我有2个不同的嵌套对象,我需要知道它们是否在其中一个嵌套属性中有不同。

var a = {};
var b = {};

a.prop1 = 2;
a.prop2 = { prop3: 2 };

b.prop1 = 2;
b.prop2 = { prop3: 3 };

对象可以更复杂,有更多嵌套的属性。但这是一个很好的例子。我可以选择使用递归函数或lodash的东西…


当前回答

作为对亚当·博杜赫的回答的补充,这个问题考虑到了性质的差异

const differenceOfKeys = (...objects) =>
  _.difference(...objects.map(obj => Object.keys(obj)));
const differenceObj = (a, b) => 
  _.reduce(a, (result, value, key) => (
    _.isEqual(value, b[key]) ? result : [...result, key]
  ), differenceOfKeys(b, a));

其他回答

一个简单而优雅的解决方案是使用_。isEqual,它执行深度比较:

Var a = {}; Var b = {}; A.prop1 = 2; A.prop2 = {prop3: 2}; B.prop1 = 2; B.prop2 = {prop3: 3}; console.log(_。isEqual (a, b));//如果不同则返回false < script src = " https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.4/lodash.min.js " > < /脚本>

然而,这个解决方案并没有显示哪个属性是不同的。

我知道这并不能直接回答OP的问题,但我是通过搜索如何删除lodash被引导到这里的。希望这能帮助到和我处境相似的人。

这要归功于@JohanPersson。我在这个答案的基础上实现了对深度嵌套值的比较,并获得对差异的键引用

getObjectDiff = (obj1, obj2) => { const obj1Props = Object.keys(obj1); const obj2Props = Object.keys(obj2); const keysWithDiffValue = obj1Props.reduce((keysWithDiffValueAccumulator, key) => { const propExistsOnObj2 = obj2.hasOwnProperty(key); const hasNestedValue = obj1[key] instanceof Object && obj2[key] instanceof Object; const keyValuePairBetweenBothObjectsIsEqual = obj1[key] === obj2[key]; if (!propExistsOnObj2) { keysWithDiffValueAccumulator.push(key); } else if (hasNestedValue) { const keyIndex = keysWithDiffValueAccumulator.indexOf(key); if (keyIndex >= 0) { keysWithDiffValueAccumulator.splice(keyIndex, 1); } const nestedDiffs = getObjectDiff(obj1[key], obj2[key]); for (let diff of nestedDiffs) { keysWithDiffValueAccumulator.push(`${key}.${diff}`); } } else if (keyValuePairBetweenBothObjectsIsEqual) { const equalValueKeyIndex = keysWithDiffValueAccumulator.indexOf(key); keysWithDiffValueAccumulator.splice(equalValueKeyIndex, 1); } return keysWithDiffValueAccumulator; }, obj2Props); return keysWithDiffValue; } const obj1 = {a0: {a1: {a2: {a3: 'Im here'}}}}; const obj2 = {a0: {a1: {a2: {a3: 'Not here', b3: 'some'}}}}; console.log('final', getObjectDiff(obj1, obj2));

这是一个简单的带有Lodash深度差异检查器的Typescript,它将生成一个新对象,只包含旧对象和新对象之间的差异。

例如,如果我们有:

const oldData = {a: 1, b: 2};
const newData = {a: 1, b: 3};

结果对象将是:

const result: {b: 3};

它还兼容多层深层对象,对于数组,它可能需要一些调整。

import * as _ from "lodash";

export const objectDeepDiff = (data: object | any, oldData: object | any) => {
  const record: any = {};
  Object.keys(data).forEach((key: string) => {
    // Checks that isn't an object and isn't equal
    if (!(typeof data[key] === "object" && _.isEqual(data[key], oldData[key]))) {
      record[key] = data[key];
    }
    // If is an object, and the object isn't equal
    if ((typeof data[key] === "object" && !_.isEqual(data[key], oldData[key]))) {
      record[key] = objectDeepDiff(data[key], oldData[key]);
    }
  });
  return record;
};

As it was asked, here's a recursive object comparison function. And a bit more. Assuming that primary use of such function is object inspection, I have something to say. Complete deep comparison is a bad idea when some differences are irrelevant. For example, blind deep comparison in TDD assertions makes tests unnecessary brittle. For that reason, I'd like to introduce a much more valuable partial diff. It is a recursive analogue of a previous contribution to this thread. It ignores keys not present in a

var bdiff = (a, b) =>
    _.reduce(a, (res, val, key) =>
        res.concat((_.isPlainObject(val) || _.isArray(val)) && b
            ? bdiff(val, b[key]).map(x => key + '.' + x) 
            : (!b || val != b[key] ? [key] : [])),
        []);

BDiff允许检查期望值,同时容忍其他属性,这正是您想要的自动检查。这允许构建各种高级断言。例如:

var diff = bdiff(expected, actual);
// all expected properties match
console.assert(diff.length == 0, "Objects differ", diff, expected, actual);
// controlled inequality
console.assert(diff.length < 3, "Too many differences", diff, expected, actual);

回到完整的解决方案。使用bdiff构建一个完整的传统diff是很简单的:

function diff(a, b) {
    var u = bdiff(a, b), v = bdiff(b, a);
    return u.filter(x=>!v.includes(x)).map(x=>' < ' + x)
    .concat(u.filter(x=>v.includes(x)).map(x=>' | ' + x))
    .concat(v.filter(x=>!u.includes(x)).map(x=>' > ' + x));
};

在两个复杂对象上运行上述函数将输出类似于下面的内容:

 [
  " < components.0.components.1.components.1.isNew",
  " < components.0.cryptoKey",
  " | components.0.components.2.components.2.components.2.FFT.min",
  " | components.0.components.2.components.2.components.2.FFT.max",
  " > components.0.components.1.components.1.merkleTree",
  " > components.0.components.2.components.2.components.2.merkleTree",
  " > components.0.components.3.FFTResult"
 ]

最后,为了了解值之间的差异,我们可能需要直接eval() diff输出。为此,我们需要一个更丑的bdiff版本,输出语法正确的路径:

// provides syntactically correct output
var bdiff = (a, b) =>
    _.reduce(a, (res, val, key) =>
        res.concat((_.isPlainObject(val) || _.isArray(val)) && b
            ? bdiff(val, b[key]).map(x => 
                key + (key.trim ? '':']') + (x.search(/^\d/)? '.':'[') + x)
            : (!b || val != b[key] ? [key + (key.trim ? '':']')] : [])),
        []);

// now we can eval output of the diff fuction that we left unchanged
diff(a, b).filter(x=>x[1] == '|').map(x=>[x].concat([a, b].map(y=>((z) =>eval('z.' + x.substr(3))).call(this, y)))));

这将输出类似于下面的内容:

[" | components[0].components[2].components[2].components[2].FFT.min", 0, 3]
[" | components[0].components[2].components[2].components[2].FFT.max", 100, 50]

“一族”许可

如果你需要知道哪些属性是不同的,使用reduce():

_.reduce(a, function(result, value, key) {
    return _.isEqual(value, b[key]) ?
        result : result.concat(key);
}, []);
// → [ "prop2" ]