javascript - 没有某些字段的深度克隆

Question

让我有下一个 javascript 对象。现在我想克隆它但没有一些字段。例如，我想要没有字段 "lastName" 和 "cars.age"
的克隆对象输入

{
   "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  
}  

你知道更简单的解决方案吗？

Answer 1

如果您不介意添加对象原型(prototype)，这是一个简单的解决方案。您可能希望对其进行一些修改以供您自己使用。

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);
  });
};

然后当你有一个像这样的对象时:

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

你可以像这样变魔术。

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

甚至像这样的魔术!

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