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Php Ajax加载幻灯片失败_Php_Ajax_Jquery - Fatal编程技术网

Php Ajax加载幻灯片失败

Php Ajax加载幻灯片失败,php,ajax,jquery,Php,Ajax,Jquery,我试图在滑动div上显示文本。要显示的文本是使用ajax动态加载的。在加载文本之前,滑块div工作正常,但在单击链接并用数据填充div之后,滑块停止工作。我不知道怎么解决这个问题。一个活的样本是。尝试单击类似事件的链接,然后单击“上一步”查看发生了什么。我的html在这里 <div class="coda-slider" id="slider-id"> <div> <?php

我试图在滑动div上显示文本。要显示的文本是使用ajax动态加载的。在加载文本之前,滑块div工作正常,但在单击链接并用数据填充div之后,滑块停止工作。我不知道怎么解决这个问题。一个活的样本是。尝试单击类似事件的链接,然后单击“上一步”查看发生了什么。我的html在这里

<div class="coda-slider"  id="slider-id">
                    <div>
                        <?php
                        $counter = 0;
                        for ($x = 0; $x < 8; $x++) {
                            $counter++;
                            ?>
                            <div class="galleryitem">
                                <a href="images/flyer1.png" target="" rel="galleryitem"> <img src="images/flyer<?php echo $counter; ?>.png" alt="Flyer <?php echo $counter; ?>" title="Click to view more"/></a>
                                <!--strong>Flyer</strong-->
                                <span class="description"><strong>Item <?php echo $counter; ?></strong> Simple description..Lorem ipsum dolor sit amet, consectetuer adipiscing elit...<a href="product.php">more</a></span>

                            </div>
                        <?php } ?>
                    </div>

                </div>
我的php后端代码在这里

<?php
if (!$_POST['page'])
    die("0");
$page = $_POST['page'];
if ($page == 'design') {
    ?>

    <div>
        <?php
        $counter = 0;
        for ($x = 0; $x < 4; $x++) {
            $counter++;
            ?>
            <div class="galleryitem">
                <a href="images/flyer1.png" target="" rel="galleryitem"> <img src="images/flyer<?php echo $counter; ?>.png" alt="Flyer <?php echo $counter; ?>" title="Click to view more"/></a>
                <!--strong>Flyer</strong-->
                <span class="description"><strong>Item <?php echo $counter; ?></strong> Simple description..Lorem ipsum dolor sit amet, consectetuer adipiscing elit...<a href="product.php">more</a></span>
            </div>
        <?php } ?>
    </div>
    <?php
} else {
    echo 'No lists in that category';
}
?>

最有可能的是,在您的ajax中更改为

$.ajax({
        type: "POST",
        url: "load_page.php",
        data: {'page':url }, /** this is syntax of POST**/
        dataType: "html",
        success: function(msg){
            if(msg.length > 0)
            {

                $('#slider-id').html(msg);
                $('#loading').css('visibility','hidden');
            }
        }

    }); 

我认为你应该做一个get请求而不是post。您不是在向服务器发布您正在获取的内容。您还必须更改load_page.php

查看您的代码,似乎在添加新内容后没有调用任何滑块函数,您可能应该在成功回调中这样做

$.ajax({
        type: "GET",
        url: "load_page.php",
        data: 'page='+url,
        dataType: "html",
        success: function(msg){
            if(parseInt(msg)!=0)
            {
                // guess you have some slider function

                $('#slider-id').html(msg);
                $('#loading').css('visibility','hidden');
                $('#slider-id').slider(); //example of slider function
            }
        }
$.ajax({
        type: "GET",
        url: "load_page.php",
        data: 'page='+url,
        dataType: "html",
        success: function(msg){
            if(parseInt(msg)!=0)
            {
                // guess you have some slider function

                $('#slider-id').html(msg);
                $('#loading').css('visibility','hidden');
                $('#slider-id').slider(); //example of slider function
            }
        }