Javascript 判断对象是否相等

 

在Javascript中相等运算包括"==","==="全等,两者不同之处,不必多数,本篇文章我们将来讲述如何判断两个对象是否相等? 你可能会认为,如果两个对象有相同的属性,以及它们的属性有相同的值,那么这两个对象就相等。那么下面我们通过一个实例来论证下:

var obj1 = {
    name: "Benjamin",
    sex : "male"
}

var obj2 = {
    name: "Benjamin",
    sex : "male"
}

//Outputs: false
console.log(obj1 == obj2);

//Outputs: false
console.log(obj1 === obj2);

通过上面的例子可以看到,无论使用"=="还是"===",都返回false。主要原因是基本类型string,number通过值来比较,而对象(Date,Array)及普通对象通过指针指向的内存中的地址来做比较。看下面一个例子:

var obj1 = {
    name: "Benjamin",
    sex : "male"
};

var obj2 = {
    name: "Benjamin",
    sex : "male"
};

var obj3 = obj1;

//Outputs: true
console.log(obj1 == obj3);

//Outputs: true
console.log(obj1 === obj3);

//Outputs: false
console.log(obj2 == obj3);

//Outputs: false
console.log(obj2 === obj3);

上例返回true,是因为obj1和ob3的指针指向了内存中的同一个地址。和面向对象的语言(Java/C++)中值传递和引用传递的概念相似。 因为,如果你想判断两个对象是否相等,你必须清晰,你是想判断两个对象的属性是否相同,还是属性对应的值是否相同,还是怎样?如果你判断两个对象的值是否相等,可以像下面这样:

function isObjectValueEqual(a, b) {
    // Of course, we can do it use for in 
    // Create arrays of property names
    var aProps = Object.getOwnPropertyNames(a);
    var bProps = Object.getOwnPropertyNames(b);

    // If number of properties is different,
    // objects are not equivalent
    if (aProps.length != bProps.length) {
        return false;
    }

    for (var i = 0; i < aProps.length; i++) {
        var propName = aProps[i];

        // If values of same property are not equal,
        // objects are not equivalent
        if (a[propName] !== b[propName]) {
            return false;
        }
    }

    // If we made it this far, objects
    // are considered equivalent
    return true;
}

var obj1 = {
    name: "Benjamin",
    sex : "male"
};

var obj2 = {
    name: "Benjamin",
    sex : "male"
};

//Outputs: true
console.log(isObjectValueEqual(obj1, obj2));

正如你所看到的,检查对象的“值相等”我们基本上是要遍历的对象的每个属性,看看它们是否相等。虽然这个简单的实现适用于我们的例子中,有很多情况下,它是不能处理。例如: 1) 如果该属性值之一本身就是一个对象吗? 2) 如果属性值中的一个是NaN(在JavaScript中,是不是等于自己唯一的价值?) 3) 如果一个属性的值为undefined,而另一个对象没有这个属性(因而计算结果为不确定?) 检查对象的“值相等”的一个强大的方法,最好是依靠完善的测试库,涵盖了各种边界情况。Underscore和Lo-Dash有一个名为_.isEqual()方法,用来比较好的处理深度对象的比较。您可以使用它们像这样:

// Outputs: true
console.log(_.isEqual(obj1, obj2));

最后附上Underscore中isEqual的部分源码:

  // Internal recursive comparison function for `isEqual`.
  var eq = function(a, b, aStack, bStack) {
    // Identical objects are equal. `0 === -0`, but they aren't identical.
    // See the [Harmony `egal` proposal](http://wiki.ecmascript.org/doku.php?id=harmony:egal).
    if (a === b) return a !== 0 || 1 / a === 1 / b;
    // A strict comparison is necessary because `null == undefined`.
    if (a == null || b == null) return a === b;
    // Unwrap any wrapped objects.
    if (a instanceof _) a = a._wrapped;
    if (b instanceof _) b = b._wrapped;
    // Compare `[[Class]]` names.
    var className = toString.call(a);
    if (className !== toString.call(b)) return false;
    switch (className) {
      // Strings, numbers, regular expressions, dates, and booleans are compared by value.
      case '[object RegExp]':
      // RegExps are coerced to strings for comparison (Note: '' + /a/i === '/a/i')
      case '[object String]':
        // Primitives and their corresponding object wrappers are equivalent; thus, `"5"` is
        // equivalent to `new String("5")`.
        return '' + a === '' + b;
      case '[object Number]':
        // `NaN`s are equivalent, but non-reflexive.
        // Object(NaN) is equivalent to NaN
        if (+a !== +a) return +b !== +b;
        // An `egal` comparison is performed for other numeric values.
        return +a === 0 ? 1 / +a === 1 / b : +a === +b;
      case '[object Date]':
      case '[object Boolean]':
        // Coerce dates and booleans to numeric primitive values. Dates are compared by their
        // millisecond representations. Note that invalid dates with millisecond representations
        // of `NaN` are not equivalent.
        return +a === +b;
    }
    if (typeof a != 'object' || typeof b != 'object') return false;
    // Assume equality for cyclic structures. The algorithm for detecting cyclic
    // structures is adapted from ES 5.1 section 15.12.3, abstract operation `JO`.
    var length = aStack.length;
    while (length--) {
      // Linear search. Performance is inversely proportional to the number of
      // unique nested structures.
      if (aStack[length] === a) return bStack[length] === b;
    }
    // Objects with different constructors are not equivalent, but `Object`s
    // from different frames are.
    var aCtor = a.constructor, bCtor = b.constructor;
    if (
      aCtor !== bCtor &&
      // Handle Object.create(x) cases
      'constructor' in a && 'constructor' in b &&
      !(_.isFunction(aCtor) && aCtor instanceof aCtor &&
        _.isFunction(bCtor) && bCtor instanceof bCtor)
    ) {
      return false;
    }
    // Add the first object to the stack of traversed objects.
    aStack.push(a);
    bStack.push(b);
    var size, result;
    // Recursively compare objects and arrays.
    if (className === '[object Array]') {
      // Compare array lengths to determine if a deep comparison is necessary.
      size = a.length;
      result = size === b.length;
      if (result) {
        // Deep compare the contents, ignoring non-numeric properties.
        while (size--) {
          if (!(result = eq(a[size], b[size], aStack, bStack))) break;
        }
      }
    } else {
      // Deep compare objects.
      var keys = _.keys(a), key;
      size = keys.length;
      // Ensure that both objects contain the same number of properties before comparing deep equality.
      result = _.keys(b).length === size;
      if (result) {
        while (size--) {
          // Deep compare each member
          key = keys[size];
          if (!(result = _.has(b, key) && eq(a[key], b[key], aStack, bStack))) break;
        }
      }
    }
    // Remove the first object from the stack of traversed objects.
    aStack.pop();
    bStack.pop();
    return result;
  };

  // Perform a deep comparison to check if two objects are equal.
  _.isEqual = function(a, b) {
    return eq(a, b, [], []);
  };

 

感谢您的阅读,希望此篇文章对您有所帮助,文中不足之处欢迎批评斧正。

 

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posted @ 2014-10-28 20:17  Benjamin-zuo  阅读(28989)  评论(2编辑  收藏  举报