function isDeepEqual(obj1, obj2, testPrototypes = false) { if (obj1 === obj2) { return true } if (typeof obj1 === "function" && typeof obj2 === "function") { return obj1.toString() === obj2.toString() } if (obj1 instanceof Date && obj2 instanceof Date) { return obj1.getTime() === obj2.getTime() } if ( Object.prototype.toString.call(obj1) !== Object.prototype.toString.call(obj2) || typeof obj1 !== "object" ) { return false } const prototypesAreEqual = testPrototypes ? isDeepEqual( Object.getPrototypeOf(obj1), Object.getPrototypeOf(obj2), true ) : true const obj1Props = Object.getOwnPropertyNames(obj1) const obj2Props = Object.getOwnPropertyNames(obj2) return ( obj1Props.length === obj2Props.length && prototypesAreEqual && obj1Props.every(prop => isDeepEqual(obj1[prop], obj2[prop])) ) } console.log(isDeepEqual({key: 'one'}, {key: 'first'})) console.log(isDeepEqual({key: 'one'}, {key: 'one'}))

我知道这有点老了，但我想添加一个解决方案，我想出了这个问题。 我有一个对象，我想知道它的数据什么时候改变。类似于Object。我所做的是:

function checkObjects(obj,obj2){
   var values = [];
   var keys = [];
   keys = Object.keys(obj);
   keys.forEach(function(key){
      values.push(key);
   });
   var values2 = [];
   var keys2 = [];
   keys2 = Object.keys(obj2);
   keys2.forEach(function(key){
      values2.push(key);
   });
   return (values == values2 && keys == keys2)
}

这里可以复制并创建另一组数组来比较值和键。 这很简单，因为它们现在是数组，如果对象大小不同将返回false。

最简单和逻辑的解决方案，比较一切像对象，数组，字符串，Int…

JSON。stringify({a: val1}) == JSON。stringify ({a: val2})

注意:

你需要用你的Object替换val1和val2 对于对象，必须对两侧对象进行递归排序(按键)

为了比较简单的键/值对对象实例的键，我使用:

function compareKeys(r1, r2) {
    var nloops = 0, score = 0;
    for(k1 in r1) {
        for(k2 in r2) {
            nloops++;
            if(k1 == k2)
                score++; 
        }
    }
    return nloops == (score * score);
};

一旦比较了键，一个简单的for. in循环就足够了。

复杂度是O(N*N)， N是键的个数。

我希望/猜测我定义的对象不会拥有超过1000个属性…

如何确定部分对象(partial <T>)等于typescript中的原始对象(T)。

function compareTwoObjects<T>(original: T, partial: Partial<T>): boolean {
  return !Object.keys(partial).some((key) => partial[key] !== original[key]);
}

附注:最初我打算提出一个有答案的新问题。但这样的问题已经存在，并被标记为重复题。

下面是一个使用ES6+的解决方案

// this comparison would not work for function and symbol comparisons
// this would only work best for compared objects that do not belong to same address in memory
// Returns true if there is no difference, and false otherwise


export const isObjSame = (obj1, obj2) => {
    if (typeof obj1 !== "object" && obj1 !== obj2) {
        return false;
    }

    if (typeof obj1 !== "object" && typeof obj2 !== "object" && obj1 === obj2) {
        return true;
    }

    if (typeof obj1 === "object" && typeof obj2 === "object") {
        if (Array.isArray(obj1) && Array.isArray(obj2)) {
            if (obj1.length === obj2.length) {
                if (obj1.length === 0) {
                    return true;
                }
                const firstElemType = typeof obj1[0];

                if (typeof firstElemType !== "object") {
                    const confirmSameType = currentType =>
                        typeof currentType === firstElemType;

                    const checkObjOne = obj1.every(confirmSameType);
                    const checkObjTwo = obj2.every(confirmSameType);

                    if (checkObjOne && checkObjTwo) {
                        // they are primitves, we can therefore sort before and compare by index
                        // use number sort
                        // use alphabet sort
                        // use regular sort
                        if (firstElemType === "string") {
                            obj1.sort((a, b) => a.localeCompare(b));
                            obj2.sort((a, b) => a.localeCompare(b));
                        }
                        obj1.sort((a, b) => a - b);
                        obj2.sort((a, b) => a - b);

                        let equal = true;

                        obj1.map((element, index) => {
                            if (!isObjSame(element, obj2[index])) {
                                equal = false;
                            }
                        });

                        return equal;
                    }

                    if (
                        (checkObjOne && !checkObjTwo) ||
                        (!checkObjOne && checkObjTwo)
                    ) {
                        return false;
                    }

                    if (!checkObjOne && !checkObjTwo) {
                        for (let i = 0; i <= obj1.length; i++) {
                            const compareIt = isObjSame(obj1[i], obj2[i]);
                            if (!compareIt) {
                                return false;
                            }
                        }

                        return true;
                    }

                    // if()
                }
                const newValue = isObjSame(obj1, obj2);
                return newValue;
            } else {
                return false;
            }
        }

        if (!Array.isArray(obj1) && !Array.isArray(obj2)) {
            let equal = true;
            if (obj1 && obj2) {
                const allKeys1 = Array.from(Object.keys(obj1));
                const allKeys2 = Array.from(Object.keys(obj2));

                if (allKeys1.length === allKeys2.length) {
                    allKeys1.sort((a, b) => a - b);
                    allKeys2.sort((a, b) => a - b);

                    allKeys1.map((key, index) => {
                        if (
                            key.toLowerCase() !== allKeys2[index].toLowerCase()
                        ) {
                            equal = false;
                            return;
                        }

                        const confirmEquality = isObjSame(obj1[key], obj2[key]);

                        if (!confirmEquality) {
                            equal = confirmEquality;
                            return;
                        }
                    });
                }
            }

            return equal;

            // return false;
        }
    }
};