PHP如何简化复杂的大数组

嗨，我想让复杂数组在视图中简单使用

然而，我有点陷入了一个循环

请看我对数组的误解

我希望这个数组简单

'collection' => [
    (int) 0 => [
        'type' => 'col',
        'name' => '2016 F/W',
        'url' => 'blablabla'
    ],
    (int) 1 => [
        'type' => 'col',
        'name' => '2015 F/W',
        'url' => 'blablabla'
    ],
    (int) 2 => [
        'type' => 'project',
        'name' => '2015 F/W',
        'url' => 'blablabla'
    ]
]

我想要这样的数组

[
(int) 0 => [
    'type' => 'col',
    'type2' => [
        (int) 0 => [
            'name' => '2016 F/W',
            'image' => [
                (int) 0 => [
                    'url' => null
                ],
                (int) 1 => [
                    'url' => null
                ]
            ]
        ],
        (int) 1 => [
            'name' => '2015 F/W',
            'image' => [
                (int) 0 => [
                    'url' => null
                ]
            ]
        ]
    ]
],
(int) 1 => [
    'type' => 'project',
    'type2' => [
        (int) 0 => [
            'name' => '2015 F/W',
            'image' => [
                (int) 0 => [
                    'url' => null
                ]
            ]
        ],
    ]
]

---

请帮忙！！我快疯了

我得说结果数组看起来不是很有用。 由于没有太多关于这项任务目标的信息，我只提供我对如何完成这项任务的看法

    $arr = array(
        'collection' => array(
            array(
                'type' => 'col',
                'name' => '2016 fw',
                'url' => 'blabla1'
            ),
            array(
                'type' => 'col',
                'name' => '2015 F/W',
                'url' => 'blabla2'
            ),
            array(
                'type' => 'project',
                'name' => '2015 F/W',
                'url' => 'blabla_project'
            ),

        ),
    );

    // $types array will hold a list of all types. type in this 
    // example is 'col' and 'project'. 
    // we will use $types to find if type X was already added to the 
    // result array.
    $types = array();

    // Every type X will be in the result array in key K, So $typesMap
    // will map between the type to the key in the result array.
    // Example: type X is located in the result array under key K
    // So I can find the type X data by looking at the 
    // $resultArr[K]
    $typesMap = array();
    $typeIndx = 0;

    // This is the result array.
    $resultArr = array();

    // run over all collections
    foreach($arr['collection'] as $collection){

        // get the current type. e.g 'col'
        $type = $collection['type'];

        // Search the $types array for 'col' to see if it was defined
        // already.
        if(!in_array($type, $types)) {
            // if it wasn't defined then define it now:

            // add it to the $types array so that we can find it on
            // the next run.
            $types[] = $type;

            // define the key (K) that it will be in the $resultArr
            $typesMap[$type] = $typeIndx;

            // Create the 'col' key in the result array.
            $resultArr[$typeIndx] = array('type' => $type, 'type2' => array());
            $typeIndx++;
        }

        // get the key for type $type.
        // remember? I can use the $typesMap to find the type's index
        // in the result array.
        $currentIndx = $typesMap[$type];

        // add the extra data to the 'type2' (not a very good name for
        // this)
        $resultArr[$currentIndx]['type2'][] = array(
            'name' => $collection['name'],
            'image' => $collection['url']
        );

    }

---

我真的不理解结果数组的url部分。

原点是第一个，结果是最后一个。你在做什么循环以使其看起来像这样，你必须提供一些代码，因为人们不能只知道你做了什么。如果第二个是结果，而不是如何从单个字符串创建多个图像url？哦，对不起，我不是这个意思。它不应该是多个urlsThanks，它帮助了我很多，但我解释得很糟糕。这很好，但我在阅读你的代码时遇到了麻烦。我一点也不懂你的代码。有什么高级关联数组练习适合我吗？理解你的代码？谢谢你的评论。你应该试着调试它。xdebug？谢谢！！！那些评论太有用了！！