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

简单使用_。isEqual方法,它将适用于所有比较…

注意:此方法支持比较数组、数组缓冲区、 布尔值、 *日期对象,错误对象,地图,数字,对象对象,正则表达式, *集合、字符串、符号和类型化数组。对象对象进行比较 *通过自身的,不可继承的,可枚举的属性。函数和DOM *节点不支持。

所以如果你有以下情况:

 const firstName = {name: "Alireza"};
 const otherName = {name: "Alireza"};

如果是:_。isEqual (firstName, otherName);

它会返回true

如果const fullName = {firstName: "Alireza", familyName: "Dezfoolian"};

如果是:_。isEqual (firstName, fullName);

将返回false

其他回答

这是基于@JLavoie,使用lodash

let differences = function (newObj, oldObj) {
      return _.reduce(newObj, function (result, value, key) {
        if (!_.isEqual(value, oldObj[key])) {
          if (_.isArray(value)) {
            result[key] = []
            _.forEach(value, function (innerObjFrom1, index) {
              if (_.isNil(oldObj[key][index])) {
                result[key].push(innerObjFrom1)
              } else {
                let changes = differences(innerObjFrom1, oldObj[key][index])
                if (!_.isEmpty(changes)) {
                  result[key].push(changes)
                }
              }
            })
          } else if (_.isObject(value)) {
            result[key] = differences(value, oldObj[key])
          } else {
            result[key] = value
          }
        }
        return result
      }, {})
    }

https://jsfiddle.net/EmilianoBarboza/0g0sn3b9/8/

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。

这段代码返回一个具有不同值的所有属性的对象,以及两个对象的值。对记录差异很有用。

var allkeys = _.union(_.keys(obj1), _.keys(obj2));
var difference = _.reduce(allkeys, function (result, key) {
  if ( !_.isEqual(obj1[key], obj2[key]) ) {
    result[key] = {obj1: obj1[key], obj2: obj2[key]}
  }
  return result;
}, {});

深度比较使用模板的(嵌套)属性进行检查

function objetcsDeepEqualByTemplate(objectA, objectB, comparisonTemplate) {
  if (!objectA || !objectB) return false

  let areDifferent = false
  Object.keys(comparisonTemplate).some((key) => {
    if (typeof comparisonTemplate[key] === 'object') {
      areDifferent = !objetcsDeepEqualByTemplate(objectA[key], objectB[key], comparisonTemplate[key])
      return areDifferent
    } else if (comparisonTemplate[key] === true) {
      areDifferent = objectA[key] !== objectB[key]
      return areDifferent
    } else {
      return false
    }
  })

  return !areDifferent
}

const objA = { 
  a: 1,
  b: {
    a: 21,
    b: 22,
  },
  c: 3,
}

const objB = { 
  a: 1,
  b: {
    a: 21,
    b: 25,
  },
  c: true,
}

// template tells which props to compare
const comparisonTemplateA = {
  a: true,
  b: {
    a: true
  }
}
objetcsDeepEqualByTemplate(objA, objB, comparisonTemplateA)
// returns true

const comparisonTemplateB = {
  a: true,
  c: true
}
// returns false
objetcsDeepEqualByTemplate(objA, objB, comparisonTemplateB)

这将在控制台中工作。如果需要,可以添加数组支持

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