获取一个数组，其中包含数组中对象的所有键

I have stuck at a point where I want to recursively traverse all the objects in an array and get the keys of those objects in a array data structure. I know how to loop over and get the keys of the object. But the problem here is I need that recursively for flexible objects. By flexible I mean it can have any level of nested properties.

我一直想要以递归方式遍历数组中的所有对象,并在数组数据结构中获取这些对象的键。我知道如何循环并获取对象的键。但这里的问题是我需要递归地为灵活的对象。灵活的我的意思是它可以有任何级别的嵌套属性。

So, what I have is an array like this:

所以,我有一个这样的数组:

let record = [{
    "province": "string",
    "city": "string",
    "type": "alternative_address",
    "address_line1": "string",
    "post_code": "5858"
  },
  {
    "province": "string",
    "city": "string",
    "type": "alternative_address",
    "post_code": "5858",
    "embedeer": {
      "veryEmbedded": {
        "veryveryEmbeded": 'yes'
      }
    }
  }
];

And with some computation I am expecting an output like:

通过一些计算,我期待输出如下:

['province','city','type','address_line1','post_code','embedeer', 'embedeer.veryEmbedded', 'embedeer.veryEmbedded.veryveryEmbeded'];

For the effort I tried on this, I used the reduce() operation on array but I am unable to get that.

对于我尝试过的努力,我在数组上使用了reduce()操作但是我无法得到它。

4 个解决方案

#1

You need to write a recursive function that takes 2 inputs

您需要编写一个需要2个输入的递归函数

object
prefix (undefined for first level keys)

前缀(未定义为第一级键)

let record = [{"province":"string","city":"string","type":"alternative_address","address_line1":"string","post_code":"5858"},{"province":"string","city":"string","type":"alternative_address","post_code":"5858","embedeer":{"veryEmbedded":{"veryveryEmbeded":"yes"}}}];

function addKeysToSet(o, p) {
  Object.keys(o).forEach(k => {
    let key = p ? p + "." + k : k; // Create the key hierarchy
    keys.add(key); // Add key to the set 
    // If the value is an object, call the function recursively 
    if(typeof o[k] === 'object') {
      addKeysToSet(o[k], key);
    }
  });
}

let keys = new Set(); // Create set of unique keys
// For each object in array, call function that adds keys to the set
record.forEach(o => addKeysToSet(o));
let result = Array.from(keys); // Create array from set
console.log(result); // logs result

#2

You could take an iterative and recursive approach and take the power of a Set for getting unique values.

您可以采用迭代和递归方法,并获取Set的强大功能以获取唯一值。

function iter(object, keys) {
    return Object
        .entries(object)
        .reduce((r, [k, v]) => r.concat(keys.concat(k).join('.'), v && typeof v === 'object'
            ? iter(v, keys.concat(k))
            : []
        ), []);        
}

var record = [{ province: "string", city: "string", type: "alternative_address", address_line1: "string", post_code: "5858" }, { province: "string", city: "string", type: "alternative_address", post_code: "5858", embedeer: { veryEmbedded: { veryveryEmbeded: 'yes' } } }],
    keys = [...record.reduce((s, o) => iter(o, []).reduce((t, v) => t.add(v), s), new Set)];

console.log(keys);

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

#3

You can Flatten the object and then get the keys.

您可以展平对象然后获取密钥。

// form https://gist.github.com/penguinboy/762197

let record = [{"province":"string","city":"string","type":"alternative_address","address_line1":"string","post_code":"5858"},{"province":"string","city":"string","type":"alternative_address","post_code":"5858","embedeer":{"veryEmbedded":{"veryveryEmbeded":"yes"}}}];

var flattenObject = function(a) {
  var b = {};
  for (var c in a)
    if (a.hasOwnProperty(c))
      if ("object" == typeof a[c]) {
        var d = flattenObject(a[c]);
        for (var e in d) d.hasOwnProperty(e) && (b[c + "." + e] = d[e]);
      } else b[c] = a[c];
  return b;
};

console.log(flattenObject(record) )

/*
 It is also taking care of index numbers of the array. ("0.province" instead of "province" If multiple entries are passed)  
 
 */
console.info( "All keys", Object.keys(flattenObject(record) ) )

// Simple
console.info( "keys", Object.keys(flattenObject(record[1]) ) )

#4

var record1 = [{"province": "string","city": "string","type": "alternative_address","address_line1": "string","post_code": "5858" },
  { "province": "string","city": "string",
    "type": "alternative_address",
    "post_code": "5858",
    "embedeer": {
      "veryEmbedded": {
        "veryveryEmbeded": 'yes'
      }
    }
  }
];

var output = [];
function getAllKeys(obj,precedor="") {
  var temp = Object.entries(obj);
  temp.forEach((el) => 
               typeof el[1] == "object" ? ( output.push(el[0]),getAllKeys(el[1],precedor==""? el[0]: precedor+"."+el[0])): output.push(precedor==""? el[0]: precedor+"."+el[0]));
}
 record1.forEach((el,i) => getAllKeys(el,""));
//To avoid duplicate entries convert array to object.
console.log(...(new Set(output)));

#1