合并两个具有不同键但相同值的无序对象？

Question

在 javascript 中合并 obj1 和 obj2 等对象以获取 obj3。

obj1 = [{fruit: 'watermelon', sweetness: 3},{fruit: 'banana', sweetness: 4},{fruit: 'apple', sweetness: 5}]
obj2 = [{fruit_name: 'apple', color: 'red'},{fruit_name: 'banana', color:'yellow'},{fruit_name: 'watermelon', color:'green'}]

obj3 = [{fruit_name: 'apple', color: 'red', sweetness: 5},{fruit_name: 'banana', color:'yellow', sweetness: 4},{fruit_name: 'watermelon', color:'green', sweetness: 3}]

Answer 1

不是一般的解决方案，但足以满足您的情况：

var obj1 = [{fruit: 'watermelon', sweetness: 3},{fruit: 'banana', sweetness: 4},{fruit: 'apple', sweetness: 5}]
var obj2 = [{fruit_name: 'apple', color: 'red'},{fruit_name: 'banana', color:'yellow'},{fruit_name: 'watermelon', color:'green'}]

var obj3 = obj2.map(o => Object.assign({}, o, {'sweetness': obj1.find(p => p.fruit === o.fruit_name).sweetness}))

console.log(obj3)

Answer 2

您可以首先通过使用#forEach 合并相似对象来创建hashtable，然后使用@987654322提取所需的数组@函数 - 见下面的演示：

var obj1 = [{fruit: 'watermelon', sweetness: 3},{fruit: 'banana', sweetness: 4},{fruit: 'apple', sweetness: 5}],
  obj2 = [{fruit_name: 'apple', color: 'red'},{fruit_name: 'banana', color:'yellow'},{fruit_name: 'watermelon', color:'green'}], hash = {};

// function to create a hashtable
function classify(e) {
   if(hash[e.fruit] || hash[e.fruit_name]) {
     Object.keys(e).forEach(function(c){
        hash[e.fruit || e.fruit_name][c] = e[c];
     });
   } else {
     hash[e.fruit_name || e.fruit] = e;
   }
}

// add to hash
obj1.forEach(classify);
obj2.forEach(classify);

// extract the result
var obj3 = Object.keys(hash).map(function(e){
  delete hash[e]['fruit'];
  return hash[e];
});

console.log(obj3);

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

Answer 3

您的数据结构不正确。不能将 "'fruit: 'watermelon'" (key, value pair) 保存在数组中。

会报错：Uncaught SyntaxError: Unexpected token :

我假设你想要做的是：

obj1 = [{fruit: 'watermelon', sweetness: 3},{fruit: 'banana', sweetness: 4},{fruit: 'apple', sweetness: 5}]
  obj2 = [{fruit_name: 'apple', color: 'red'},{fruit_name: 'banana', color:'yellow'},{fruit_name: 'watermelon', color:'green'}]
  obj3 = [];
  
for (i = 0; i < obj1.length; i++) {
  for (j = 0; j < obj2.length; j++) {
    if (obj1[i].fruit === obj2[j].fruit_name) {
      var temp = {
        fruit_name: obj2[j].fruit_name,
        color: obj2[j].color,
        sweetness: obj1[i].sweetness
      }
      obj3.push(temp);
    }
  }
}

console.log(obj3);

Answer 4

将两个具有共享键的对象数组合并在一起的通用函数相当简单。最简单的方法是使用基于关联数组的映射，如下所示。请注意，您可以使用此例程来解决任何类似类型的问题，但它绝对适用于您的数据——请参阅最后链接的 JSFiddle。

（ETA：共享密钥仅添加一次，名称为 key1；如果您希望第二个密钥在输出中结束，只需将参数对交换给函数。）

obj1 = [{fruit: 'watermelon', sweetness: 3},{fruit: 'banana', sweetness: 4},{fruit: 'apple', sweetness: 5}];
obj2 = [{fruit_name: 'apple', color: 'red'},{fruit_name: 'banana', color:'yellow'},{fruit_name: 'watermelon', color:'green'}];

function mergeObjectArrays(array1, key1, array2, key2) {
  var map = []; // an associative array/hashtable
  var arrayValue, mapValue, propertyNames, propertyName, propertyValue;

  // 1. Loop over one array, populating the map by each object's specified key
    for(var x = 0; x < array1.length; x++) {
    array1Value = array1[x];
    map[array1Value[key1]] = array1Value;
    map.push(array1Value);
  }

  // 2. Loop over the other array, matching on the provided keys
  for(var x = 0; x < array2.length; x++) {
    arrayValue = array2[x];
    mapValue = map[arrayValue[key2]];

    if (typeof(mapValue) != 'undefined') { // add all missing non-keyed properties to the mapped/merged object
      propertyNames = Object.keys(arrayValue);
      for (var y = 0; y < propertyNames.length; y++) {
          propertyName = propertyNames[y];
          if (propertyName != key1 && propertyName != key2) { // .. as that shared value is already added
            propertyValue = arrayValue[propertyName];
            mapValue[propertyName] = propertyValue;
          }
      }      
    }
    else { // it's missing from the mapping, so at least add it though it will be missing array1 data
      map[arrayValue[key2]] = arrayValue;
      map.push(arrayValue);
    }
  }

  return map;
}

var mergedArrays = mergeObjectArrays(obj1, 'fruit', obj2, 'fruit_name');

Here's a working sample.

Answer 5

let obj1 = [{fruit: 'watermelon', sweetness: 3},{fruit: 'banana', sweetness: 4},{fruit: 'apple', sweetness: 5}];
let obj2 = [{fruit_name: 'apple', color: 'red'},{fruit_name: 'banana', color:'yellow'},{fruit_name: 'watermelon', color:'green'}];

function regulate(a) {
  return a.map(v => {
    if (v.fruit) {
        v.fruit_name = v.fruit;
        delete v.fruit;
     }
     return v;
  });
}


let a = [].concat(regulate(obj1), regulate(obj2));

let merged = [];
a.reduce((m, v) => {
  let f = m.filter(v2 => v2.fruit_name === v.fruit_name);
  if (f.length > 0) {
    Object.assign(f[0], v);
  } else {
    m.push(v);
  }
  return m;
}, merged);

console.log(merged);