比較兩個JavaScript對象的鍵

Question

我有兩組JavaScript對象。我想比較object1和object2，然後獲取object1中所有鍵的列表，但不是object2中的所有鍵。我一直在尋找資源來幫助我，但我只是最終找到了簡單對象的比較函數。我想比較的對象有很多嵌套。我在底部包含了一個例子。

我該如何去做一個比較這兩個對象的函數？是否可以創建一個靈活的函數，如果對象要更改幷包含更多的嵌套，這也可以工作？

const object1 = { 
    "gender": "man", 
    "age": 33, 
    "origin": "USA", 
    "jobinfo": { 
     "type": "teacher", 
     "school": "Wisconsin" 
    }, 
    "children": [ 
     { 
      "name": "Daniel", 
      "age": 12, 
      "pets": [ 
       { 
        "type": "cat", 
        "name": "Willy", 
        "age": 2 
       }, 
       { 
        "type": "dog", 
        "name": "jimmie", 
        "age": 5 
       } 
      ] 
     }, 
     { 
      "name": "Martin", 
      "age": 14, 
      "pets": [ 
       { 
        "type": "bird", 
        "name": "wagner", 
        "age": 12 
       } 
      ] 
     } 
    ], 
    "hobbies": { 
     "type": "football", 
     "sponsor": { 
      "name": "Pepsi", 
      "sponsorAmount": 1000, 
      "contact": { 
       "name": "Leon", 
       "age": 59, 
       "children": [ 
        { 
         "name": "James", 
         "pets": [ 
          { 
           "type": "dog", 
           "age": 4 
          } 
         ] 
        } 
       ] 
      } 
     } 
    } 
} 

const object2 = { 
    "gender": "man", 
    "jobinfo": { 
     "type": "teacher" 
    }, 
    "children": [ 
     { 
      "name": "Daniel", 
      "age": 12, 
      "pets": [ 
       { 
        "type": "cat", 
        "name": "Willy", 
        "age": 2 
       }, 
       { 
        "type": "dog", 
        "name": "jimmie", 
        "age": 5 
       } 
      ] 
     } 
    ] 
} 

所以，我想通過比較這兩個對象要實現什麼，在這種情況下有一個包含了在object1的按鍵陣列的回報，而不是對象2。所以數組看起來像這樣。

["age", "hobbies", "type", "sponsor", "name", "sponsorAmount", "contact", "name", "age", "children", "name", "pets", "type", "age"]. 

這是我到目前爲止得到的結果。這是有點工作。但是它不會打印出年齡，因爲age是一個存在於多個嵌套對象中的屬性。 的jsfiddle：https://jsfiddle.net/rqdgojq2/

我有一個看看下面的資源：使用設置對象和自定義getAllKeyNames()遞歸函數

• https://stackoverflow.com/a/25175871/4623493
• https://stackoverflow.com/a/21584651/4623493

Answer 1

感謝您的反饋意見。 我最終解決了這個問題，並從羅馬人的回答中獲得了很多靈感。

const compareObjects = (obj1, obj2) => { 
    function getAllKeyNames(o, arr, str){ 
     Object.keys(o).forEach(function(k){ 
      if (Object.prototype.toString.call(o[k]) === "[object Object]") { 
       getAllKeyNames(o[k], arr, (str + '.' + k)); 
      } else if (Array.isArray(o[k])) { 
       o[k].forEach(function(v){ 
        getAllKeyNames(v, arr, (str + '.' + k)); 
       }); 
      } 
      arr.push(str + '.' + k); 
     }); 
    } 

    function diff(arr1, arr2) { 
     for(let i = 0; i < arr2.length; i++) { 
      arr1.splice(arr1.indexOf(arr2[i]), 1); 
     } 
     return arr1; 
    } 

    const o1Keys = []; 
    const o2Keys = []; 
    getAllKeyNames(obj1, o1Keys, ''); // get the keys from schema 
    getAllKeyNames(obj2, o2Keys, ''); // get the keys from uploaded file 

    const missingProps = diff(o1Keys, o2Keys); // calculate differences 
    for(let i = 0; i < missingProps.length; i++) { 
     missingProps[i] = missingProps[i].replace('.', ''); 
    } 
    return missingProps; 
} 

