vue form表单post请求结合Servlet实现文件上传功能

Sahar ·
更新时间:2024-09-20
· 823 次阅读

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软件开发网。如有错误或未考虑完全的地方,望不吝赐教。



VUE post请求 文件上传 form Servlet post

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