React-router跟踪活动链接

时间:2017-04-05 18:30:06

标签: javascript reactjs react-router

由于react-router没有为链接或li添加活动类,我希望将活动链接保持在菜单状态。当我记录this.state.activeLink时,我得到正确的值,但由于某种原因,类'activeLink'没有添加到li。为什么是这样?当我记录它时,isActive()正确返回true或false,但仍然没有将类添加到li中。这是怎么回事?

import React, { Component } from 'react';
import { Link } from 'react-router-dom';

import logo from '../images/logo.png';

export default class Navbar extends Component {
    constructor(props){
        super(props);

        this.state = { activeLink : '' };
        this.setActive = this.setActive.bind(this);
    }

    setActive(event){
        this.setState({ activeLink: event.target.id });
    }

    isActive(link){
        console.log('isActive:', link == this.state.activeLink);
        link == this.state.activeLink ? 'activeLink' : '';
    }

    render(){
        return (
            <div className="header">
                <div className="row">
                    <div className="header-left">
                        <div className="logo">
                            <img src={logo} alt=""/> 

                        </div>
                        <div className="header-title">Physical Twist</div>

                        <div className="menu">
                            <ul className="nav">
                                <li className={this.isActive('home')}><Link to="/" id="home" onClick={this.setActive}>Home</Link></li>
                                <li className={this.isActive('store')}><Link to="/store" id="store" onClick={this.setActive}>Store</Link></li>
                                <li className={this.isActive('catalogue')}> <Link to="#" id="catalogue" onClick={this.setActive}>Catalogue</Link></li>
                                <li className={this.isActive('contact')} ><Link to="#" id="contact" onClick={this.setActive}>Contact</Link></li>
                            </ul>
                        </div>                              
                    </div>
                </div>
            </div>
        );
    };
}

2 个答案:

答案 0 :(得分:1)

isActive没有返回任何内容。

isActive(link){
    link == this.state.activeLink ? 'activeLink' : '';
}

应该是

isActive(link){
    return link == this.state.activeLink ? 'activeLink' : '';
}

答案 1 :(得分:1)

由于您忘记在return功能中添加isActive(),请使用此功能:

isActive(link){
   console.log('isActive:', link == this.state.activeLink);
   return link == this.state.activeLink ? 'activeLink' : '';
}