javascript - 如何过滤对象的所有属性？

Question

我正在尝试创建一个仅包含非对象属性的对象的副本。但子对象会随之被复制。

var testObject = {
  stringProperty: "hi",
  intProperty: 4,
  objectProperty: {},
  nullProperty: null
};

console.log(removeChildObjects(testObject));

function removeChildObjects(object) {
  var keys = Object.keys(object);
  var newObject = {};
  keys.forEach(function(key) {
    console.log(key, object[key], typeof object[key]);
    if (typeof object[key] != "object") {
      newObject[key] = object[key];
    }
  });
  return object;
}

也可以在这里查看 https://jsfiddle.net/uss94sc3/1/

Answer 1

如果你想严格过滤掉对象属性(保留null和undefined属性)，那么你不能依赖损坏的typeof一元运算符。

typeof null
// "object"

您可以将代码更改为:

function removeChildObjects(object) {
    var keys = Object.keys(object);
    var newObject = {};
    keys.forEach(function(key) {
        if (typeof object[key] != "object" || object[key] == null) {
            newObject[key] = object[key];
        }
    });
    return newObject;
}

或更简洁地使用下划线:

function removeChildObjects(object) {
    return _.omit(object, _.isObject);
}