我正在尝试使用React创建一个切换内容按钮。但是我只能使其打开,而再次单击时则无法关闭。有人可以看看一下,让我知道要在代码中进行更改以完成此操作吗?
这是我到目前为止所拥有的:
class Test extends React.Component {
constructor(props) {
super(props)
this.state = {
activeLocation: 0,
}
}
changeActiveLocation = (activeLocation) => {
this.setState({
activeLocation: activeLocation,
});
}
render() {
const activeLocation = company.locations[this.state.activeLocation];
return (
{company.locations.map((location, index) => (
<div className="test-item">
<div className="test-item-container" onClick={() => {this.changeActiveLocation(index)}}>
<div className="test-item-header">
<h3>Text goes here!</h3>
<a><FontAwesomeIcon icon={(this.state.activeLocation === index) ? 'times' : 'chevron-right'} /></a>
</div>
</div>
</div>
))}
)
}
}
谢谢!
答案 0 :(得分:3)
您正在将活动位置设置为与您已经单击的位置相同的位置,因此this.state.activeLocation === index
始终为true。我将使用isOpen
状态值将位置重构为它们自己的组件,该状态值在单击位置时会更新。因此,如下所示:
// test class
class Test extends React.Component {
constructor(props) {
super(props)
this.state = {
activeLocation: 0,
}
}
changeActiveLocation = (activeLocation) => {
this.setState({
activeLocation: activeLocation,
});
}
render() {
const activeLocation = company.locations[this.state.activeLocation];
return (
{company.locations.map((location, index) => (
<LocationItem location={location} onClick={() => this.changeActiveLocation(index)} />
))}
)
}
}
// LocationItem
class LocationItem extends React.Component {
state = { isOpen: false };
handleClick = () => {
this.setState(prevState => { isOpen: !prevState.isOpen});
// call parent click to set new active location if that's still needed
if(this.props.onClick) this.props.onClick;
}
render() {
return <div className="test-item">
<div className="test-item-container" onClick={this.handleClick}>
<div className="test-item-header">
<h3>Text goes here!</h3>
<a><FontAwesomeIcon icon={(this.state.isOpen ? 'times' : 'chevron-right'} /></a>
</div>
</div>
</div>
}
}