JavaScript:深层检查对象具有相同的密钥

Asi*_*K T 5 javascript key javascript-objects

问题类似于:如何检查两个对象是否具有相同的属性名称集?但只有一个区别

我想查一下:

var objOne = {"a":"one","b":"two","c":{"f":"three_one"}};
var objTwo = {"a":"four","b":"five","c":{"f":"six_one"}};
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在所有级别都有相同的密钥集?

例如deepCheckObjKeys(objOne, objTwo)将返回true那里deepCheckObjKeys(objOne, objThree)回报false,如果:

var objThree = {"a":"four","b":"five","c":{"g":"six_one"}};
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因为objThree.a.c.fundefinedobjThree.

像这样的函数:

'使用严格';

function objectsHaveSameKeys() {
   for (var _len = arguments.length, objects = Array(_len), _key = 0; _key < _len; _key++) {
      objects[_key] = arguments[_key];
   }

   var allKeys = objects.reduce(function (keys, object) {
      return keys.concat(Object.keys(object));
   }, []);
   var union = new Set(allKeys);
   return objects.every(function (object) {
      return union.size === Object.keys(object).length;
   });
}
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只检查第一级.

PS:objectsHaveSameKeys()ES6等效:

function objectsHaveSameKeys(...objects):boolean {
   const allKeys = objects.reduce((keys, object) => keys.concat(Object.keys(object)), []);
   const union = new Set(allKeys);
   return objects.every(object => union.size === Object.keys(object).length);
}
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T.J*_*der 8

我只是做一个递归检查,如果属性的值是一个对象; 看评论:

const deepSameKeys = (o1, o2) => {
  // Get the keys of each object
  const o1keys = Object.keys(o1).sort();
  const o2keys = Object.keys(o2).sort();
  // Make sure they match
  // If you don't want a string check, you could do
  // if (o1keys.length !== o2keys.length || !o1keys.every((key, index) => o2keys[index] === key)) {
  if (o1keys.join() !== o2keys.join()) {
    // This level doesn't have the same keys
    return false;
  }
  // Check any objects
  return o1keys.every(key => {
    const v1 = o1[key];
    const v2 = o2[key];
    if (v1 === null) {
      return v2 === null;
    }
    const t1 = typeof v1;
    const t2 = typeof v2;
    if (t1 !== t2) {
      return false;
    }
    return t1 === "object" ? deepSameKeys(v1, v2) : true;
  });
};
var objOne = {"a":"one","b":"two","c":{"f":"three_one"}};
var objTwo = {"a":"four","b":"five","c":{"f":"six_one"}};
var objThree = {"a":"four","b":"five","c":{"g":"six_one"}};

console.log("objOne, objTwo: " + deepSameKeys(objOne, objTwo));
console.log("objTwo, objThree: " + deepSameKeys(objTwo, objThree));
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  • @ThirueswaranRajagopalan:好吧,我和其他许多人.但是,是的,它几乎是一种风格的东西,虽然*可能*有助于编译器知道该变量不会改变而不必进行必要的静态分析来证明它... (3认同)
  • `deepSameKeys({ a: null }, { a: 17 })` 和 `deepSameKeys({ a: '1' }, { a: 1 })` 都返回 `false` (2认同)