给定两个对象 destination
和 source
,Lodash merge()
函数 将第二个对象 自己的属性 和继承的属性复制到第一个对象中。
const destination = { name: Will Riker, rank: Commander };
const source = { ship: USS Enterprise };
_.merge(destination, source);
destination.name; // Will Riker
destination.rank; // Commander
destination.ship; // USS Enterprise
听起来很像 Object.assign()
, 正确的? 尽管 merge()
非常相似 Object.assign()
和 _.assign()
,有一些细微的差别。
merge()
和 assign()
之间的差异
第一个细节是 merge()
是 递归 复制对象,所以 _.merge()
是 深拷贝 ,而 _.assign()
是浅拷贝。
const obj = {
name: {
first: Will,
last: Riker
}
};
const deepClone = _.merge({}, obj);
deepClone.name === obj.name; // false
deepClone.name.first = Thomas;
obj.name.first; // Will
const shallowClone = _.assign({}, obj);
shallowClone.name === obj.name; // true
shallowClone.name.first = Thomas;
obj.name.first; // Thomas
第二个细节是如何 merge()
处理 undefined
,如果 source
有一个值严格等于的键 undefined
,merge()
不会覆盖该键 destination
。
let destination = {
firstName: Will,
lastName: Riker,
rank: Commander
};
// Since `source.rank` is undefined, `merge()` wont overwrite
// `destination.rank`.
_.merge(destination, { firstName: Thomas, rank: undefined });
destination.firstName; // Thomas
destination.rank; // Commander
destination = {
firstName: Will,
lastName: Riker,
rank: Commander
};
// But `_.assign()` and `Object.assign()` overwrite `destination.rank`.
_.assign(destination, { firstName: Thomas, rank: undefined });
destination.firstName; // Thomas
destination.rank; // undefined
当您考虑如何 merge()
处理 类 。
class Ship {};
Ship.prototype.shipName = USS Enterprise;
const ship = new Ship();
// `merge()` copies inherited properties, so it will copy
// `shipName`
const merged = _.merge({ name: Will Riker, rank: Commander }, ship);
merged.shipName; // USS Enterprise
// `assign()` does **not** copy inherited properties.
const assigned = Object.assign({ name: Will Riker, rank: Commander }, ship);
assigned.shipName; // undefined
© 版权声明
文章版权归作者所有,未经允许请勿转载。
THE END
请登录后查看评论内容