vue form表单post请求结合Servlet实现文件上传功能

编程学习 2021-07-04 14:07www.dzhlxh.cn编程入门
这篇文章主要介绍了vue form表单post请求结合Servlet实现文件上传功能,具有很好的参考价值,希望对大家有所帮助。一起跟随长沙网络推广过来看看吧

前端测试页面代码:

<template>
 <div>
  <input type="file" name="file" @change="change($event)">
 </div>
</template>
<script>
 export default {
  created(){
   this.path = this.$route.query;
   for (let i in this.path) {
    this[i] = decodeURIComponent(this.path[i]);
   }
  },
  methods:{
   change(ev){
    let file = ev.target.files[0];
    let size = file.size;
    let name = file.name;
    if(size > 314572800){
     this.$message.warning('上传文件不能超过300M');
     return;
    }
    let formData = new FormData();
    formData.append('file',file,name)
    this.$axios.post('/JT3'+this.getddRecordDelete,formData,{
     headers:{"Content-Type":"multipart/form-data"}
    }).then(data=>{
     console.log(data);
    })
   }
  }
 }
</script>
<style scoped>
</style>

后端servlet接收代码

package jt3.control.zygkh;
import java.io.File;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.util.List;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
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.FileUploadException; 
import org.apache.commons.fileupload.disk.DiskFileItemFactory; 
import org.apache.commons.fileupload.servlet.ServletFileUpload;
import jtacc.filter.JTKit;
import jtacc.jtpub.DT; 
@WebServlet(urlPatterns = "/upfile/file") 
public class UploadServlet extends HttpServlet {
	private static final long serialVersionUID = 1L;
	protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
		System.out.println(11);
		this.doPost(request, response);
 }
 protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
 	String uri="/u/file/"+DT.getFormatDate("yyyyMMdd")+"/"; //定义路径
 	String tmpPath=JTKit.getBaseDIR()+uri;//此处为个人项目路径,根据需求定义路径
 	DiskFileItemFactory factory = new DiskFileItemFactory();
		factory.setRepository(new File(tmpPath));//临时文件存储路径
		ServletFileUpload fileUpload = new ServletFileUpload(factory);//核心操作对象
		fileUpload.setHeaderEncoding("utf-8");//防乱码
		try {
			//此处如果要实时强行转换则需要下载jar包(commons-fileupload-1.3.3.jar)
			List<FileItem> list = fileUpload.parseRequest(request);
			for (FileItem fileItem : list) {
				InputStream in = fileItem.getInputStream();
				String filename = fileItem.getName();
				if (fileItem != null) {
					System.out.println(filename);
					int len = 0;
					byte[] array = new byte[1024];
					FileOutputStream fos = new FileOutputStream(tmpPath+filename);
					while((len = in.read(array))!=-1){//表示每次最多读1024个字节
						fos.write(array,0,len);
						fos.flush();
					}
					fos.close();
					in.close();
					fileItem.delete();
					response.setCharacterEncoding("UTF-8");
					String realPath = uri+filename;
					response.getWriter().append(realPath);
				}
			}
		} catch (FileUploadException e) {
			// TODO Auto-generated catch block
			e.printStackTrace();
		}
 }
 
}

测试结果

补充:Servlet获取表单提交过来的数据

在Servlet的doPost方法:

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

中获取表单数据,首先,为了防止出现中文乱码问题,需要给request设置编码为“UTF-8”:

request.setCharacterEncoding("utf-8");

获取单个字符串的方式:

<pre style="font-family: 宋体; font-size: 12pt; background-color: rgb(255, 255, 255);"><span style="font-size:18px; font-family: Arial, Helvetica, sans-serif; background-color: rgb(240, 240, 240);">String username = request.getParameter("username");</span>

获取字符串数组的方式:

String[] favorites = request.getParameterValues("favorite");

以上为个人经验,希望能给大家一个参考,也希望大家多多支持狼蚁SEO。如有错误或未考虑完全的地方,望不吝赐教。

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