<?php$userData = [ [ "UID" => "5f10482574d83d4b726fe5", "name" => "Yug Gill", "orgID" => "5f10481d74d83d4b726", "imageURL" => "female.png" ]];$userProductsData = [ [ "UPID" => "5f10482574d83d4b6fe007", "UID" => "5f10482574d83d4b726fe5", ]];$userDetailsResult = [];foreach ($userData as $key => $value) { $userData[$key]["UPID"] = $userProductsData[$value["UID"]] ?? [];}預(yù)期輸出$userData = [ [ "UID" => "5f10482574d83d4b726fe5", "name" => "Yug Gill", "orgID" => "5f10481d74d83d4b726", "imageURL" => "female.png", "UPID" => "5f10482574d83d4b6fe007" ]];我有兩個(gè)UID數(shù)組通用的數(shù)組，現(xiàn)在我想從中獲取 UPID$userProductsData并推入$userData，我嘗試過(guò)無(wú)法正常工作，請(qǐng)有人更新我的代碼嗎？>
2 回答
猛跑小豬
TA貢獻(xiàn)1858條經(jīng)驗(yàn) 獲得超8個(gè)贊
試試這個(gè)。
$userData = [
[
"UID" => "5f10482574d83d4b726fe5",
"name" => "Yug Gill",
"orgID" => "5f10481d74d83d4b726",
"imageURL" => "female.png"
]
];
$userProductsData = [
[
"UPID" => "5f10482574d83d4b6fe007",
"UID" => "5f10482574d83d4b726fe5",
]
];
$userDetailsResult = [];
foreach ($userProductsData as $key => $value) {
$userData[$key]["UPID"] = $value['UPID'];
}
print_r($userData);
慕工程0101907
TA貢獻(xiàn)1887條經(jīng)驗(yàn) 獲得超5個(gè)贊
您$userProductsData是一個(gè)常規(guī)數(shù)組，它有索引 0、1、2 等...然后您嘗試通過(guò)字符串 key 從此數(shù)組中獲取項(xiàng)目"5f10482574d83d4b726fe5"。
$userProductsData應(yīng)該是這樣的鍵數(shù)組：
$userProductsData = [
"5f10482574d83d4b726fe5" => [
"UPID" => "5f10482574d83d4b6fe007",
"UID" => "5f10482574d83d4b726fe5",
],
];
然后你可以通過(guò) key 從這個(gè)數(shù)組中獲取一個(gè)項(xiàng)目"5f10482574d83d4b726fe5"。
- 2 回答
- 0 關(guān)注
- 201 瀏覽
添加回答
舉報(bào)
0/150
提交
取消