javascript – 没有一些字段的深克隆

让我们下一个 javascript对象.现在我想克隆它,但没有一些领域.例如我想要克隆对象没有字段“lastName”和“cars.age”
输入

{
   "firstName":"Fred","lastName":"McDonald","cars":[
           {
              "type":"mersedes","age":5
           },{
              "model":"bmw","age":10
           }
       ]
}

输出(克隆)

{
   "firstName":"Fred","cars":[
       {
          "model":"mersedes"
       },{
          "model":"bmw"
       }
   ]
}

我可以做一些类似的事情

var human = myJson   
var clone = $.extend(true,{},human)  
delete clone.lastName  
_.each(clone.cars,function(car))  
{  
   delete car.age  
}

你知道更容易的解决方案吗？

解决方法

如果您不介意添加对象原型,这是一个简单的解决方案.您可能需要修改一些以供自己使用.

Object.prototype.deepOmit = function(blackList) {
  if (!_.isArray(blackList)) { 
    throw new Error("deepOmit(): argument must be an Array");
  }

  var copy = _.omit(this,blackList);
  _.each(blackList,function(arg) {
    if (_.contains(arg,'.')) {
      var key  = _.first(arg.split('.'));
      var last = arg.split('.').slice(1);
      copy[key] = copy[key].deepOmit(last);
    }
  });
  return copy;
};

Array.prototype.deepOmit = function(blackList) {
  if (!_.isArray(blackList)) { 
    throw new Error("deepOmit(): argument must be an Array");
  }

  return _.map(this,function(item) {
    return item.deepOmit(blackList);
  });
};

那么当你有一个对象,如：

var personThatOwnsCars = {
   "firstName":"Fred","age":10
           }
       ]
};

你可以这样做魔术.

personThatOwnsCars.deepOmit(["firstName","cars.age"]);

甚至这样的魔法！

[person1,person2].deepOmit(["firstName","cars.age"]);