我有一個(gè)具有以下結(jié)構(gòu)的數(shù)組。它有一些重復(fù)的值，我需要根據(jù)正確的制造商和型號(hào)對(duì)其進(jìn)行排序。2 => array:3 [ "id_111" => "ALFA ROMEO" "id_112" => "4C" "id_8702" => "P"]3 => array:3 [ "id_111" => "ALFA ROMEO" "id_112" => "GIULIA" "id_8702" => "D"]4 => array:3 [ "id_111" => "ALFA ROMEO" "id_112" => "GIULIA" "id_8702" => "U"]5 => array:3 [ "id_111" => "ALFA ROMEO" "id_112" => "GIULIETTA" "id_8702" => "D"]6 => array:3 [ "id_111" => "ALFA ROMEO" "id_112" => "GIULIETTA" "id_8702" => "U"]7 => array:3 [ "id_111" => "ALFA ROMEO" "id_112" => "STELVIO" "id_8702" => "D"]從這個(gè)查詢中，我必須使結(jié)構(gòu)化數(shù)組如下所示：manufacturers: [0: { name: "ALFA ROMEO", models: [ 0: { name: "4C", ids: [ 0: { name: "p" } ] }, 1: { name: "GIULIA", ids: [ 0: {name: "D"}, 1: {name: "U"} ] }, 2: { name: "GIULIETTA", ids: [ 0: {name: "U"}, 1: {name: "D"} ] } ]}1: {}]到目前為止，我已經(jīng)嘗試過以下代碼。但出于某種原因，我有點(diǎn)迷路了。 $data = array(); foreach ($records as $record) { $id1 = $record["id_111"]; $data["manufacturer"] = array(); if (!in_array("id1", $data["manufacturer"])) { $data["manufacturer"] = array( "name" => $id1 ); } $id2 = $record["id_112"]; $data["manufacturer"][] = array("name" => $id2); }任何人都可以幫助我。謝謝你。
1 回答
牛魔王的故事
TA貢獻(xiàn)1830條經(jīng)驗(yàn) 獲得超3個(gè)贊
定義一個(gè) groupBy 函數(shù)：
function groupBy($items, $key)
{
$array = array();
foreach($items as $item) {
foreach($item as $k => $v) {
if($key == $k) $array[$v] []= $item;
}
}
return $array;
}
接著
$names = groupBy($records, 'id_111');
$name_array = [];
foreach($names as $name => $name_items) {
$models = groupBy($name_items, 'id_112');
$model_array = [];
foreach($models as $model => $model_items) {
$ids = array_map(function($item) {
return ['name' => $item['id_8702']];
},$model_items);
$model_array []= ['name' => $model, 'ids' => $ids];
}
$name_array []= ['name' => $name, 'models' => $model_array];
}
$results = ['manufacturers' => $name_array];
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