php如何为数组中未找到的行设置值

Question

我是 Php 7 的新手，我正在尝试编写一个函数来检查数据库中产品是否在某个月份有 qte（从日期字段中提取），但如果产品日期没有有这个月，所以所有数据都必须添加到数组中，但 qte 必须设置为零到那个月的值，我只得到了月份中现有 qte 的结果，但是产品做的月份的结果没有任何 qte 不在我的最终数组中

这是我尝试过的：

   public function getProductsStatsByMonths() {
    $today =  date("Y-m-d"); 
    $date_arr = explode("-", $today);  
    $year = $date_arr[0];
    $month = $date_arr[1];
    $day = $date_arr[2];
 
    $prods = array();
    $months = array(
    1 ,
    2 ,
    3 ,
    4 ,
    5 ,
    6 ,
    7 ,
    8 ,
    9 ,
    10 ,
    11 ,
    12
 );

    foreach ($months as $month){
        $stmt = $this->conn->prepare("SELECT sum(qte) as total,ligne, produit,date,heure,qte FROM production where YEAR(date) = ? and MONTH(date) = ? group by LOWER(ligne) ");
        $stmt->execute([$year,$month]);

        while( $row = $stmt->fetch(pdo::FETCH_ASSOC)){
            array_push($prods, new ProductionByMonth(0,$row["ligne"],$row["produit"],$row["date"],$row["heure"],$row["qte"],$month,$year));
        }
    }

    echo json_encode($prods);
   
   }

我得到的结果：

[
    {
        "id": 0,
        "ligne": "Biscuit",
        "produit": "Major",
        "date": "2021-08-10",
        "heure": "10:00",
        "qte": "130",
        "month": 8,
        "year": "2021"
    },
    {
        "id": 0,
        "ligne": "Eau",
        "produit": "Safia 1.5",
        "date": "2021-08-10",
        "heure": "14:00",
        "qte": "200",
        "month": 8,
        "year": "2021"
    },
    {
        "id": 0,
        "ligne": "Lait",
        "produit": "Vitalait 1/2",
        "date": "2021-08-10",
        "heure": "8:00",
        "qte": "80",
        "month": 8,
        "year": "2021"
    },
    {
        "id": 0,
        "ligne": "Salami",
        "produit": "Mazraa",
        "date": "2021-08-10",
        "heure": "8:00",
        "qte": "100",
        "month": 8,
        "year": "2021"
    },
    {
        "id": 0,
        "ligne": "Yaourt",
        "produit": "Delice",
        "date": "2021-08-10",
        "heure": "12:00",
        "qte": "150",
        "month": 8,
        "year": "2021"
    }
]

例如，此行包含此产品，该产品在第 8 个月有一个 qte：

{
            "id": 0,
            "ligne": "Biscuit",
            "produit": "Major",
            "date": "2021-08-10",
            "heure": "10:00",
            "qte": "130",
            "month": 8,
            "year": "2021"
        },

我的目标是在剩下的几个月里为同一产品添加额外的行，但将它们的 qte 设置为“0”

我的目标是确保我的数组最后还包含没有 qte (quantity) 的月份，并且 qte 的值必须设置为零，并且数据必须像所有存在的数据一样显示，除了字段qte的值

Answer 1

这里描述了适用于此类问题的一般方法。

1. 从数据库数据中获取唯一产品。
2. 创建包含所有所需日期的日历。
3. 在 PHP 中对来自 1 和 2 的数据集执行相当于数据库“交叉连接”的操作，并将生成的所有组合的 qte 设置为 0。
4. 用非零数量覆盖数据库中实际存在的那些组合的 qte。

“组合”是指产品和日期。因此，如果您创建一个包含 2021 年 12 个月并有 5 个产品的日历，那么第 3 步将生成 60 个组合。

另外，如果$day 没有在代码中的任何地方使用，则无需创建它。

编辑 - 这是代码：

// this is modified sample data
// changes made:
// set date values to first day of each month
// added "productId" with fabricated values since "id" was 0 for everything in original data and therefore not useful
// added another entry for productID = 12 with a different date
$prods = '[
    {
        "id": 0,
        "ligne": "Biscuit",
        "produit": "Major",
        "date": "2021-08-01",
        "heure": "10:00",
        "qte": "130",
        "month": 8,
        "year": "2021",
        "productId": 12
    },
    {
        "id": 0,
        "ligne": "Biscuit",
        "produit": "Major",
        "date": "2021-09-01",
        "heure": "12:00",
        "qte": "158",
        "month": 9,
        "year": "2021",
        "productId": 12
    },
    {
        "id": 0,
        "ligne": "Eau",
        "produit": "Safia 1.5",
        "date": "2021-08-01",
        "heure": "14:00",
        "qte": "200",
        "month": 8,
        "year": "2021",
        "productId": 16
    },
    {
        "id": 0,
        "ligne": "Lait",
        "produit": "Vitalait 1/2",
        "date": "2021-08-01",
        "heure": "8:00",
        "qte": "80",
        "month": 8,
        "year": "2021",
        "productId": 38
    },
    {
        "id": 0,
        "ligne": "Salami",
        "produit": "Mazraa",
        "date": "2021-08-01",
        "heure": "8:00",
        "qte": "100",
        "month": 8,
        "year": "2021",
        "productId": 27
    },
    {
        "id": 0,
        "ligne": "Yaourt",
        "produit": "Delice",
        "date": "2021-08-01",
        "heure": "12:00",
        "qte": "150",
        "month": 8,
        "year": "2021",
        "productId": 22
    }
]';

// create an associative array of products - assign a unique value as key (this will be useful in a later step)
$products = [];
foreach (json_decode($prods, true) as $product) {
    $key = $product['productId'] . '_' . $product['date'];
    $products[$key] = $product;
}

// get unique product IDs from database data
$productIds = [];
foreach ($products as $product) {
    $productIds[] = $product['productId'];
}
$uniqueProductIds = array_unique($productIds);

// create a calendar of all possible dates needed (just did a year's worth in this example)
$year = date("Y");
$buckets = [];
for ($month = 1; $month <= 12; $month++) {
    $buckets[] = date($year . '-' . str_pad($month, 2, '0', STR_PAD_LEFT) . '-01');
}

// do cross join equivalent to create all combinations of bucket and unique product ID (in this example: 12 buckets x 5 unique products IDs = 60 combinations)
$cartesianProduct = [];
foreach ($buckets as $bucket) {
    foreach ($uniqueProductIds as $uniqueProductId) {
        $data = [];
        $data['date'] = $bucket;
        $data['id'] = $uniqueProductId;
        $data['qte'] = 0;
        $key = $uniqueProductId . '_' . $bucket;
        $cartesianProduct[$key] = $data;
    }
}

// overwrite combination data (in $cartesianProduct) with data from database (in $products) - this is where $key in $products and in $cartesianProduct is critical
$result = array_merge($cartesianProduct, $products);

// look at results    
echo '<pre>' . print_r($result, 1) . '</pre>';