Ajax 上传图片并预览的简单实现_AJAX相关教程-查字典教程网

1. 直接上最简单的一种 ajax 异步上传图片，并预览

html：

<!DOCTYPE html> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"> <title>图片上传 | cookie</title> </head> <body> file: <input type="file" id="images" name="image" /><br><br> desc: <input type="text" id="desc" name="desc" /><br><br> <input type="button" value="upload"> <div></div> <script type="text/javascript" src="js/jquery-1.12.4.min.js"></script> <script type="text/javascript" src="js/upload.js"></script> <script type="text/javascript"> function upload() { $.ajaxFileUpload({ url : 'upload.htm', fileElementId : 'images', dataType : 'json', data : {desc : $("#desc").val()}, success : function(data) { var html = $(".images").html(); html += '<img width="100" height="100" src="http://www.jb51.netHotelManager/upload/' + data.url + '">' $(".images").html(html); } }) return false; } </script> </body> </html>

servlet：

protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { DiskFileItemFactory factory = new DiskFileItemFactory(); ServletFileUpload upload = new ServletFileUpload(factory); String path = request.getServletContext().getRealPath("/upload"); String name = null; try { List<FileItem> items = upload.parseRequest(request); for (FileItem item : items) { if(item.isFormField()){ System.out.println(item.getFieldName() + ": " + item.getString()); } else { name = item.getName(); item.write(new File(path,name)); } } PrintWriter out = response.getWriter(); out.print("{"); out.print("url:"" + name +"""); out.print("}"); } catch (Exception e) { e.printStackTrace(); } }

2. 这里会用到一个 ajaxupload.js，网上多得很。

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