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
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"]);
