获取JSON解析对象中的所有键/数组键？

Question

我有这样的示例JSON

`{
  values:[{
    "name": "Base Url",
    "url": "https://kubemanagement-prod.kohls.com"
  },
  {
    "name": "Base Url newwww",
    "url": "https://batman.com"
  }]
}`

当前，当我将此特殊的JSON添加到lodash _.keys ，结果为["0", "1"] ，这基本上是第一个和第二个对象0和1的索引。

我真正想要的是检索JSON对象的所有键，包括子对象属性。 在这种情况下， ["values","0", "1","name","url"]

有谁知道lodash方法或将复杂JSON对象中给定的所有键检索到n级的机制？

语言：角度+打字稿

Answer 1

此函数使用_.keys()和_.flatMap()以及_.union()来递归地从对象树中获取键，以组合键列表，并仅获取唯一值：

 const getAllKeys = obj => _.union( _.keys(obj), _.flatMap(obj, o => _.isObject(o) ? getAllKeys(o) : []) ) const arr = {"values": [{"name":"Base Url","url":"https://kubemanagement-prod.kohls.com"},{"name":"Base Url newwww","url":"https://batman.com"}]} const result = getAllKeys(arr) console.log(result) 

 <script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.js"></script> 

同样，没有lodash的想法是，使用Object.values()和Array.flatMap()迭代当前对象（或数组），并使用Array.concat()和Set来使键唯一：

 const getAllKeys = obj => [...new Set([].concat( Object.keys(obj), Object.values(obj).flatMap(o => typeof o === 'object' ? getAllKeys(o) : []) ))] const arr = {"values": [{"name":"Base Url","url":"https://kubemanagement-prod.kohls.com"},{"name":"Base Url newwww","url":"https://batman.com"}]} const result = getAllKeys(arr) console.log(result) 

 <script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.js"></script> 

没有数组索引：

 const getAllKeys = obj => _.union( _.isArray(obj) ? [] : _.keys(obj), _.flatMap(obj, o => _.isObject(o) ? getAllKeys(o) : []) ) const arr = {"values": [{"name":"Base Url","url":"https://kubemanagement-prod.kohls.com"},{"name":"Base Url newwww","url":"https://batman.com"}]} const result = getAllKeys(arr) console.log(result) 

 <script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.js"></script> 

Answer 2

您不需要lodash， Object.keys就足够了。 您只需编写一个递归函数，写一个Set ，完成后就可以转换为数组：

 const array = [ { "name": "Base Url", "url": "https://kubemanagement-prod.kohls.com" }, { "name": "Base Url newwww", "url": "https://batman.com" } ]; function addAll(set, entries) { for (const entry of entries) { set.add(entry); } return set; } function allKeys(obj/*: object|array*/, keys/*: Set<string>*/ = new Set()) { addAll(keys, Object.keys(obj)); for (const entry of Object.values(obj)) { if (entry && typeof entry === "object") { allKeys(entry, keys); } } return keys; } console.log([...allKeys(array)]); 

或在您的编辑中使用结构：

 const array = { values:[{ "name": "Base Url", "url": "https://kubemanagement-prod.kohls.com" }, { "name": "Base Url newwww", "url": "https://batman.com" }] } function addAll(set, entries) { for (const entry of entries) { set.add(entry); } return set; } function allKeys(obj/*: object|array*/, keys/*: Set<string>*/ = new Set()) { addAll(keys, Object.keys(obj)); for (const entry of Object.values(obj)) { if (entry && typeof entry === "object") { allKeys(entry, keys); } } return keys; } console.log([...allKeys(array)]); 

Answer 3

怎么样

 const arr = {"values": [{"name":"Base Url","url":"https://kubemanagement-prod.kohls.com"},{"name":"Base Url newwww","url":"https://batman.com"}]} let result1 = Object.keys(arr) // get the object keys which is "values" let result2 = Object.keys(arr[result1]); // get the keys of the object name "0,1" here let result3 = Object.keys(arr[result1][0]); // get property names of one object "name and url" here let resutltant = result1.concat(result2, result3) // "merge array" console.log(resutltant) 

假设您的对象属性名称将保持不变

Answer 4

如果使用打字稿，为什么不使用Object.entries或Object.values呢？

您需要将以下行添加到tsconfig中：

lib：[......“ es2018”，]