为什么 useState 在输出中返回 undefined?
Why does useState is returning undefined in output?
我在这里提供isFollowing=true
但在 useStatefollowButtonState
中返回 undefined 作为输出。
const [followButtonState, setFollowButtonState] = useState(isFollowing);
console.log(`profile info component: ${isFollowing}`); //output :true
console.log(followButtonState); //output :undefined
如果您将 isFollowing
作为 prop 传递给您的组件,则在 useEffect 中设置 followButtonState
的值,而不是直接分配它:
const [followButtonState, setFollowButtonState] = useState('');
React.useEffect(() => {
setFollowButtonState(isFollowing)
}, [isFollowing)
希望这能解决您的问题。
我在这里提供isFollowing=true
但在 useStatefollowButtonState
中返回 undefined 作为输出。
const [followButtonState, setFollowButtonState] = useState(isFollowing);
console.log(`profile info component: ${isFollowing}`); //output :true
console.log(followButtonState); //output :undefined
如果您将 isFollowing
作为 prop 传递给您的组件,则在 useEffect 中设置 followButtonState
的值,而不是直接分配它:
const [followButtonState, setFollowButtonState] = useState('');
React.useEffect(() => {
setFollowButtonState(isFollowing)
}, [isFollowing)
希望这能解决您的问题。