如何在单击React Native时更改状态中所选项目的css

import React, { Component } from 'react';
import { View, TouchableOpacity, StyleSheet } from 'react-native';

export default class MyComponent extends Component {
  constructor(props) {
    super(props);
    this.state = {
      selectedProject: null,
    };
  }

  handleItemClick = (project) => {
    this.setState({ selectedProject: project });
  }

  render() {
    const { selectedProject } = this.state;

    return (
      <View>
        <TouchableOpacity
          style={[styles.projectItem, selectedProject === 'project1' && styles.selectedItem]}
          onPress={() => this.handleItemClick('project1')}
        >
          {/* Project 1 */}
        </TouchableOpacity>

        <TouchableOpacity
          style={[styles.projectItem, selectedProject === 'project2' && styles.selectedItem]}
          onPress={() => this.handleItemClick('project2')}
        >
          {/* Project 2 */}
        </TouchableOpacity>
      </View>
    );
  }
}

const styles = StyleSheet.create({
  projectItem: {
    // 默认项目样式
  },
  selectedItem: {
    // 所选项目的样式
  },
});