输出引用表的递归函数

Question

假设我有一个这样的数组

Array(
    Array("id_1" => 1,"id_2" => 1,"name" => "test1","type" => "A","ref_1" => 0,"ref_2" => 0,),
    Array("id_1" => 1,"id_2" => 2,"name" => "test2","type" => "B","ref_1" => 1,"ref_2" => 1,),
    Array("id_1" => 1,"id_2" => 3,"name" => "test3","type" => "B","ref_1" => 1,"ref_2" => 1,),
    Array("id_1" => 2,"id_2" => 1,"name" => "test4","type" => "B","ref_1" => 1,"ref_2" => 1,),
    Array("id_1" => 2,"id_2" => 3,"name" => "test5","type" => "C","ref_1" => 1,"ref_2" => 2,),
    Array("id_1" => 2,"id_2" => 15,"name" => "test6","type" => "C","ref_1" => 1,"ref_2" => 3,),
    Array("id_1" => 5,"id_2" => 22,"name" => "test7","type" => "B","ref_1" => 4,"ref_2" => 9,),
    Array("id_1" => 4,"id_2" => 9,"name" => "test8","type" => "C","ref_1" => 1,"ref_2" => 1,),
    Array("id_1" => 1,"id_2" => 7,"name" => "test9","type" => "C","ref_1" => 2,"ref_2" => 1,),
    Array("id_1" => 5,"id_2" => 20,"name" => "test10","type" => "B","ref_1" => 4,"ref_2" => 9,),
    Array("id_1" => 5,"id_2" => 5,"name" => "test11","type" => "B","ref_1" => 4,"ref_2" => 9,),
    Array("id_1" => 5,"id_2" => 4,"name" => "test12","type" => "B","ref_1" => 1,"ref_2" => 1,),
);

“id_1”和“id_2”两个“主键”，两者的组合不能重复。每一行都可以被另一个使用“ref_1”和“ref_2”引用，被引用的行将包含来自referer（父亲）的“id_1”和“id_2”。

无法引用 A 类行。类型 B 和 C 可以被 A 引用。同样，B 可以被 C 引用，反之亦然。树并不总是完整的，它可能只有 TYPE A 和 B 或 TYPE A 和 C，或者只有 TYPE A。

我正在尝试创建一个返回如下表格的函数：

|   TYPE A NAME |   TYPE B NAME |   TYPE C NAME |
|   ------------------------------------------- |
|   test1       |   test2       |   test5       |
|   test1       |   test3       |   test6       |
|   test1       |   test7       |   test8       |
|   test1       |   test4       |   test9       |
|   test1       |   test10      |   test8       |
|   test1       |   test11      |   test8       |
|   test1       |   test12      |               |

每行引用 2 个（或更多）其他行，数据是父亲的数据被拆分并为每个引用输出一行。如果一个 A 类引用 2 个 B 类，每个 B 类引用其他 2 个 C 类，则输出表将有 4 行。

不止三种，这只是一个PoC。我试图尽可能详细地解释这一点，但可能有点难以理解。我知道它可能涉及一些递归，但我在过去 8 小时里一直在尝试，但无法完成这项工作。

PS：输出也可以是一个数组。

如果需要任何进一步的信息，请告诉我。

Answer 1

好吧，我终于让它工作了。我添加了一些信息，让我的生活更轻松。

我为每一行添加了一个标志，说明该行是否没有子行（所以它是最后一个）。 “最后”行数是输出的行数，所以我从那里开始。

此外，我还为每一行添加了一个“key”，即“id_1-id_2”。

这里是完整的代码：

$Adata = Array(
    "1-1" =>    Array("id_1" => 1,"id_2" => 1,"name" => "test1",    "type" => "A","ref_1" => 0,"ref_2" => 0,"last" => false),
    "1-2" =>    Array("id_1" => 1,"id_2" => 2,"name" => "test2",    "type" => "B","ref_1" => 1,"ref_2" => 1,"last" => false),
    "1-3" =>    Array("id_1" => 1,"id_2" => 3,"name" => "test3",    "type" => "B","ref_1" => 1,"ref_2" => 1,"last" => false),
    "2-1" =>    Array("id_1" => 2,"id_2" => 1,"name" => "test4",    "type" => "B","ref_1" => 1,"ref_2" => 1,"last" => false),
    "2-3" =>    Array("id_1" => 2,"id_2" => 3,"name" => "test5",    "type" => "C","ref_1" => 1,"ref_2" => 2,"last" => true),
    "2-15" =>   Array("id_1" => 2,"id_2" => 15,"name" => "test6",   "type" => "C","ref_1" => 1,"ref_2" => 3,"last" => true),
    "5-22" =>   Array("id_1" => 5,"id_2" => 22,"name" => "test7",   "type" => "B","ref_1" => 4,"ref_2" => 9,"last" => true),
    "4-9" =>    Array("id_1" => 4,"id_2" => 9,"name" => "test8",    "type" => "C","ref_1" => 1,"ref_2" => 1,"last" => false),
    "1-7" =>    Array("id_1" => 1,"id_2" => 7,"name" => "test9",    "type" => "C","ref_1" => 2,"ref_2" => 1,"last" => true),
    "5-20" =>   Array("id_1" => 5,"id_2" => 20,"name" => "test10",  "type" => "B","ref_1" => 4,"ref_2" => 9,"last" => true),
    "5-5" =>    Array("id_1" => 5,"id_2" => 5,"name" => "test11",   "type" => "B","ref_1" => 4,"ref_2" => 9,"last" => true),
    "5-4" =>    Array("id_1" => 5,"id_2" => 4,"name" => "test12",   "type" => "B","ref_1" => 1,"ref_2" => 1,"last" => true),
);

$Atable = Array();
foreach ($Adata as $key=>$data){
    if (!$data['last']){
        continue;
    }
    $output = Array('A'=>'','B'=>'','C'=>'');
    echo "SAVE {$data['type']} - {$data['name']}<br>";
    $output[$data['type']] = $data['name'];

    $referer = $data;
    do {
        $referer = $Adata["{$referer['ref_1']}-{$referer['ref_2']}"];
        $output[$referer['type']] = $referer['name'];
    } while (!empty($referer['ref_1']) && !empty($referer['ref_2']));

    $Atable[] = $output;
}

foreach ($Atable as $row){
    echo "{$row['A']}-{$row['B']}-{$row['C']}<br>";
}