用php代码将引导旋转木马链接到数据库

Question

<div class="container slider">
    <div id="carouselExampleCaptions" class="carousel slide" data-bs-ride="carousel">
        <div class="carousel-inner">
            <?php 
            $slidersor=$db->prepare("SELECT * FROM slider");
            $slidersor->execute();

            while($slidercek=$slidersor->fetch(PDO::FETCH_ASSOC)) {
            ?>      
                <div class="carousel-item active">
                    <a href="blog-page"> <img src="<?php echo $slidercek['slider_resimyol'] ?>" class="d-block w-100" alt="Yeni Blog"> </a>
                    <div class="carousel-caption d-none d-md-block">
                        <a href="blog-page" class="a-none">  <h5>First slide label</h5></a>
                    </div>
                </div>
                
            <?php } ?>
                
        </div>

        <a class="carousel-control-prev" href="#carouselExampleCaptions" role="button" data-bs-slide="prev">
            <span class="carousel-control-prev-icon" aria-hidden="true"></span>
            <span class="visually-hidden">Geri</span>
        </a>
        
        <a class="carousel-control-next" href="#carouselExampleCaptions" role="button" data-bs-slide="next">
            <span class="carousel-control-next-icon" aria-hidden="true"></span>
            <span class="visually-hidden">İleri</span>
        </a>
    </div>
</div>

图像来自数据库，但不发生转换。看起来在php代码中没有问题。当我将php代码写入不同的div中时，来自数据库的图像会出现在另一个下面。我想知道问题在哪里

Answer 1

您只需要将active类放到第一个元素中。

<?php $active = true; ?>
    //your foreach
  <div class="carousel-item <?php echo ($active == true)?"active":"" ?>"> 
    ...
  </div>
    //your foreach end
<?php $active = false; ?>