【发布时间】:2020-02-28 18:14:23
【问题描述】:
<?php
$con=mysqli_connect('localhost','root');
mysqli_select_db($con,'tims');
$query = "SELECT * FROM products p JOIN categories c ON p.c_id = c.c_id";
$result=mysqli_query($con,$query);
$num=mysqli_num_rows($result);
if ($num>0) {
// output data of each row
while($row=mysqli_fetch_array($result)) {
?>
<div class="col-12 col-sm-6 col-md-12 col-xl-6">
<div class="single-product-wrapper">
<!-- Product Image -->
<div class="product-img">
<img src="<?php echo $row['image'] ?>" alt="" />;
<img src="<?php echo $row['image2'] ?>" alt="" class="hover-img" />
</div>
<!-- Product Description -->
<div class="product-description d-flex align-items-center justify-content-between">
<!-- Product Meta Data -->
<div class="product-meta-data">
<div class="line"></div>
<p class="product-price"> ₹<?php echo $row['price']; ?> </p>
<a href="product-details.php?action=add&p_id=<?php echo $row['p_id']; ?> ">
<h6> <?php echo $row['title'] ?> </h6>
</a>
</div>
<!-- Ratings & Cart -->
<div class="ratings-cart text-right">
<div class="ratings">
<i class="fa fa-star" aria-hidden="true"></i>
<i class="fa fa-star" aria-hidden="true"></i>
<i class="fa fa-star" aria-hidden="true"></i>
<i class="fa fa-star" aria-hidden="true"></i>
</div>
<div class="cart">
<a href="cart.html" name="add-to-cart" data-toggle="tooltip" data-placement="left" title="Add to Cart"><img src="img/core-img/cart.png" alt=""></a>
</div>
</div>
</div>
</div>
</div>
<?php
}
} else { echo "0 results";}
?>
这是我当前的代码。它显示表格中的每个项目。我只想要指定的类别。
【问题讨论】:
标签: php html css mysql database