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...
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。
数据存储得不好,因为这是一个相当古老的网站。
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。
| products.id | products.sku |
|---|---|
| 1 | a |
| options.id | options.product_id |
|---|---|
| 1 | 1 |
| 2 | 1 |
| 3 | 1 |
| 4 | 1 |
| option_values.id | option_values.option_id | option_values.value |
|---|---|---|
| 1 | 1 | b |
| 2 | 1 | c |
| 3 | 1 | d |
| 4 | 2 | e |
| 5 | 2 | f |
| 6 | 3 | g |
| 7 | 3 | h |
| 8 | 4 | i |
| 9 | 4 | j |
| 10 | 4 | k |
迭代每个选项的每个可能组合,并输出生成的SKU;
| 可能的SKUs |
|---|
| abegi |
| acegi |
| adegi |
| abfgi |
| acfgi |
| adfgi |
| abehi |
| acehi |
| adehi |
| abegj |
| acegj |
| adegj |
| abegk |
| acegk |
| adegk |
| [等等] |
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...
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。
DB Fiddle - https://www.db-fiddle.com/f/vHWiKsKi9WUvvDwAa6pqw6/0
谢谢!
The function that creates the Cartesian product of all options is inspired bythe answer to this question.
$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);If you are only interested in the sku string, you can simplify the function to:
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; }I'm sure someone can provide a more efficient answer, but this answer produces the desired output based on your example data.