Twiiter Bootstrap旋转木马用PHP制作动态

Question

我正在尝试制作一个使用Twitter Bootstrap默认JavaScript和CSS的旋转木马,但应该从MySQL数据库中获取图像.

我试图用PHP动态的旋转木马的HTML是这样的:

<div class="carousel-inner">
   <div class="active item">
                <img src="images/sampleAd.jpg" class="pull-left marginTop">
                <img src="images/sampleAd.jpg" class="pull-left marginTop">
                <img src="images/sampleAd.jpg" class="pull-left marginTop">
                <img src="images/sampleAd.jpg" class="pull-left marginLast">
   </div>
   <a class="carousel-control left" href="#myCarousel" data-slide="prev">&lsaquo;</a>
   <a class="carousel-control right" href="#myCarousel" data-slide="next">&rsaquo;</a>
</div>

我期待的PHP文件的最终结果是这样的

<div class="carousel-inner">
   <?php renderAds(); ?>
</div>

我写的PHP函数是这样的:

function renderAds(){
    $query = queryAds();
    $query_exe = mysql_query($query);

    if(mysql_num_rows($query_exe) == 0){
        echo '<img src="images/sampleAd.jpg" >';
    }else{

        $numb =1;
        $flag =1;
        while($fetched_data = mysql_fetch_array($query_exe)){

            if($numb == 1){
                if($flag == 1){
                    echo '<div class="active item">';
                }else{
                    echo '<div class="item">';
                }
            }elseif($numb == 4){
                echo '<a href="'.$fetched_data['url'].'"  class="pull-left marginLast" ><img src="'.$fetched_data['image_url'].'"></a>';
                echo '</div>';
                $numb =0;
            }else{
                echo '<a href="'.$fetched_data['url'].'" class="pull-left marginTop" ><img src="'.$fetched_data['image_url'].'"></a>';
            }
            $numb++;
            $flag++;
        }


        echo '<a class="carousel-control left" href="#myCarousel" data-slide="prev">&lsaquo;</a>';
        echo '<a class="carousel-control right" href="#myCarousel" data-slide="next">&rsaquo;</a>';
    }


}

在这里,通过与twitter bootstrap的默认轮播不同,我用4张图片制作每张幻灯片,我希望每张幻灯片转换包含4张图片,而不是默认情况下一张幻灯片的单张图片.在HTML中它可以工作,但是当我用我的PHP函数替换它时,第一张幻灯片可以工作,但如果我按下旋转木马中的下一个箭头整个div崩溃.

我怎样才能使它工作？

注意:此处最后一张图片必须包含marginLast类,另一张图片必须包含marginTop和pull-left类.

我期望在HTML中呈现的是这样的:

<div class="carousel-inner">

            <div class="active item">
                <img src="images/sampleAd.jpg" class="pull-left marginTop">
                <img src="images/sampleAd.jpg" class="pull-left marginTop">
                <img src="images/sampleAd.jpg" class="pull-left marginTop">
                <img src="images/sampleAd.jpg" class="pull-left marginLast">
            </div>

            <div class="item">
                <img src="images/sampleAd.jpg" class="pull-left marginTop">
                <img src="images/sampleAd.jpg" class="pull-left marginTop">
                <img src="images/sampleAd.jpg" class="pull-left marginTop">
                <img src="images/sampleAd.jpg" class="pull-left marginLast">
            </div>

            <div class="item">
                <img src="images/sampleAd.jpg" class="pull-left marginTop">
                <img src="images/sampleAd.jpg" class="pull-left marginTop">
                <img src="images/sampleAd.jpg" class="pull-left marginTop">
                <img src="images/sampleAd.jpg" class="pull-left marginLast">
            </div>

            <a class="carousel-control left" href="#myCarousel" data-slide="prev">&lsaquo;</a>
            <a class="carousel-control right" href="#myCarousel" data-slide="next">&rsaquo;</a>
   </div>

如何使我的PHP函数显示如上所示的结果标记？

编辑:我得到的输出是这样的:

<div id="myCarousel" class="carousel slide">
          <!-- Carousel items -->
          <div class="carousel-inner">

                <div class="active item">
                    <a href="xx2" class="pull-left marginTop" ><img src="images/sampleAd.jpg"></a>
                    <a href="xx" class="pull-left marginTop" ><img src="images/sampleAd.jpg"></a>
                    <a href="sample.com"  class="pull-left marginLast" ><img src="images/sampleAd.jpg"></a>
                </div>

                <a class="carousel-control left" href="#myCarousel" data-slide="prev">&lsaquo;</a>
                <a class="carousel-control right" href="#myCarousel" data-slide="next">&rsaquo;</a>             
          </div><!-- end carousel-inner -->

  </div><!-- myCarousel -->

根据标记,这应该工作,但它不起作用,默认情况下,如果没有更多的幻灯片应该被禁用,但它是可点击的箭头.

Answer 1

你有MySQL查询只返回一个数组.这样,您的查询可以重复使用(然后可以成为一个函数),您可以根据自己的需要呈现数据.

<?php
$mysqli = new mysqli('localhost', 'username', 'password', 'database');

$sql = "SELECT * FROM table";
$res = $mysqli->query($sql);

$rows = array();
while ($row = $res->fetch_assoc()) {
    $rows[] = $row;
}
?>
<div class="carousel">
  <div class="carousel-inner">
<?php $i = 1; ?>
<?php foreach ($rows as $row): ?>
<?php $item_class = ($i == 1) ? 'item active' : 'item'; ?>
    <div class="<?php echo $item_class; ?>">
      <a href="<?php echo $row['url']; ?>">
        <img src="<?php echo $row['image']; ?>" alt="<?php echo $row['title']; ?>" />
      </a>
    </div>
<?php $i++; ?>
<?php endforeach; ?>
  </div>
</div>

您显然需要更改数据库查询以及foreach循环中使用的值.我也使用了这些mysqli函数,因为这些mysql函数现在已被弃用,而有利于MySQLi(改进了MySQL).