1. 直接上最简单的 一种 ajax 异步上传图片，并预览
html:
1 <!DOCTYPE html>
2 <html>
3 <head>
4 <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
5 <title>图片上传 | cookie</title>
6 </head>
7 <body>
8 file: <input type="file" id="images" name="image" /><br><br>
9 desc: <input type="text" id="desc" name="desc" /><br><br>
10 <input type="button" value="upload" onclick="upload();">
11
12 <div class="images"></div>
13
14 <script type="text/javascript" src="js/jquery-1.12.4.min.js"></script>
15 <script type="text/javascript" src="js/upload.js"></script>
16 <script type="text/javascript">
17 function upload() {
18 $.ajaxFileUpload({
19 url : ‘upload.htm‘,
20 fileElementId : ‘images‘,
21 dataType : ‘json‘,
22 data : {desc : $("#desc").val()},
23 success : function(data) {
24 var html = $(".images").html();
25 html += ‘<img width="100" height="100" src="/HotelManager/upload/‘ + data.url + ‘">‘
26 $(".images").html(html);
27 }
28 })
29 return false;
30 }
31 </script>
32 </body>
33 </html>
servlet:
1 protected void doPost(HttpServletRequest request, HttpServletResponse response)
2 throws ServletException, IOException {
3 DiskFileItemFactory factory = new DiskFileItemFactory();
4
5 ServletFileUpload upload = new ServletFileUpload(factory);
6
7 String path = request.getServletContext().getRealPath("/upload");
8 String name = null;
9 try {
10 List<FileItem> items = upload.parseRequest(request);
11 for (FileItem item : items) {
12 if(item.isFormField()){
13 System.out.println(item.getFieldName() + ": " + item.getString());
14 } else {
15 name = item.getName();
16 item.write(new File(path,name));
17 }
18 }
19 PrintWriter out = response.getWriter();
20 out.print("{");
21 out.print("url:\"" + name +"\"");
22 out.print("}");
23
24 } catch (Exception e) {
25 e.printStackTrace();
26 }
27 }
3. 这里会 用到一个 ajaxupload.js， 网上多得很，实在找不到，也可以私我，给你们
时间： 2024-12-24 05:18:14