0

在 jquery 插件成功的情况下,我需要操作 DOM 并创建/使一些新的 html 控件可见...

HTML..

<input type="file" name="fileupload" id="fileupload" />

<%=Html.ActionLink("TestA", "TestA", "ControllerA", new { Id = Id})%> 

<%=Ajax.ActionLink("TestB","TestB","ControllerB", new { Id= Id})%>

JS:

<script type="text/javascript">
$(function () {

    $('#fileupload').uploadify({
        'buttonText': 'Upload',
        'auto': true,
        'multi': false,
        'swf': '/Uploadify/uploadify.swf',
        'uploader': '/ControllerA/LinkMethod',
        'formData': { 'Id': Id},
        'debug': false,          
        'onUploadSuccess': function (file, data, response) {

         //Here I want to show two action links and disable/invisible the plugin.

        }
    });
}); 

4

1 回答 1

3

只需调用具有相应 id 值的各个元素即可show()hide()

 'onUploadSuccess': function (file, data, response) {

        $('#MyIdForLink1').show();
        $('#MyIdForLink2').show();
        $('#pluginId').hide();

 }
于 2012-08-27T21:50:34.050 回答