JQuery的 - 编码问题

Question

嗨，我工作的一个“状态” -Updater。它的工作原理，只有一个问题，发送状态我必须手动刷新页面来再次让脚本后做事。你能帮助我吗？下面的代码：

<script language="javascript">
$(document).ready(function(){
    $("form#status_form").submit(function(){
        var s_autor =   $('#s_autor').attr('value');
        var s_status    =   $('#s_status').attr('value');
        $.ajax({
            type: "POST",
            url: "/admin/request.php",
            data: "s_autor="+ s_autor +"& s_status="+ s_status,
            success: function(){
                $('#show').load("/admin/request.php").fadeIn("slow", function(){
                    setTimeout(function(){
                        $(function() {
                             $("#show").fadeTo("slow", 0.01, function(){
                                 $(this).slideUp("slow", function() {
                                     $(this).remove();
                                 });
                             });
                        });
                    }, 2000);
                });
            },
        });
        return false;
    });
});
</script>

所以，你知道如何编写代码的时候有可能要重复我多久点击提交按钮？脚本

顺便说一句，这里的HTML表格：

<form id="status_form" method="post" action="request.php" onsubmit="return false;">
    <input type="text" id="s_autor" name="s_autor" value="<?= $user_id; ?>" style="display: none;" /><input type="text" id="s_status" name="s_status" value="Statusnachricht" /><input type="submit" value="" class="submit" id="status_submit" />
</form>

Answer 1

它看起来像你第一次调用此代码后，你摧毁#show，因此，接下来的时间，没有更多的#show？我要一些评论添加到您的代码，看看我是这个正确理解：

success: function() {
    // File has been successfully posted to request.php
    // We are now going to GET the contents of request.php (ignoring any response from the previous POST)
    // Also note, should probably put remainder of code in a callback to "load", so we can be sure it has had a chance to load!...
    $('#show').load("/admin/request.php").fadeIn("slow", function(){
        setTimeout(function(){
            $(function() {
                 $("#show").fadeTo("slow", 0.01, function(){
                     $(this).slideUp("slow", function() {
                         // Note! #show is being removed from the document here. It won't be available for future events.
                         $(this).remove();
                     });
                 });
            });
        }, 2000);
    });
},

祝你好运！

修改在此寻找更多一些，我注意到您在您的setTimeout后创建文档“准备就绪”事件？

我试图找出你想要的效果。它看起来像你想拥有的东西淡入上传完成（在#show DIV）后，等待几秒钟，然后再次淡出出来？ （你slideUp充当一个淡出的元件上，所以效果将不可见。）

怎么样这样呢？

success: function() {
    // File has been successfully posted to request.php
    // We are now going to GET the contents of request.php (ignoring any response from the previous POST)
    $('#show').load("/admin/request.php", function() {
        $(this).fadeIn("slow", function() {
            setTimeout(function() {

                $("#show").fadeOut("slow", function() {
                    // don't remove #show. Just empty it for using again next time.
                    $(this).empty()
                });

            }, 2000);
        });
    }); 
},