一、文件上传
???1.对表单的要求
???????（1）提交方式必须是post;
???????（2）上传时，enctype = "multipart/form-data";
???????（3）必须要有<input type="file">
???2.对servlet的要求
???????request.getInputStream();
???3.fileupload组件
???????comments-fileupload.jar
???????comments-io.jar
???????三个核心类
boolean ismultipartContent = ServletFileUpload.isMultipartContent(request); ???????if (!ismultipartContent){ ???????????throw new RuntimeException("你的表单不是enctype=multipart/form-data"); ???????} ???????//1.创建DiskFileFactory工厂类 ???????DiskFileItemFactory dfif = new DiskFileItemFactory(); ???????//2.创建servletFileUpload核心对象 ???????ServletFileUpload sfu = new ServletFileUpload(dfif); ???????//3.解析request对象，并得到一个表单项的集合 ???????try { ???????????List<FileItem> fileItems = sfu.parseRequest(request); ???????????for(FileItem fileItem:fileItems){ ???????????????if(fileItem.isFormField()){ ???????????????????//是普通表单 ???????????????????processFormFiled(fileItem); ???????????????}else{ ???????????????????//上传表单项 ???????????????????processUploadFiled(fileItem); ???????????????} ???????????} ???????} catch (FileUploadException e) { ???????????e.printStackTrace(); ???????} ???}
???4.FileItem提供的方法
???????getFeildName(); //得到字段名 ???
???????getString();//得到字段的值
???????getName();//得到文件名
???????getInputStream();
???????isFormFeild();//判断是否是普通表单项
???????getContentType();//获取上传文件的MIME类型
???????getSize();//获取上传文件的大小
//普通表單項 ???private void processFormFiled(FileItem fileItem) { ???????String fieldName = fileItem.getFieldName(); ???????String filedValue = fileItem.getString(); ???????System.out.println(fieldName + " = " + filedValue); ???}private void processUpload(FileItem fileItem){ ???????//得到上传的名字 ???????String fileName = fileItem.getName(); ???????//得到文件流 ???????try { ???????????InputStream is = fileItem.getInputStream(); ???????????//创建存盘的路径 ???????????String realPath = this.getServletContext().getRealPath("/day18/upload"); ???????????File storeDirectory = new File(realPath ); ???????????if(!storeDirectory.exists()){ ???????????????storeDirectory.mkdirs(); ???????????} ???????????//处理文件名 ???????????//fileName = fileName.substring(fileName.lastIndexOf(File.separator)+1); ???????????if(fileName!=null){ ???????????????fileName = FilenameUtils.getName(fileName); ???????????}文件上传
原文地址：https://www.cnblogs.com/q-1993/p/10526333.html