需要幫助，使用PHP將關聯數組轉換為嵌套JSON

Question

在過去的一周中，我一直在為此苦苦掙扎，這讓我發瘋，所以如果有人可以幫助我，我將永遠感激不已。

查詢數據庫后，我使用以下方法遍歷數據：

while ($row=mysqli_fetch_assoc($result)) {...

這是行的構造方式：

示例行1：

(
"countryId" => "2",
"countryDescription" => "Canada",
"cityId" => "3",
"cityDescription" => "Montreal",
"restaurantFranchiseId" => "2",
"restaurantFranchiseDescription" => "Kentucky Fried Chicken"
)

示例行2：

(
"countryId" => "2",
"countryDescription" => "Canada",
"cityId" => "3",
"cityDescription" => "Montreal",
"restaurantFranchiseId" => "3",
"restaurantFranchiseDescription" => "Taco Bell"
)

請注意，只有餐廳專營權在上述兩行中有所不同。 國家和城市在兩行中都相同。

我想將行變成一個嵌套的JSON文件，如下所示。 如下所示，每個國家/地區都是唯一的對象。 每個城市都是一個唯一的對象，並且它是相應國家對象的子元素。 然而，餐廳特許經營並不是唯一的，因為它們並不與特定的國家或城市掛鈎。

如何根據我的數據在下面創建如上所述的JSON文件？

謝謝！！！

{
"Countries": [{
    "countryId": "1",
    "countryDescription": "USA",
    "cities": [{
        "cityId": "1",
        "cityDescription": "Houston",
        "restaurantFranchises": [{
            "restaurantFranchiseId": "1",
            "restaurantFranchiseDescription": "Mc Donald's"
        }, {
            "restaurantFranchiseId": "2",
            "restaurantFranchiseDescription": "Kentucky Fried Chicken"
        }, {
            "restaurantFranchiseId": "4",
            "restaurantFranchiseDescription": "Pizza Hut"
        }]
    }, {
        "cityId": "2",
        "cityDescription": "New york",
        "restaurantFranchises": [{
            "restaurantFranchiseId": "1",
            "restaurantFranchiseDescription": "Mc Donald's"
        }, {
            "restaurantFranchiseId": "4",
            "restaurantFranchiseDescription": "Pizza Hut"
        }]
    }]
}, {
    "countryId": "2",
    "countryDescription": "Canada",
    "cities": [{
        "cityId": "3",
        "cityDescription": "Montreal",
        "restaurantFranchises": [{
            "restaurantFranchiseId": "1",
            "restaurantFranchiseDescription": "Mc Donald's"
        }, {
            "restaurantFranchiseId": "3",
            "restaurantFranchiseDescription": "Taco Bell"
        }, {
            "restaurantFranchiseId": "4",
            "restaurantFranchiseDescription": "Pizza Hut"
        }]
    }, {
        "cityId": "4",
        "cityDescription": "Ottawa",
        "restaurantFranchises": [{
            "restaurantFranchiseId": "2",
            "restaurantFranchiseDescription": "Kentucky Fried Chicken"
        }, {
            "restaurantFranchiseId": "3",
            "restaurantFranchiseDescription": "Taco Bell"
        }, {
            "restaurantFranchiseId": "4",
            "restaurantFranchiseDescription": "Pizza Hut"
        }]
    }]
}]

}

Answer 1

您可以使用以下代碼：

$result = [];
$lastCity = [ "cityId" => null ];
$lastCountry = [ "countryId" => null ];
while ($row=mysqli_fetch_assoc($result)) {
    if ($row["countryId"] !== $lastCountry["countryId"]) {
        // Country is not the same as in previous row, so create
        // a new entry for it in the first level of the result array. 
        // The city and franchises data will be inserted further down.
        $result[] = [
            "countryId" => $row["countryId"],
            "countryDescription" => $row["countryDescription"],
            "cities" => []
        ];
        // Get a reference (`&`) to the new country entry added to `$result`.
        // Whatever is later changed in `$lastCountry` will change inside the
        // `$result` data structure.
        $lastCountry = &$result[count($result)-1];
    }
    if ($row["cityId"] !== $lastCity["cityId"]) {
        // City is not the same as in previous row, so create
        // a new entry for it in the second level of `$result`.
        // We use the `$lastCountry` "shortcut" to manipulate `$result`.
        $lastCountry["cities"][] = [
            "cityId" => $row["cityId"],
            "cityDescription" => $row["cityDescription"],
            "restaurantFranchises" => []
        ];
        // Get a reference (`&`) to the new city entry added to `$result`.
        // Whatever is later changed in `$lastCity` will change inside the
        // `$result` data structure (and `$lastCountry`).
        $lastCity = &$lastCountry["cities"][count($lastCountry["cities"])-1];
    }
    // Create a new entry for the franchise in the third level of `$result`.
    // We use the `$lastCity` "shortcut" to manipulate `$result`.
    $lastCity["restaurantFranchises"][] = [
        "restaurantFranchiseId" => $row["restaurantFranchiseId"],
        "restaurantFranchiseDescription" => $row["restaurantFranchiseDescription"],
    ];
}

看到它在eval.in上運行。

關於輔助變量

$lastCity和$lastCountry這兩個變量是對$result數據結構中位置的引用（循環的開頭除外，當它們是偽值時）。 要在$result數組中獲取此類引用，請使用&運算符 。 可以在沒有這兩個變量的情況下完成此操作，但是這會使賦值語句變得很長，因為每次$result數組中的最后一個元素從它的cities數組中的最后一個元素..引用時，您都需要引用。 。等等。

假設條件

該算法需要將查詢結果集按國家和城市排序，即城市不應該首先是“ New York”，然后是“ Los Angeles”，再是“ New York”。

此外，假定cityId值是唯一的。 例如，美國的城市不應與加拿大的城市具有相同的cityId 。 如果是這種情況，則上面的代碼應稍作修改。