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php - 使用来自 mysql 数据库的 php 填充多个变量

转载 作者:行者123 更新时间:2023-11-30 21:59:22 25 4
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我的 mysql 数据库中有各种产品的数据。例如

ID    Name     price    detail    fact
1 test 20.00 text example
2 testing 30.00 apple orange
3 tester 5.43 plum grape

我有类似于下面的 html 代码,然后重复大约 30 次,因为 mysql 中有 30 种不同的产品数据

<!-- This is product 1 code -->
<div class="col-sm-4">
<div class="product-image-wrapper">
<div class="single-products">
<div class="productinfo text-center">

<img src="sample image 1" alt="" />
<h2>"price 1"</h2>
<p>"detail 1"</p>
<a href="#" class="btn btn-default add-to-cart"><i class="fa fa-shopping-cart"></i>Add to cart</a>
</div>
<div class="product-overlay">
<div class="overlay-content">
<h2>"price 1"</h2>
<h2>"ID"</h2>
<p>"detail 1"</p>
<a href="#" class="btn btn-default add-to-cart"><i class="fa fa-shopping-cart"></i>Add to cart</a>
</div>
</div>
</div>
<div class="choose">
<ul class="nav nav-pills nav-justified">
<li><a href=""><i class="fa fa-plus-square"></i>Add to wishlist</a></li>
<li><a href=""><i class="fa fa-plus-square"></i>Add to compare</a></li>
</ul>
</div>
</div>

<!-- This is product 2 code -->
<div class="col-sm-4">
<div class="product-image-wrapper">
<div class="single-products">
<div class="productinfo text-center">

<img src="sample image 2" alt="" />
<h2>"price 2"</h2>
<p>"Detail 2"</p>
<a href="#" class="btn btn-default add-to-cart"><i class="fa fa-shopping-cart"></i>Add to cart</a>
</div>
<div class="product-overlay">
<div class="overlay-content">
<h2>"price 2"</h2>
<p>"ID2"</p>
<p>"Detail 2"</p>
<a href="#" class="btn btn-default add-to-cart"><i class="fa fa-shopping-cart"></i>Add to cart</a>
</div>
</div>
</div>
<div class="choose">
<ul class="nav nav-pills nav-justified">
<li><a href=""><i class="fa fa-plus-square"></i>Add to wishlist</a></li>
<li><a href=""><i class="fa fa-plus-square"></i>Add to compare</a></li>
</ul>
</div>
</div>

例如,我想要完成的是,在产品 1 代码中输入“价格 1”将是我数据库中第一行中价格的变量,“详细信息 1”将是第一行详细信息数据库。依此类推第四。然后,当我们到达产品 2 代码时,跳到第二行数据并在其正确字段中插入匹配数据。重复此过程,直到数据库中的所有产品都与代码匹配。另请注意,并非 mysql 数据行中的所有字段都像“事实”一样使用。

我不确定如何完成这个。

最佳答案

您需要通过查询您的MySQL数据库来获取数据。执行此操作后,您可以逐行遍历结果集并输出。

<?php

// Some code here to obtain MySQL data result set, like...
$result_set = mysqli_query( 'SELECT * FROM products_table_name WHERE 1 ORDER BY name ASC' );

$x = 1;
foreach( $result_set as $row ){

?>
<!-- This is product <?php echo($x); ?> code -->
<div class="col-sm-4">
<div class="product-image-wrapper">
<div class="single-products">
<div class="productinfo text-center">

<img src="sample image 1" alt="" />
<h2><?php echo $row['price']; ?></h2>
<p><?php echo $row['detail']; ?></p>
<a href="#" class="btn btn-default add-to-cart"><i class="fa fa-shopping-cart"></i>Add to cart</a>
</div>
<div class="product-overlay">
<div class="overlay-content">
<h2><?php echo $row['price']; ?></h2>
<h2><?php echo $row['ID']; ?></h2>
<p><?php echo $row['detail']; ?></p>
<a href="#" class="btn btn-default add-to-cart"><i class="fa fa-shopping-cart"></i>Add to cart</a>
</div>
</div>
</div>
<div class="choose">
<ul class="nav nav-pills nav-justified">
<li><a href=""><i class="fa fa-plus-square"></i>Add to wishlist</a></li>
<li><a href=""><i class="fa fa-plus-square"></i>Add to compare</a></li>
</ul>
</div>
</div>
<?php
$x++;
}

关于php - 使用来自 mysql 数据库的 php 填充多个变量,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/43928710/

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