文件上传下载之FileUpload

    技术2022-05-20  92

     

    FileUpload依赖于commons-io.jar

    The simplest case

    // Check that we have a file upload request boolean isMultipart = ServletFileUpload.isMultipartContent(request); // Create a factory for disk-based file items FileItemFactory factory = new DiskFileItemFactory(); // Create a new file upload handler ServletFileUpload upload = new ServletFileUpload(factory); upload.setHeaderEncoding("utf-8");  //转码为你需要的格式,后面就不要转码了。后面得到对应的字符串后再转码还是有可能遇到乱码,所以就在这儿转码一半我是转成utf-8, //好像跟我的页面编码也没关系,因为我的页面是gbk的,是不是和操作系统有关系呢。 // Parse the request List< FileItem> items = upload.parseRequest(request); //fileItem的一些操作 fileItem.getInputStream()==========>InputStream fileItem.get() ==========>byte[] fileItem.getName() fileItem.getFieldName() for(FileItem item:items){ byte[] b=item.get(); FileOutputStream out=new FileOutputStream("d:/"+item.getName()); out.write(b); out.close(); } HTML端的书写 <form action="run" enctype="multipart/form-data" method="post"> <input type="file" name="a"/> <input type="file" name="b"/> <input type="submit" value="提交"/> </form> 注意 要上传的file必须要有name=“a”即使你不需要,也得有,要不然解析不出来的。

     


    最新回复(0)