps/imgs.jsp").forward(request, response);
    }catch(Exception e){
      e.printStackTrace();
    }
  
  }
 
}

以上实现了java多文件上传,解决了文件重名问题,希望对大家的学习有所帮助。