React Native 弹出菜单移动选项框 [英] React Native Popup Menu Moving the Options Box

问题描述

我有一个 React Native 弹出菜单实现如下:

I have a React Native Popup Menu implemented as follows:

import React, { Component } from 'react';
import { Text } from 'react-native';
import { Icon, Divider } from 'react-native-elements';
import {
  Menu,
  MenuTrigger,
  MenuOptions,
  MenuOption
} from 'react-native-popup-menu';
import { connect } from 'react-redux';
import firebase from 'firebase';
import { STATUS_BAR_HEIGHT } from '../constants';

class PopUpMenu extends Component {
  render() {
    const { menuStyle, menuOptionsStyle, menuTriggerStyle } = styles;

    return (
      <Menu style={menuStyle}>
        <MenuTrigger style={menuTriggerStyle}>
          <Icon
            name="menu"
            color="white"
            size={30}
          />
        </MenuTrigger>
        <MenuOptions style={menuOptionsStyle}>
          <MenuOption>
            <Text>{this.props.user.email}</Text>
          </MenuOption>
          <MenuOption>
            <Divider />
          </MenuOption>
          <MenuOption text="Log Out" onSelect={() => this.signOutUser()} />
        </MenuOptions>
      </Menu>
    );
  }
}

const styles = {
  menuStyle: {
    marginTop: STATUS_BAR_HEIGHT,
    marginRight: 12
  },
  menuTriggerStyle: {},
  menuOptionsStyle: {}
};

现在它看起来像这样关闭:

Now currently it looks like this closed:

这打开了:

我想让打开的框向下移动到触发器按钮下方，同时仍将触发器保持在同一位置.我如何通过样式实现这一点?

I would like to have the opened box move down below the trigger button while still keeping the trigger in the same place. How can I accomplish that with styles?

推荐答案

您可以通过将 optionsContainerStyle 道具传递给 MenuOptions 组件而不是样式道具来实现这一点.

You can achieve that by passing optionsContainerStyle props to MenuOptions component, not style props.

类似的东西.

<MenuOptions optionsContainerStyle={{ marginTop: 40 }}>
...
</MenuOptions>

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