JavaScript - 合并两个对象数组并根据属性值进行重复数据删除

Question

我想用另一个数组中的对象更新（替换）我的数组中的对象。每个对象具有相同的结构。例如

var origArr = [
  {name: 'Trump', isRunning: true},
  {name: 'Cruz', isRunning: true},
  {name: 'Kasich', isRunning: true}
];
var updatingArr = [
  {name: 'Cruz', isRunning: false},
  {name: 'Kasich', isRunning: false}
];
// desired result:
NEWArr = [
  {name: 'Trump', isRunning: true},
  {name: 'Cruz', isRunning: false},
  {name: 'Kasich', isRunning: false}
];

我已经尝试过concat（）和Underscore的_.uniq函数，但它总是转储较新的对象并返回原来的数组。

有没有办法用origArr中的对象覆盖（替换）updatingArr - 匹配name属性？

Answer 1

使用双循环和拼接，你可以这样做：

for(var i = 0, l = origArr.length; i < l; i++) {
    for(var j = 0, ll = updatingArr.length; j < ll; j++) {
        if(origArr[i].name === updatingArr[j].name) {
            origArr.splice(i, 1, updatingArr[j]);
            break;
        }
    }
}

例子here

Answer 2

尝试使用ES-6设置数据结构的这种方法：const result = [...new Set([...origArr, ...updatingArr])]

Answer 3

我来到这里正是在寻找这个，看到@Gruff Bunny的技巧，并想知道'lodash'是否也不是一个优越的选择，即使是'下划线'？

瞧，看哪：

let result = _.unionBy(updatingArr, origArr, 'name');

Answer 4

你可以将Array#map与Array#reduce结合使用

var origArr = [{ name: 'Trump', isRunning: true }, { name: 'Cruz', isRunning: true }, { name: 'Kasich', isRunning: true }],
    updatingArr = [{ name: 'Cruz', isRunning: false }, { name: 'Kasich', isRunning: false }],
    NEWArr = origArr.map(function (a) {
        return this[a.name] || a;
    }, updatingArr.reduce(function (r, a) {
        r[a.name] = a;
        return r;
    }, Object.create(null)));

document.write('<pre>' + JSON.stringify(NEWArr, 0, 4) + '</pre>');

Answer 5

您可以使用通过名称提供索引的哈希值，并使用Object.assign进行更新。

var hash = origArr.reduce(function(hash, obj, index) {
  hash[obj.name] = index;
  return hash;
}, Object.create(null));
for(var obj of updatingArr) {
  Object.assign(origArr[hash[obj.name]], obj);
}

Answer 6

你可以尝试一下。

var origArr = [
  {name: 'Trump', isRunning: true},
  {name: 'Cruz', isRunning: true},
  {name: 'Kasich', isRunning: true}
];
var updatingArr = [
  {name: 'Cruz', isRunning: false},
  {name: 'Kasich', isRunning: false}
];

var origLength = origArr.length;
var updatingLength = updatingArr.length;

//Traverse the original array and replace only if the second array also has the same value
for(i = origLength-1; i >= 0; i--) {
    for(j = updatingLength -1; j >= 0; j--) {
    if(origArr[i].name === updatingArr[j].name) {
        origArr[i] = updatingArr[j];
    }
  }
}

console.log(origArr);

Answer 7

这是使用下划线的解决方案：

var result = _.map(origArr, function(orig){
    return _.extend(orig, _.findWhere(updatingArr, {name: orig.name}));
});

Answer 8

这将满足您的需求：

var origArr = [
  {name: 'Trump', isRunning: true},
  {name: 'Cruz', isRunning: true},
  {name: 'Kasich', isRunning: true}
];

var updatingArr = [
  {name: 'Cruz', isRunning: false},
  {name: 'Kasich', isRunning: false}
];

for (var i = 0; i < updatingArr.length; ++i) {
  var updateItem = updatingArr[i];
  for (var j = 0; j < origArr.length; ++j) {
    var origItem = origArr[j];
    if (origItem.name == updateItem.name) {
      origItem.isRunning = updateItem.isRunning;
      break;    
    }
  }
}

document.write('<pre>' + JSON.stringify(origArr, 0, 4) + '</pre>');

Answer 9

const origArr = [
  {name: 'Trump', isRunning: true},
  {name: 'Cruz', isRunning: true},
  {name: 'Kasich', isRunning: true}
];

const updatingArr = [
  {name: 'Cruz', isRunning: false},
  {name: 'Kasich', isRunning: false}
];

let hash = {};

for(let i of origArr.concat(updatingArr)) {
  if(!hash[i]) {
    hash[i.name] = i;
  }
}

let newArr = [];

for(let i in hash) {
  newArr.push(hash[i])
}

console.log(newArr);

Answer 10

此版本允许您定义将对象定义为重复的selector。

• forEach迭代新数据
• 如果两个选择器相等，则findIndex返回索引>= 0。如果没有，则返回-1
• 如果有重复，我们使用slice来替换原来的new。
• 如果没有重复，我们将其推入原始数组。

const origArr = [
  {name: 'Trump', isRunning: true},
  {name: 'Cruz', isRunning: true},
  {name: 'Kasich', isRunning: true}
];

const updatingArr = [
  {name: 'Cruz', isRunning: false},
  {name: 'Kasich', isRunning: false}
];

const mergeArrayOfObjects = (original, newdata, selector = 'key') => {
	newdata.forEach(dat => {
		const foundIndex = original.findIndex(ori => ori[selector] == dat[selector]);
		if (foundIndex >= 0) original.splice(foundIndex, 1, dat);
        else original.push(dat);
	});

	return original;
};

const result = mergeArrayOfObjects(origArr, updatingArr, "name")
console.log('RESULT -->', result)