转：一个简单的jQuery插件ajaxfileupload实现ajax上传文件例子

时间：2014-09-05 12:35:21 收藏：0 阅读：263

页面代码：

<html>
    
    <script type="text/javascript" src="js/jquery.js"></script>
    <script type="text/javascript" src="js/ajaxfileupload.js"></script>

    
    <script type="text/javascript">
        function ajaxFileUpload(){
            $.ajaxFileUpload(
                 {
                url:‘update.do?method=uploader‘,            //需要链接到服务器地址
                secureuri:false,
                fileElementId:‘houseMaps‘,                        //文件选择框的id属性
                dataType: ‘xml‘,                                     //服务器返回的格式，可以是json
                success: function (data, status)            //相当于java中try语句块的用法
                {
                    $(‘#result‘).html(‘添加成功‘);
                },
                error: function (data, status, e)            //相当于java中catch语句块的用法
                {
                    $(‘#result‘).html(‘添加失败‘);
                }
            }

            );

        }
    </script>
</head>

<body>
    <form method="post" action="update.do?method=uploader" enctype="multipart/form-data">
        <input type="file" id="houseMaps" name="houseMaps"/>
        <input type="button" value="提交" onclick="ajaxFileUpload()"/>
    </form>
    <div id="result"></div>

</body>
</html>

服务器代码：

public class UpdateAction extends DispatchAction {

    public ActionForward uploader(ActionMapping mapping, ActionForm form,
            HttpServletRequest request, HttpServletResponse response) {
        UpFormForm upFormForm = (UpFormForm) form;
        FormFile ff = upFormForm.getHouseMaps();
        try {
            InputStream is = ff.getInputStream();
            File file = new File("D:/" + ff.getFileName());            //指定文件存储的路径和文件名
            OutputStream os = new FileOutputStream(file);

            byte[] b = new byte[1024];
            int len = 0;
            while((len = is.read(b)) != -1){
                os.write(b, 0, len);
            }
            os.close();
            is.close();
        } catch (Exception e) {
            e.printStackTrace();

        }

        return null;
    }
}