我有一个包含团队成员数组的对象,最小的对象类型实现了interface Member或interface Manager。管理器对象可以包含其他interface Member对象。每个成员/经理都有自己的key: string。
正如您从代码中看到的,我正在尝试实现一个函数setProfession(key, profession),它根据某个成员/经理的个人key: string为他们设置一个profession: string属性。
在我的setProfession函数中有一个错误:
Property 'members' does not exist on type 'Member | Manager'.我不明白这个错误,因为我已经在错误所在的线前设置了一个安全防护装置。尽管interface Manager显然包含members属性:误差印刷机,但它仍然引发错误。
interface Member {
key: string;
name: string;
profession: string;
}
interface Manager {
key: string;
name: string;
profession: string;
members: Array<Member>
}
interface Team {
members: Array<Member | Manager>
}
const TEAM: Team = {
members: [
{
key: 'RodneyMcKay1',
name: 'Rodney McKay',
profession: 'Scientist'
}, {
key: 'JohnShephard1',
name: 'John Shephard',
profession: 'Soldier',
members: [
{
key: 'TeylaEmmagan1',
name: 'Teyla Emmagan',
profession: 'Soldier'
}, {
key: 'RononDex1',
name: 'Ronon Dex',
profession: 'Soldier'
}
]
}
]
};
const isManager = (member: Member | Manager): member is Manager => {
return member.hasOwnProperty('members');
}
const setProfession = (key: string, profession: string): boolean => {
for (let i = 0; i < TEAM.members.length; i++) {
if (isManager(TEAM.members[i])) { // SAFE GUARD
for (let j = 0; j < TEAM.members[i].members.length; j++) { // ERROR
if (TEAM.members[i].members[j].key === key) { // ERROR
TEAM.members[i].members[j].profession = profession; // ERROR
return true;
}
}
} else if (TEAM.members[i].key === key) {
TEAM.members[i].profession = profession;
return true;
}
}
return false;
}发布于 2022-07-07 23:17:52
打字稿不能很好地追踪所有的训练。您可以通过将测试过的值赋值给一个变量来帮助它:
for (let i = 0; i < TEAM.members.length; i++) {
const person = TEAM.members[i]
if (isManager(person)) {
for (let j = 0; j < person.members.length; j++) {
if (person.members[j].key === key) {
person.members[j].profession = profession;
return true;
}
}
} else if (person.key === key) {
person.profession = profession;
return true;
}
}如果使用迭代而不是索引循环,这就容易得多。然后就可以免费创建这个变量,作为循环的一部分。而且您的代码更容易阅读和理解,这是一个额外的好处。
for (const person of TEAM.members) {
if (isManager(person)) {
for (const member of person.members) {
if (member.key === key) {
member.profession = profession;
return true;
}
}
} else if (person.key === key) {
person.profession = profession;
return true;
}
}https://stackoverflow.com/questions/72905123
复制相似问题