Question

让我们说我有一个像这样的数组

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,),
);

两个＆＃34;主键＆＃34;这是＆＃34; id_1＆＃34;和＆＃34; id_2＆＃34;，两者的组合不能重复。每一行都可以使用＆＃34; ref_1＆＃34;和＆＃34; ref_2＆＃34;，引用的行将包含＆＃34; id_1＆＃34;和＆＃34; id_2＆＃34;形成引用者（父亲）。

无法引用A类行。 B型和B型C可以由A引用。另外，B可以从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.

我试图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      |               |

每当一行引用2个（或更多）其他行时，数据就会被父数据分割，并为每个引用输出一行。如果类型A引用2类型B，并且每个类型B引用其他2类型C，则输出表将具有4行。

有三种以上的类型，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。我知道它可能会涉及一些递归，但我在过去的8个小时内一直在努力，但是无法完成这项工作。

PS：输出也可以是数组。

如果需要进一步的信息，请告诉我。

Answer 1

好吧，最后我开始工作了。我添加了一些信息，让我的生活更轻松。

我在每一行都添加了一个标志，表示该行是否没有子节点（所以它是最后一行）。 “last”行的数量是输出的行数，所以我从那里开始。

另外，我已经为每一行添加了一个“密钥”，即“id_1-id_2”。

以下是完整代码：

$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>";
}

递归函数，输出引用表

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