Ajax 上传图片并预览的简单实现

2024-09-01 08:27:44

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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" 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. 这里会用到一个 ajaxupload.js，网上多得很。

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