重新表述的标题:列出所有可能的产品选项
P粉976737101
2023-08-20 10:31:11
[PHP讨论组]
<p>以前,我可以轻松处理这样的事情,但自从7年前离开开发世界以来,我失去了开发思维...</p>
<p>我的情况是,我正在尝试输出客户可以选择的每个可能的产品选项以及生成的SKU - 由每个选项的SKU附加到产品的SKU。</p>
<p>数据存储得不好,因为这是一个相当古老的网站。</p>
<p>下面是MySQL中存储数据的示例,以及我在PHP中尝试实现的目标。我将其限制为一个具有多个选项的产品。</p>
<table class="s-table">
<thead>
<tr>
<th>products.id</th>
<th>products.sku</th>
</tr>
</thead>
<tbody>
<tr>
<td>1</td>
<td>a</td>
</tr>
</tbody>
</table>
<table class="s-table">
<thead>
<tr>
<th>options.id</th>
<th>options.product_id</th>
</tr>
</thead>
<tbody>
<tr>
<td>1</td>
<td>1</td>
</tr>
<tr>
<td>2</td>
<td>1</td>
</tr>
<tr>
<td>3</td>
<td>1</td>
</tr>
<tr>
<td>4</td>
<td>1</td>
</tr>
</tbody>
</table>
<table class="s-table">
<thead>
<tr>
<th>option_values.id</th>
<th>option_values.option_id</th>
<th>option_values.value</th>
</tr>
</thead>
<tbody>
<tr>
<td>1</td>
<td>1</td>
<td>b</td>
</tr>
<tr>
<td>2</td>
<td>1</td>
<td>c</td>
</tr>
<tr>
<td>3</td>
<td>1</td>
<td>d</td>
</tr>
<tr>
<td>4</td>
<td>2</td>
<td>e</td>
</tr>
<tr>
<td>5</td>
<td>2</td>
<td>f</td>
</tr>
<tr>
<td>6</td>
<td>3</td>
<td>g</td>
</tr>
<tr>
<td>7</td>
<td>3</td>
<td>h</td>
</tr>
<tr>
<td>8</td>
<td>4</td>
<td>i</td>
</tr>
<tr>
<td>9</td>
<td>4</td>
<td>j</td>
</tr>
<tr>
<td>10</td>
<td>4</td>
<td>k</td>
</tr>
</tbody>
</table>
<p>迭代每个选项的每个可能组合,并输出生成的SKU;</p>
<table class="s-table">
<thead>
<tr>
<th>可能的SKUs</th>
</tr>
</thead>
<tbody>
<tr>
<td>abegi</td>
</tr>
<tr>
<td>acegi</td>
</tr>
<tr>
<td>adegi</td>
</tr>
<tr>
<td>abfgi</td>
</tr>
<tr>
<td>acfgi</td>
</tr>
<tr>
<td>adfgi</td>
</tr>
<tr>
<td>abehi</td>
</tr>
<tr>
<td>acehi</td>
</tr>
<tr>
<td>adehi</td>
</tr>
<tr>
<td>abegj</td>
</tr>
<tr>
<td>acegj</td>
</tr>
<tr>
<td>adegj</td>
</tr>
<tr>
<td>abegk</td>
</tr>
<tr>
<td>acegk</td>
</tr>
<tr>
<td>adegk</td>
</tr>
<tr>
<td>[等等]</td>
</tr>
</tbody>
</table>
<p>当我像这样写出来时,它似乎非常简单,这让我想知道我是否遗漏了什么...</p>
<p>我目前正在迭代每个产品,对于每个产品的每个选项,然后对于每个选项的每个值,但显然这不能满足每种可能的情况。</p>
<p>DB Fiddle - https://www.db-fiddle.com/f/vHWiKsKi9WUvvDwAa6pqw6/0</p>
<p>谢谢!</p>
创建所有选项的笛卡尔积的函数受到这个问题的答案的启发。
<?php function all_skus(array $product) { $skus = []; $result = [[]]; foreach ($product['options'] as $key => $option) { $append = []; foreach ($result as $options) { foreach ($option as $value) { $append[] = $options + [$key => $value]; } } $result = $append; } foreach ($result as $option_set) { $skus[] = $product['sku'] . implode($option_set); } return $skus; } $pdo = new PDO(/* your stuff here */); $pdo->setAttribute(PDO::ATTR_DEFAULT_FETCH_MODE, PDO::FETCH_OBJ); $res = $pdo->query(' SELECT `o`.`product_id`, `p`.`sku`, `ov`.`option_id`, `ov`.`value` FROM `products` `p` JOIN `options` `o` ON `p`.`id` = `o`.`product_id` JOIN `option_values` `ov` ON `o`.`id` = `ov`.`option_id`' ); $nested = []; foreach ($res as $row) { $nested[$row->product_id]['options'][$row->option_id][] = $row->value; $nested[$row->product_id]['sku'] = $row->sku; } $skus = []; foreach ($nested as $i => $product) { $skus = array_merge($skus, all_skus($product)); unset($nested[$i]); } var_dump($skus);如果你只对sku字符串感兴趣,你可以简化函数为:
function all_skus(array $product) { $result = [$product['sku']]; foreach ($product['options'] as $option) { $append = []; foreach ($result as $options) { foreach ($option as $value) { $append[] = $options . $value; } } $result = $append; } return $result; }我相信有人可以提供一个更高效的答案,但这个答案根据你的示例数据产生了所需的输出。