如何访问数组元素中的子数组？

Question

这是一个 api 响应。我遇到的问题是这里有几个机组人员。我可以通过他们的基本索引找到任何人。我该怎么做才能只获取作者姓名，或者我应该说如何使用 Writer 作为参数访问数组并从该数组中获取他/她的姓名。

  "crew": [
        {
          "credit_id": "53f52157c3a36833fa002225",
          "department": "Writing",
          "gender": 2,
          "id": 1930,
          "job": "Writer",
          "name": "Truman Capote",
          "profile_path": "/kmhddndt7hI38aN6NzbhPcnjKSh.jpg"
        },
        {
          "credit_id": "52fe4a6ec3a368484e152e99",
          "department": "Directing",
          "gender": 1,
          "id": 96199,
          "job": "Director",
          "name": "Liz Garbus",
          "profile_path": "/1Z5a6djeDX57yyLe2ixX3VzIaLJ.jpg"
        },
        {
          "credit_id": "53f5216dc3a36833fd0021bd",
          "department": "Writing",
          "gender": 1,
          "id": 96199,
          "job": "Writer",
          "name": "Liz Garbus",
          "profile_path": "/1Z5a6djeDX57yyLe2ixX3VzIaLJ.jpg"
        },
        {
          "credit_id": "53f52180c3a36834000023b2",
          "department": "Writing",
          "gender": 0,
          "id": 1355119,
          "job": "Writer",
          "name": "Ralph Greenson",
          "profile_path": null
        }
      ]

在此先感谢您的帮助。

Answer 1

所以根据大家的建议，我尝试调整一些东西，虽然速度不快但效果很好。（PHP 代码）

 $data = json_decode($response,true);

   $n = sizeof($data['crew']);

    for ($i=0; $i < $n; $i++) { 

        if($data['crew'][$i]['job']=='Writer')
        {
            echo $data['crew'][$i]['name'];
        }
    }

Answer 2

如果您想在 php 中处理 JSON 对象，请尝试以下代码

<?php
  // JSON string
  $someJSON = '[{"name":"Jonathan Suh","gender":"male"},{"name":"William Philbin","gender":"male"},{"name":"Allison McKinnery","gender":"female"}]';

  // Convert JSON string to Array
  $someArray = json_decode($someJSON, true);
  print_r($someArray);        // Dump all data of the Array
  echo $someArray[0]["name"]; // Access Array data

  // Convert JSON string to Object
  $someObject = json_decode($someJSON);
  print_r($someObject);      // Dump all data of the Object
  echo $someObject[0]->name; // Access Object data
?>

或者如果你想在 Javascript（客户端）中处理数据

obj = [{
          "credit_id": "53f52157c3a36833fa002225",
          "department": "Writing",
          "gender": 2,
          "id": 1930,
          "job": "Writer",
          "name": "Truman Capote",
          "profile_path": "/kmhddndt7hI38aN6NzbhPcnjKSh.jpg"
        },
                {
          "credit_id": "53f52157c3a36833fa002225",
          "department": "Writing",
          "gender": 2,
          "id": 1930,
          "job": "Writer",
          "name": "Truman Capote 2",
          "profile_path": "/kmhddndt7hI38aN6NzbhPcnjKSh.jpg"
        },
]

obj.forEach(function(v,i){console.log(v.name)})

Answer 3

使用 PHP，您可以使用 FilterIterator 类，如下例所示...

class CrewFilter extends FilterIterator {
    protected $department;

    public function __construct(Iterator $iterator, $department) {
        $this->department = $department;
        parent::__construct($iterator);
    }

    public function accept(): bool
    {
        $current = $this->getInnerIterator()->current();
        return $current->department === $department;
    }
}

$data = json_decode($response);
$iterator = new ArrayIterator($data['crew']);

$filter = new CrewFilter($iterator, 'Writing');
foreach ($filter as $crew) {
    // output the name
    var_dump($crew->name);
}

上面的代码是做什么的。 FilterIterator 实例采用$department 参数，该参数为您想要查看的船员设置部门。第一个参数是一个迭代器的实例。在这种情况下，这将是来自 api 的 json 解码数据。对接收到的json数据进行解码是基本的。

foreach 循环只会回馈部门Writing 的船员。

Answer 4

如果您想从浏览器端搜索，假设您的 api 响应 json 在 obj 中，这里是 javascript 代码

var results = [];
var searchField = "job";
var searchVal = "Writer";
for (var i=0 ; i < obj.crew.length ; i++)
{
    if (obj.list[i][searchField] == searchVal) {
        results.push(obj.list[i]);
    }
}

那么结果变量将包含所有作者数据 希望这是您正在寻找的解决方案