严格相等运算符将告诉您两个对象类型是否相等。然而,是否有一种方法来判断两个对象是否相等,就像Java中的哈希码值一样?

堆栈溢出问题JavaScript中有hashCode函数吗?类似于这个问题,但需要一个更学术的答案。上面的场景说明了为什么有必要有一个,我想知道是否有等效的解决方案。


当前回答

虽然这个问题已经有很多答案了。我只是想提供另一种实现方法:

const primitveDataTypes = ['number', 'boolean', 'string', 'undefined']; const isDateOrRegExp = (value) => value instanceof Date || value instanceof RegExp; const compare = (first, second) => { let agg = true; if(typeof first === typeof second && primitveDataTypes.indexOf(typeof first) !== -1 && first !== second){ agg = false; } // adding support for Date and RegExp. else if(isDateOrRegExp(first) || isDateOrRegExp(second)){ if(first.toString() !== second.toString()){ agg = false; } } else { if(Array.isArray(first) && Array.isArray(second)){ if(first.length === second.length){ for(let i = 0; i < first.length; i++){ if(typeof first[i] === 'object' && typeof second[i] === 'object'){ agg = compare(first[i], second[i]); } else if(first[i] !== second[i]){ agg = false; } } } else { agg = false; } } else { const firstKeys = Object.keys(first); const secondKeys = Object.keys(second); if(firstKeys.length !== secondKeys.length){ agg = false; } for(let j = 0 ; j < firstKeys.length; j++){ if(firstKeys[j] !== secondKeys[j]){ agg = false; } if(first[firstKeys[j]] && second[secondKeys[j]] && typeof first[firstKeys[j]] === 'object' && typeof second[secondKeys[j]] === 'object'){ agg = compare(first[firstKeys[j]], second[secondKeys[j]]); } else if(first[firstKeys[j]] !== second[secondKeys[j]]){ agg = false; } } } } return agg; } console.log('result', compare({a: 1, b: { c: [4, {d:5}, {e:6}]}, r: null}, {a: 1, b: { c: [4, {d:5}, {e:6}]}, r: 'ffd'})); //returns false.

其他回答

虽然这个问题已经有很多答案了。我只是想提供另一种实现方法:

const primitveDataTypes = ['number', 'boolean', 'string', 'undefined']; const isDateOrRegExp = (value) => value instanceof Date || value instanceof RegExp; const compare = (first, second) => { let agg = true; if(typeof first === typeof second && primitveDataTypes.indexOf(typeof first) !== -1 && first !== second){ agg = false; } // adding support for Date and RegExp. else if(isDateOrRegExp(first) || isDateOrRegExp(second)){ if(first.toString() !== second.toString()){ agg = false; } } else { if(Array.isArray(first) && Array.isArray(second)){ if(first.length === second.length){ for(let i = 0; i < first.length; i++){ if(typeof first[i] === 'object' && typeof second[i] === 'object'){ agg = compare(first[i], second[i]); } else if(first[i] !== second[i]){ agg = false; } } } else { agg = false; } } else { const firstKeys = Object.keys(first); const secondKeys = Object.keys(second); if(firstKeys.length !== secondKeys.length){ agg = false; } for(let j = 0 ; j < firstKeys.length; j++){ if(firstKeys[j] !== secondKeys[j]){ agg = false; } if(first[firstKeys[j]] && second[secondKeys[j]] && typeof first[firstKeys[j]] === 'object' && typeof second[secondKeys[j]] === 'object'){ agg = compare(first[firstKeys[j]], second[secondKeys[j]]); } else if(first[firstKeys[j]] !== second[secondKeys[j]]){ agg = false; } } } } return agg; } console.log('result', compare({a: 1, b: { c: [4, {d:5}, {e:6}]}, r: null}, {a: 1, b: { c: [4, {d:5}, {e:6}]}, r: 'ffd'})); //returns false.

