从嵌套数组中检索数据

Question

如何从嵌套数组中检索 ID？
最初我得到这样的 json 响应，我需要从所有嵌套数组中获取所有 ID。有人可以帮我吗？我应该使用过滤器还是任何查找功能？一个例子或解释可能很棒。

{
    "id": 271,
    "name": "anything",
    "description": null,
    "entries": [
        {
            "id": "fda2afe0-dfc4-4373-9e50-8b140a46f25e",
            "name": "first occurence",
            "runs": [
                {
                    "id": 284,
                    "name": "the element from which I want to get id",
                    "description": null,
                    "created_on": 1530627823,
                    "created_by": 2
                },
                {
                    "id": 285,
                    "name": "element for id 2",
                    "created_by": 2
                },
                {
                    "id": 296,
                    "name": "element for id 3",
                    "created_on": 1530710993,
                    "created_by": 2
                }
            ]
        },
        {
            "id": "a65dd3f0-3fc1-4f93-9123-f5a05ae50703",
            "name": "second occurence",
            "runs": [
                {
                    "id": 272,
                    "name": "element for id 4",
                    "created_by": 2,
                },
                {
                    "id": 273,
                    "created_by": 2,
                },
                {
                    "id": 274,
                    "created_by": 2,
                }
            ]
        }
    ]
}

Answer 1

假设您将 json 存储在某个变量中，比如 json,

const ids = json.entries.map(entry => entry.runs.map(run => run.id));

应该给你一个嵌套的 ID 数组，看起来像

[ [ 284, 285, 296 ], [ 272, 273, 274 ] ].

如果您希望它位于单一维度列表中，即

[ 284, 285, 296, 272, 273, 274 ],

你可以使用 concat.apply.

const concatIds = [].concat.apply([], ids);

Answer 2

假设您正在寻找最深的 ID（运行个 ID），这里是一个如何完成的示例。

let response = {"id":271,"name":"anything","description":null,"entries":[{"id":"fda2afe0-dfc4-4373-9e50-8b140a46f25e","name":"first occurence","runs":[{"id":284,"name":"the element from which I want to get id","description":null,"created_on":1530627823,"created_by":2},{"id":285,"name":"element for id 2","created_by":2},{"id":296,"name":"element for id 3","created_on":1530710993,"created_by":2}]},{"id":"a65dd3f0-3fc1-4f93-9123-f5a05ae50703","name":"second occurence","runs":[{"id":272,"name":"element for id 4","created_by":2,},{"id":273,"created_by":2,},{"id":274,"created_by":2,}]}]}

function getRunIDs() {

  let entries = response['entries'] || []
  let runIDs = []

  entries.forEach(entry => {
    entry['runs'].forEach(run => {
      runIDs.push(run['id'])
    })
  })
  
  return runIDs
}

console.log({ runIDs: getRunIDs() })

Answer 3

过滤和查找功能是多余的，因为您获取的是所有 ID，而不是特定的 ID。要解决您的问题，您只需使用带有 map 的函数即可。对于此示例，您的对象将被称为 exampleObj，您可以在其中简单地执行此操作以获取所有 ID 的数组：

exampleObj.entries.map(entryObj => {
  return entryObj.runs.map(runObj=>{return runObj.id});
});

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