因此...
var outObj = people[0];
outObj.oAuthID = null;
delete outObj.oAuthID;
给我...
{
"uuid": "39b2b45f-1dde-4c9a-8765-1bc76f55848f",
"oAuthID": null,
"date": "2013-10-21T16:48:47.079Z",
"updated": "2013-10-21T16:48:47.079Z",
"id": "52655aefcc81bb9adc000001"
}
But this...
function clone(obj) {
// Handle the 3 simple types, and null or undefined
if (null == obj || "object" != typeof obj) return obj;
// Handle Date
if (obj instanceof Date) {
var copy = new Date();
copy.setTime(obj.getTime());
return copy;
}
// Handle Array
if (obj instanceof Array) {
var copy = [];
for (var i = 0, len = obj.length; i < len; i++) {
copy[i] = clone(obj[i]);
}
return copy;
}
// Handle Object
if (obj instanceof Object) {
var copy = {};
for (var attr in obj) {
if (obj.hasOwnProperty(attr)) copy[attr] = clone(obj[attr]);
}
return copy;
}
throw new Error("Unable to copy obj! Its type isn't supported.");
}
var outObj = clone(people[0]);
outObj.oAuthID = null;
delete outObj.oAuthID;
给我...
{
"uuid": "39b2b45f-1dde-4c9a-8765-1bc76f55848f",
"date": "2013-10-21T16:48:47.079Z",
"updated": "2013-10-21T16:48:47.079Z",
"id": "52655aefcc81bb9adc000001"
}
我真的不想每次都要克隆一切来隐藏我的结果中的一个属性。发生了什么?为什么会这样?如何修复它以使其正常工作?
people[0]
是一个原型包含oAuthID
的自定义对象,那么你将得到这个结果(http://jsfiddle.net/qw3UV/)。但是,“gives me”是什么意思?你是如何得到那个结果的?在控制台中吗?也许如果你想要将该对象传输到其他地方,那么将它JSON序列化/反序列化就足够了。 - freakishJSON.stringify
不会查看原型。然而,这种行为可以被库覆盖,在这种情况下,你唯一的选择就是复制。 - freakish