Question

美好的一天，

我有一个数组（请求）；

$appliedFilters = [
    "6",
    "37",
    "40",
    "41"
]

然后我将所有产品按“过滤器类别”分组（我知道这是json输出，但更易于共享）；

{
   "6":[
      {
         "product_id":"12",
      }
   ],
   "7":[
      {
         "product_id":"12",
      }
   ],
   "12":[],
   "48":[
      {
         "product_id":"12",
      }
   ],
   "33":[],
   "34":[],
   "35":[],
   "36":[
      {
         "product_id":"12",
      }
   ],
   "37":[
      {
         "product_id":"12",
      }
   ],
   "38":[],
   "39":[],
   "40":[],
   "41":[]
}

如何退回所有$ appliedFilters与过滤器完全相同的产品？

我只需要$ appliedFilters与filterCategories完全相同的产品。

例如，给定的$appliedFilters应该返回 0个产品，因为不存在带有过滤器40或41的产品。

第二个示例

$appliedFilters = [
    "6",
    "36",
    "48"
]

在这种情况下，应该返回product_id = 12的产品，因为所有$appliedFilters都有该产品。该产品是否还具有更多的过滤器（例如7和48）也没关系。

到目前为止，我已经尝试过，但是预期结果不正确：

// fetch all filter items
$productsGroupByFilterItems = [];
$rows = $this->getFilters();
foreach ($rows as $row) {
    $productsGroupByFilterItems[$row['filterItem_id']] = [];
}

// group all products by filter ids
foreach ($products as $product) {
    $productFilters = explode(',', $product['chosenFilterIds']);
    foreach ($productsGroupByFilterItems as $filterItemId => $filterItem) {
        foreach ($productFilters as $filterId) {
            if ($filterId == $filterItemId) {
                $productsGroupByFilterItems[$filterItemId][] = $product;
            }
        }
    }
}

// remove all products that are not part of the applied filters
foreach($productsGroupByFilterItems as $key => $val)
{
    if(array_search($key, $appliedFilters) === false)
    {
        unset($productsGroupByFilterItems[$key]);
    }
}

// finally return only the products that exist in all the filters items
return call_user_func_array('array_intersect_key', $productsGroupByFilterItems);

Answer 1

代码太复杂，无法解析，但这可能会让您有所了解。

假设产品以$products形式存在于PHP数组中，只需翻转$appliedFilters即可创建密钥，然后计算密钥的交集：

$result = array_intersect_key(array_flip($appliedFilters), $products);

这将为您提供$products中密钥在$appliedFilters中的所有项目。相反，请查看array_diff_key。

Answer 2

您应该执行以下操作（我已将JSON转换为简单数组）。

$filteredData = [
    6 => [
        ['product_id' => 12]
    ],
    7 => [
        ['product_id' => 12]
    ],
    12 => [],
    48 => [
        ['product_id' => 12]
    ],
    33 => [],
    34 => [],
    35 => [],
    36 => [
        ['product_id' => 12]
    ],
    37 => [
        ['product_id' => 12]
    ],
    38 => [],
    39 => [],
    40 => [],
    41 => []
];

//case 1
$appliedFilters = [
    "6",
    "37",
    "40",
    "41"
];

$out = array_intersect_key($filteredData, array_flip($appliedFilters));
$products = null;

foreach ($out as $filteredProducts) {
    if ($products === null) {
        $products = $filteredProducts;
        continue;
    }

    $products = @array_intersect($products, $filteredProducts);
}

var_dump($products);

//case 1
$appliedFilters = [
    "6",
    "36",
    "48"
];

$out = array_intersect_key($filteredData, array_flip($appliedFilters));
$products = null;

foreach ($out as $filteredProducts) {
    if ($products === null) {
        $products = $filteredProducts;
        continue;
    }

    $products = @array_intersect($products, $filteredProducts);
}

var_dump($products);

结果是：

//case 1
array(0) {
}

//case 2
array(1) {
  [0]=>
  array(1) {
    ["product_id"]=>
    int(12)
  }
}

您可以在此处查看它的运行情况：https://3v4l.org/JEjcB

基本上，我们将已过滤数组的键和过滤键相交。然后将所有乘积数组相交成一个循环。

根据过滤器ID过滤产品

2 个答案: