由于两个组件同属一个导航下,如果这样写,只能渲染一个组件。如果换到主路由上去,导航的激活就不见,这种情况如何处理?
render () {
let {match} = this.props;
let list = this.state.list.map((item, index) =>
<div className="feedback-item feedback-question" key={index}>
<div className="feedback-item-info">
<div className="info-head">
<span className="phone">{item.phone.replace(/^(\d{4})\d{4}(\d+)/,"$1****$2")}</span><span>{formatTime(item.date)}</span>发布
</div>
<div className="info-con">{item.title}</div>
</div>
<div className="feedback-item-answer" onClick={this.toChild(`${match.url + '/' + item.id}`)}>去回答</div>
</div>
)
return(
<div className="feedback">
<Scroll click={true} pullUpLoad pullUpLoadMoreData={this.loadMoreData} isPullUpTipHide={ false }>
<div className="feedback-lits">
<Feedlabel changeTags={this.changeTags.bind(this)} />
<div className="quiz">
<div className="quiz-btn" onClick={this.toChild('ask')}>
<i className="feed-icon-add"></i>
我要提问
</div>
</div>
<div className="feedback-list">
{list}
</div>
</div>
</Scroll>
<Route path={`${match.url + '/:id'}`} component={Answer}/>
<Route path={`${match.url + '/ask'}`} component={Ask}/>
</div>
)
}