英文:
How to display the php row with the same value as one in HTML?
问题
从数据库中获取多个具有相同值的数据。如何构建可以作为子类别工作的数据以合并所有具有相同名称的行？
网站应该只显示一个 "Pampanga"。这应该作为一个子类别工作。
以下是此特定页面的代码：
<section class="page-section bg-dark" id="home">
<div class="container">
<h2 class="text-center">旅游套餐</h2>
<div class="d-flex w-100 justify-content-center">
<hr class="border-warning" style="border:3px solid" width="15%">
</div>
<div class="d-flex w-100">
<?php
$packages = $conn->query("SELECT * FROM `packages` order by rand() ");
while($row = $packages->fetch_assoc() ):
$cover='';
if(is_dir(base_app.'uploads/package_'.$row['id'])){
$img = scandir(base_app.'uploads/package_'.$row['id']);
$k = array_search('.',$img);
if($k !== false)
unset($img[$k]);
$k = array_search('..',$img);
if($k !== false)
unset($img[$k]);
$cover = isset($img[2]) ? 'uploads/package_'.$row['id'].'/'.$img[2] : ""
}
$row['description'] = strip_tags(stripslashes(html_entity_decode($row['description'])));
?>
<div class="card w-100 rounded-0">
<img class="card-img-top" src="<?php echo validate_image($cover) ?>" alt="<?php echo $row['title'] ?>" height="200rem" style="object-fit:cover">
<div class="card-body">
<h5 class="card-title truncate-1"><?php echo $row['title'] ?></h5>
<p class="card-text truncate"><?php echo $row['description'] ?></p>
<div class="w-100 d-flex justify-content-end">
<a href="./?page=packages&id=<?php echo md5($row['id']) ?>" class="btn btn-sm btn-flat btn-warning">查看套餐 <i class="fa fa-arrow-right"></i></a>
</div>
</div>
</div>
<?php endwhile; ?>
</div>
<div class="d-flex w-100 justify-content-end">
<a href="./?page=packages" class="btn btn-flat btn-warning mr-4">探索套餐 <i class="fa fa-arrow-right"></i></a>
</div>
</div>
</section>
英文:
Multiple data is fetched from the database with the same value. How do I construct this that will work as a sub category. How can I combine all the row with the same name into one?
The site should display only one pampanga. This should work as a sub category.
Here is the code for this specific page.
<section class="page-section bg-dark" id="home">
<div class="container">
<h2 class="text-center">Tour Packages</h2>
<div class="d-flex w-100 justify-content-center">
<hr class="border-warning" style="border:3px solid" width="15%">
</div>
<div class="d-flex w-100">
<?php
$packages = $conn->query("SELECT * FROM `packages` order by rand() ");
while($row = $packages->fetch_assoc() ):
$cover='';
if(is_dir(base_app.'uploads/package_'.$row['id'])){
$img = scandir(base_app.'uploads/package_'.$row['id']);
$k = array_search('.',$img);
if($k !== false)
unset($img[$k]);
$k = array_search('..',$img);
if($k !== false)
unset($img[$k]);
$cover = isset($img[2]) ? 'uploads/package_'.$row['id'].'/'.$img[2] : "";
}
$row['description'] = strip_tags(stripslashes(html_entity_decode($row['description'])));
?>
<div class="card w-100 rounded-0">
<img class="card-img-top" src="<?php echo validate_image($cover) ?>" alt="<?php echo $row['title'] ?>" height="200rem" style="object-fit:cover">
<div class="card-body">
<h5 class="card-title truncate-1"><?php echo $row['title'] ?></h5>
<p class="card-text truncate"><?php echo $row['description'] ?></p>
<div class="w-100 d-flex justify-content-end">
<a href="./?page=packages&id=<?php echo md5($row['id']) ?>" class="btn btn-sm btn-flat btn-warning">View Package <i class="fa fa-arrow-right"></i></a>
</div>
</div>
</div>
<?php endwhile; ?>
</div>
<div class="d-flex w-100 justify-content-end">
<a href="./?page=packages" class="btn btn-flat btn-warning mr-4">Explore Package <i class="fa fa-arrow-right"></i></a>
</div>
</div>
</section>
答案1
得分: 2
如果您只想获取每个旅游地点的一个实例，您可以在请求中使用GROUP BY语句。
SELECT tour_location FROM `your_table_name` GROUP BY tour_location;
更多信息请参考 https://sql.sh/cours/group-by
另一个解决方法可以是使用SQL的DISTINCT。
SELECT DISTINCT tour_location FROM `your_table_name`
英文:
If you want only one instance of each tour location you car use a group by statement in your request.
SELECT tour_location FROM `your_table_name` GROUP BY tour_location;
See more on https://sql.sh/cours/group-by
Another solution could be the use of sql DISTINCT.
SELECT DISTINCT tour_location FROM `your_table_name`
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