在 ES6 深度嵌套的 javascript 个对象数组中查找值

Question

在一个对象数组中，我需要找到一个 value——其中 key 是 activity：但是 activity key 可能很深像这样嵌套在数组中：

const activityItems = [
    {
        name: 'Sunday',
        items: [
            {
                name: 'Gym',
                activity: 'weights',
            },
        ],
    },
    {
        name: 'Monday',
        items: [
            {
                name: 'Track',
                activity: 'race',
            },
            {
                name: 'Work',
                activity: 'meeting',
            },
            {
                name: 'Swim',
                items: [
                    {
                        name: 'Beach',
                        activity: 'scuba diving',
                    },
                    {
                        name: 'Pool',
                        activity: 'back stroke',
                    },
                ],
            },
        ],    
    },
    {} ...
    {} ...
];

所以我写了一个递归算法来查找某个activity是否在数组中：

let match = false;
const findMatchRecursion = (activity, activityItems) => {
    for (let i = 0; i < activityItems.length; i += 1) {
        if (activityItems[i].activity === activity) {
            match = true;
            break;
        }

        if (activityItems[i].items) {
            findMatchRecursion(activity, activityItems[i].items);
        }
    }

    return match;
};

是否有 ES6 方法来确定 activity 是否存在于这样的数组中？

我试过这样的事情：

const findMatch(activity, activityItems) {
    let obj = activityItems.find(o => o.items.activity === activity);
    return obj;
}

但这不适用于深度嵌套的活动。

谢谢

Answer 1

您可以使用 some() 方法和递归来查找 activity 是否存在于任何级别，结果是 return true/false。

const activityItems = [{"name":"Sunday","items":[{"name":"Gym","activity":"weights"}]},{"name":"Monday","items":[{"name":"Track","activity":"race"},{"name":"Work","activity":"meeting"},{"name":"Swim","items":[{"name":"Beach","activity":"scuba diving"},{"name":"Pool","activity":"back stroke"}]}]}]

let findDeep = function(data, activity) {
  return data.some(function(e) {
    if(e.activity == activity) return true;
    else if(e.items) return findDeep(e.items, activity)
  })
}

console.log(findDeep(activityItems, 'scuba diving'))

Answer 2

首先，您的函数可以通过递归调用找到匹配项后停止来改进。此外，您既在外部声明 match，又在外部声明 return。可能比 return.

更好

const findMatchRecursion = (activity, activityItems) => {
    for (let i = 0; i < activityItems.length; i += 1) {
        if (activityItems[i].activity === activity) {
            return true;
        }

        if (activityItems[i].items && findMatchRecursion(activity, activityItems[i].items) {
            return true;
        }
    }

    return false;
};

没有内置深度搜索，但您可以根据需要将 .find 与命名函数一起使用。

var result = !!activityItems.find(function fn(item) {
  return item.activity === "Gym" || (item.items && item.items.find(fn));
});

Answer 3

虽然不如递归算法优雅，但您可以 JSON.stringify() 数组，它给出了这个：

[{"name":"Sunday","items":[{"name":"Gym","activity":"weights"}]},{"name":"Monday","items":[{"name":"Track","activity":"race"},{"name":"Work","activity":"meeting"},{"name":"Swim","items":[{"name":"Beach","activity":"scuba diving"},{"name":"Pool","activity":"back stroke"}]}]}]

然后您可以使用 template literal 搜索模式：

`"activity":"${activity}"`

完整功能：

findMatch = (activity, activityItems) =>
  JSON.stringify(activityItems).includes(`"activity":"${activity}"`);

const activityItems = [{
    name: 'Sunday',
    items: [{
      name: 'Gym',
      activity: 'weights',
    }, ],
  },
  {
    name: 'Monday',
    items: [{
        name: 'Track',
        activity: 'race',
      },
      {
        name: 'Work',
        activity: 'meeting',
      },
      {
        name: 'Swim',
        items: [{
            name: 'Beach',
            activity: 'scuba diving',
          },
          {
            name: 'Pool',
            activity: 'back stroke',
          },
        ],
      },
    ],
  }
];

findMatch = (activity, activityItems) =>
  JSON.stringify(activityItems).includes(`"activity":"${activity}"`);

console.log(findMatch('scuba diving', activityItems)); //true
console.log(findMatch('dumpster diving', activityItems)); //false

Answer 4

我们现在使用 object-scan 来处理像这样的简单数据处理任务。一旦您全神贯注于如何使用它，它真的很棒。以下是如何回答您的问题

// const objectScan = require('object-scan');

const find = (activity, input) => objectScan(['**'], {
  abort: true,
  rtn: 'value',
  filterFn: ({ value }) => value.activity === activity
})(input);

const activityItems = [{"name":"Sunday","items":[{"name":"Gym","activity":"weights"}]},{"name":"Monday","items":[{"name":"Track","activity":"race"},{"name":"Work","activity":"meeting"},{"name":"Swim","items":[{"name":"Beach","activity":"scuba diving"},{"name":"Pool","activity":"back stroke"}]}]}]

console.log(find('scuba diving', activityItems));
// => { name: 'Beach', activity: 'scuba diving' }

.as-console-wrapper {max-height: 100% !important; top: 0}

<script src="https://bundle.run/object-scan@13.8.0"></script>

免责声明：我是object-scan

的作者

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