按照 Ionic 文档,我试图让弹出框粘在按钮上( like on their own example )。
不幸的是,我不知道如何实现这一目标......
谢谢
import React, { useState } from 'react';
import { IonPopover, IonButton } from '@ionic/react';
export const PopoverExample: React.FC = () => {
const [showPopover, setShowPopover] = useState(false);
return (
<>
<IonPopover
isOpen={showPopover}
onDidDismiss={e => setShowPopover(false)}
>
<p>This is popover content</p>
</IonPopover>
<IonButton onClick={() => setShowPopover(true)}>Show Popover</IonButton>
</>
);
};
最佳答案
您还需要在 showPopover Hook 中包含一个事件 -
const [showPopover, setShowPopover] = useState<{open: boolean, event: Event | undefined}>({
open: false,
event: undefined,
});
<IonPopover
isOpen={showPopover.open}
event={showPopover.event}
onDidDismiss={e => setShowPopover({open: false, event: undefined})}
>
<p>This is popover content</p>
</IonPopover>
<IonButton onClick={(e) => setShowPopover({open: true, event: e.nativeEvent})}>Click</IonButton>
关于reactjs - 带有 React 的 Ionic popover - 如何使其粘在按钮上,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/59707783/