我写这个方法只是为了确保数组和对象都能被清晰地比较。

这应该也能做到!:)

public class Objects {
    /**
     * Checks whether a value is of type Object
     * @param value the value
     */
    public static isObject = (value: any): boolean => {
        return value === Object(value) && Object.prototype.toString.call(value) !== '[object Array]'
    }

    /**
     * Checks whether a value is of type Array
     * @param value the value
     */
    public static isArray = (value: any): boolean => {
        return Object.prototype.toString.call(value) === '[object Array]' && !Objects.isObject(value)
    }

    /**
     * Check whether two values are equal
     */
    public static isEqual = (objectA: any, objectB: any) => {
        // Objects
        if (Objects.isObject(objectA) && !Objects.isObject(objectB)) {
            return false
        }
        else if (!Objects.isObject(objectA) && Objects.isObject(objectB)) {
            return false
        }
        // Arrays
        else if (Objects.isArray(objectA) && !Objects.isArray(objectB)) {
            return false
        }
        else if (!Objects.isArray(objectA) && Objects.isArray(objectB)) {
            return false
        }
        // Primitives
        else if (!Objects.isArray(objectA) && !Objects.isObject(objectA)) {
            return objectA === objectB
        }
        // Object or array
        else {
            const compareObject = (objectA: any, objectB: any): boolean => {
                if (Object.keys(objectA).length !== Object.keys(objectB).length) return false

                for (const propertyName of Object.keys(objectA)) {
                    const valueA = objectA[propertyName]
                    const valueB = objectB[propertyName]

                    if (!Objects.isEqual(valueA, valueB)) {
                        return false
                    }
                }

                return true
            }
            const compareArray = (arrayA: any[], arrayB: any[]): boolean => {
                if (arrayA.length !== arrayB.length) return false

                for (const index in arrayA) {
                    const valueA = arrayA[index]
                    const valueB = arrayB[index]

                    if (!Objects.isEqual(valueA, valueB)) {
                        return false
                    }
                }

                return true
            }
            if (Objects.isObject(objectA)) {
                return compareObject(objectA, objectB)
            } else {
                return compareArray(objectA, objectB)
            }
        }
    }
}

我不是Javascript专家,但这里有一个简单的解决方法。我检查三件事:

它是一个对象,而且它不是null,因为typeof null是对象。 如果两个对象的属性计数相同?否则它们就不相等。 遍历一个对象的属性,并检查对应的属性在第二个对象中是否具有相同的值。

function deepEqual (first, second) { // Not equal if either is not an object or is null. if (!isObject(first) || !isObject(second) ) return false; // If properties count is different if (keys(first).length != keys(second).length) return false; // Return false if any property value is different. for(prop in first){ if (first[prop] != second[prop]) return false; } return true; } // Checks if argument is an object and is not null function isObject(obj) { return (typeof obj === "object" && obj != null); } // returns arrays of object keys function keys (obj) { result = []; for(var key in obj){ result.push(key); } return result; } // Some test code obj1 = { name: 'Singh', age: 20 } obj2 = { age: 20, name: 'Singh' } obj3 = { name: 'Kaur', age: 19 } console.log(deepEqual(obj1, obj2)); console.log(deepEqual(obj1, obj3));

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)

需要一个比已经发布的更通用的对象比较函数,我炮制了以下。批判赞赏……

Object.prototype.equals = function(iObj) {
  if (this.constructor !== iObj.constructor)
    return false;
  var aMemberCount = 0;
  for (var a in this) {
    if (!this.hasOwnProperty(a))
      continue;
    if (typeof this[a] === 'object' && typeof iObj[a] === 'object' ? !this[a].equals(iObj[a]) : this[a] !== iObj[a])
      return false;
    ++aMemberCount;
  }
  for (var a in iObj)
    if (iObj.hasOwnProperty(a))
      --aMemberCount;
  return aMemberCount ? false : true;
}