要显示与QUOT;装载....."在PHP页面 [英] Want to show "loading....." in PHP page

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问题描述

我开发一个网页中,需要下列功能的PHP: 1.当用户点击表示感谢,应以改变完成!。 2.同时,我想打电话给在indexController的动作。 3.在这个时候,我想显示加载...... 4.当前页面有大量的动态内容,也应该不会改变。

I am developing a web-page in PHP that needs following functionality: 1. When User click on "Say Thanks" it should be changed with "Done!". 2. At the same time I want to call an action in indexController. 3. At this time I want to show "loading...." 4. The current page has a lot of dynamic contents that should also not change.

请建议我应该怎么做才能完成上述任务......

Please suggest me what should I do to complete above tasks......

推荐答案

我想你需要一个AJAX调用。我通常做的装载意见和这样的,当你preSS更多。就我而言,有一个空的div和< A>在链接的评论视图(带有单独行动,OFC)标记。然后我使用jQuery的AJAX魔法:

I figure you need an AJAX call. I usually do that for loading comments and such when you press "more". In my case, there's an empty div and an <a> tag with the link to the comments view (with a separate action, ofc). Then I use jQuery for the AJAX magic:

$(function() { 
    $("a.CommentsListBtn").click(function() { 
        var tmpHref = $(this).attr("href");
        var tmpLayer = $(this).parent().children("div.CommentsList");
        tmpLayer.load(tmpHref, function() {
            tmpLayer.stop().slideDown("medium");
        });
        return false;
    });
});

我希望这有助于。

I hope this helps.

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