的jsfiddle這裏：https://jsfiddle.net/p9Lm8b53/

Answer 2

複雜的解決方案從指定對象獲取所有唯一鍵名稱：

var object1 = {"gender":"man","age":33,"origin":"USA","jobinfo":{"type":"teacher","school":"Wisconsin"},"children":[{"name":"Daniel","age":12,"pets":[{"type":"cat","name":"Willy","age":2},{"type":"dog","name":"jimmie","age":5}]},{"name":"Martin","age":14,"pets":[{"type":"bird","name":"wagner","age":12}]}],"hobbies":{"type":"football","sponsor":{"name":"Pepsi","sponsorAmount":1000,"contact":{"name":"Leon","age":59,"children":[{"name":"James","pets":[{"type":"dog","age":4}]}]}}}}, 
 
    object2 = {"gender":"man","age":33,"origin":"USA","jobinfo":{"type":"teacher","school":"Wisconsin"},"children":[{"name":"Daniel","age":12,"pets":[{"type":"cat","name":"Willy","age":2},{"type":"dog","name":"jimmie","age":5}]}]}; 
 
    
 
    function getAllKeyNames(o, res){ 
 
     Object.keys(o).forEach(function(k){ 
 
      if (Object.prototype.toString.call(o[k]) === "[object Object]") { 
 
       getAllKeyNames(o[k], res); 
 
      } else if (Array.isArray(o[k])) { 
 
       o[k].forEach(function(v){ 
 
        getAllKeyNames(v, res); 
 
       }); 
 
      } 
 
      res.add(k); 
 
     }); 
 
    } 
 
    
 
    var o1Keys = new Set(), o2Keys = new Set(); 
 
    getAllKeyNames(object1, o1Keys); // unique keys of object1 
 
    getAllKeyNames(object2, o2Keys); // unique keys of object2 
 
    
 
    // get a list of all the keys that are in object1, but not in object2 
 
    var diff = [...o1Keys].filter((x) => !o2Keys.has(x)); 
 
    console.log(diff); 
 

Answer 3

你可以使用一個對象計數。

function getCount(object, keys, inc) { 
 
    Object.keys(object).forEach(function (k) { 
 
     if (!Array.isArray(object)) { 
 
      keys[k] = (keys[k] || 0) + inc; 
 
      if (!keys[k]) { 
 
       delete keys[k]; 
 
      } 
 
     } 
 
     if (object[k] && typeof object[k] === 'object') { 
 
      getCount(object[k], keys, inc) 
 
     } 
 
    }); 
 
} 
 

 
var object1 = { gender: "man", age: 33, origin: "USA", jobinfo: { type: "teacher", school: "Wisconsin" }, children: [{ name: "Daniel", age: 12, pets: [{ type: "cat", name: "Willy", age: 2 }, { type: "dog", name: "jimmie", age: 5 }] }, { name: "Martin", age: 14, pets: [{ type: "bird", name: "wagner", age: 12 }] }], hobbies: { type: "football", sponsor: { name: "Pepsi", sponsorAmount: 1000, contact: { name: "Leon", age: 59, children: [{ name: "James", pets: [{ type: "dog", age: 4 }] }] } } } }, 
 
    object2 = { gender: "man", jobinfo: { type: "teacher" }, children: [{ name: "Daniel", age: 12, pets: [{ type: "cat", name: "Willy", age: 2 }, { type: "dog", name: "jimmie", age: 5 }] }] }, 
 
    count = {}; 
 

 
getCount(object1, count, 1); 
 
getCount(object2, count, -1); 
 

 
console.log(count);

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