使用下面的代码,我试图打开子菜单点击,但它会自动打开,每次我悬停在它上。我需要帮助找到它。
代码框:https://codesandbox.io/s/falling-field-10ilwd?file=/src/index.js
代码:
import React from "react";
import ReactDOM from "react-dom";
import "antd/dist/antd.css";
import { Menu, Icon } from "antd";
const { SubMenu } = Menu;
class Sider extends React.Component {
state = {
mode: "vertical",
theme: "light"
};
render() {
return (
<div>
<br />
<br />
<Menu
style={{ width: 256 }}
mode={this.state.mode}
theme={this.state.theme}
>
<SubMenu
key="sub1"
title={
<span>
<Icon type="appstore" />
<span>Navigation Three</span>
</span>
}
>
<Menu.Item key="3">Option 3</Menu.Item>
<Menu.Item key="4">Option 4</Menu.Item>
<SubMenu key="sub1-2" title="Submenu">
<Menu.Item key="5">Option 5</Menu.Item>
<Menu.Item key="6">Option 6</Menu.Item>
</SubMenu>
</SubMenu>
<SubMenu
key="sub2"
title={
<span>
<Icon type="setting" />
<span>Navigation Four</span>
</span>
}
>
<Menu.Item key="7">Option 7</Menu.Item>
<Menu.Item key="8">Option 8</Menu.Item>
<Menu.Item key="9">Option 9</Menu.Item>
<Menu.Item key="10">Option 10</Menu.Item>
</SubMenu>
</Menu>
</div>
);
}
}
ReactDOM.render(<Sider />, document.getElementById("container"));子菜单应该只在我点击导航三或四个时打开,当我在上面悬停时不应该打开。谢谢。
发布于 2022-11-29 10:09:07
您可以使用"triggerSubMenuAction"参数更改子菜单触发器选项。在菜单组件中添加此参数。
<Menu
style={{ width: 256 }}
mode={this.state.mode}
theme={this.state.theme}
triggerSubMenuAction={"click"}
>
.....https://codesandbox.io/s/awesome-waterfall-ml8vyj?file=/src/index.js
https://stackoverflow.com/questions/74612067
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