som*_*491 1 typescript reactjs
我想从一个组件访问另一个组件的状态。为此,我只想将 contextprovider 包装到单击按钮时状态发生变化的组件,并从 usehook 返回状态,以便另一个组件可以访问该状态。
下面是在没有应用上下文的情况下组件的外观,
function UploadButton () { //this is where state is set
const [isDialogOpen, setIsDialogOpen] = React.useState(false);
const handleClick = () => {
setIsDialogOpen(!isDialogOpen);
}
return (
<>
<Button onClick={handleClick}/>
{isDialogOpen && <Upload/>}
</>
);
}
function UserButton() { //this is where state is accessed
return (
<Icon/> //this icon should be displayed only if !isDialogOpen
);
}
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上下文如下所示,我在 UploadButton 所在的同一文件中有 DialogContext。
interface DialogCtxState {
isDialogOpen: boolean;
setIsDialogOpen: React.Dispatch<React.SetStateAction<boolean>>;
}
const initialDialogState: DialogCtxState = {
isDialogOpen: false,
setIsDialogOpen: () => {},
};
const DialogContext = React.createContext<DialogCtxState>(
initialDialogState
);
export const DialogContextProvider: React.FC = ({ children }) => {
const [isDialogOpen, setIsDialogOpen] = React.useState<boolean>(false);
return (
<DialogContext.Provider
value={{
isDialogOpen,
setIsDialogOpen,
}}
>
{children}
</DialogContext.Provider>
);
}
function UploadButton () {
const {isDialogOpen, setIsDialogOpen} = React.useContext(DialogContext);
const handleClick = () => {
setIsDialogOpen(!isDialogOpen);
console.log('isDialogOpen', isDialogOpen) //prints false here.
}
return (
<DialogContextProvider>
<>
<Button onClick={handleClick}/>
{isDialogOpen && <Upload/>} //this doesnt render on clicking button as isDialogOpen
//is false
</>
</DialogContextProvider>
);
}
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上面的代码片段不会渲染上传组件,因为无论我是否单击按钮,isDialogOpen 始终为 false。
我不知道出了什么问题。有人可以帮我吗?谢谢。
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