[ { "status": true, "data": [ { "id": 2, "name": "Algeria", "country_code": "DZ", "regions": [ { "id": 2, "country_id": 2, "region_code": "RG01", "depots": [ { "id": 5, "region_id": 2, "depot_code": "DP04", "depot_name": "North Depot", "area": [ { "id": 1, "depot_id": 5, "area_code": "AR1", "area_name": "Area-1" }, { "id": 2, "depot_id": 5, "area_code": "AR2", "area_name": "Area-2" } ] }, { "id": 6, "region_id": 2, "depot_code": "DP06", "depot_name": "east Depot", "area": [ { "id": 3, "depot_id": 6, "area_code": "AR3", "area_name": "Area-3" } ] } ] }, { "id": 3, "country_id": 2, "region_code": "RG02", "depots": [] } ] } ] } ] 我想將區域合并到單個數組中 "area": [ { "id": 1, "depot_id": 5, "area_code": "AR1", "area_name": "Area-1" },我不想使用多個 foreach。提前致謝。
2 回答
qq_笑_17
TA貢獻1818條經驗 獲得超7個贊
您可以使用以下代碼遞歸迭代所有結構并僅提取您需要的數據（按鍵）。我假設 $data 具有您的原始數據結構，并且在 $result 中您將獲得區域數組：
$it = new RecursiveIteratorIterator(new RecursiveArrayIterator($data), RecursiveIteratorIterator::SELF_FIRST);
$result = [];
foreach ($it as $k => $v) {
if ($k === 'area') {
$result = array_merge($result, $v);
}
}
元芳怎么了
TA貢獻1798條經驗 獲得超7個贊
我不會使事情過于復雜，遍歷您的數據結構。將所有區域添加到集合中。準備您的數據結構，我添加了一個檢查以確保沒有重復項unique()。
$data = 'your-data';
$areas = collect();
foreach($data->data as $country) {
foreach($country->regions as $region) {
foreach ($region->depots as $depot) {
$areas->concat($depot->area);
}
}
}
$result = ['area' => $areas->unique('id')->all()];
- 2 回答
- 0 關注
- 145 瀏覽
添加回答
舉報
0/150
提交
取消