1. 直接上最簡單的 一種 ajax 異步上傳圖片，并預覽
html：
<!DOCTYPE html><html><head><meta http-equiv="Content-Type" content="text/html; charset=UTF-8"><title>圖片上傳 | 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" onclick="upload();"> <div class="images"></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="/HotelManager/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. 這里會 用到一個 ajaxupload.js， 網上多得很。
以上就是小編為大家帶來的Ajax 上傳圖片并預覽的簡單實現的全部內容了，希望對大家有所幫助，多多支持錯新站長站~
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