PHP 中的图片库

Question

我想通过 php 和 mysqli 在我的数据库中的画廊中显示一些图像，我有一个画廊模板和我的 php 代码也从我的画廊中获取照片。 我提供了画廊代码和 php，请指导我如何合并这两个代码。

画廊代码是

<body>
   <section class="gallery-block cards-gallery">
        <div class="container">
            <div class="heading">
              <h2>Cards Gallery</h2>
            </div>
            <div class="row">
                <div class="col-md-6 col-lg-4">
                    <div class="card border-0 transform-on-hover">
                        <a class="lightbox" href="img/image2.jpg">
                            <img src="img/image2.jpg" alt="Card Image" class="card-img-top">
                        </a>
                        <div class="card-body">
                            <h6><a href="#">Lorem Ipsum</a></h6>
                            <p class="text-muted card-text">Lorem ipsum dolor sit amet, consectetur adipiscing elit. Nunc quam urna.</p>
                        </div>
                    </div>
                </div>              
            </div>
        </div>
    </section>
    <script src="css/baguetteBox.min.js"></script>
    <script>
        baguetteBox.run('.cards-gallery', { animation: 'slideIn'});
    </script>
</body>
</html>


<PHP CODE IS --> 


<?php                                   
$img="SELECT * FROM professionaldetails WHERE professionalid=0";
$query=mysqli_query($con,$img);
while($row=mysqli_fetch_assoc($query))
{
?>

<img  src="images/<?php   echo $row['topimg'] ?>" id="changeimage1"  width="150" height="227" style="padding:5px;">
<?php } ?>```

Answer 1

你可以在 HTML 标签之间写 PHP。

<?php    
$query = "SELECT * FROM professionaldetails WHERE professionalid=0";

$result = $mysqli->query($query)                               

?>

<body>
   <section class="gallery-block cards-gallery">
        <!-- begin loop -->
        <?php while ($row = $result->fetch_assoc()) { ?>
        <div class="container">
            <div class="heading">
              <h2>Cards Gallery</h2>
            </div>
            <div class="row">
                <div class="col-md-6 col-lg-4">
                    <div class="card border-0 transform-on-hover">
                        <a class="lightbox" href="img/image2.jpg">
                            <img src="<?php echo $row['topimg'] ?>" alt="Card Image" class="card-img-top">
                        </a>
                        <div class="card-body">
                            <h6><a href="#">Lorem Ipsum</a></h6>
                            <p class="text-muted card-text">Lorem ipsum dolor sit amet, consectetur adipiscing elit. Nunc quam urna.</p>
                        </div>
                    </div>
                </div>              
            </div>
        </div>
        <!-- end loop -->
        <?php } ?>
    </section>
    <script src="css/baguetteBox.min.js"></script>
    <script>
        baguetteBox.run('.cards-gallery', { animation: 'slideIn'});
    </script>
</body>

PHP 中的图片库

问题描述

1 个解决方案

解决方案1
0 已采纳 2021-01-16 13:29:37

PHP 中的图片库

问题描述

1 个解决方案

解决方案1 0 已采纳 2021-01-16 13:29:37

解决方案1
0 已采纳 2021-01-16 13:29:37