我需要添加课程"活跃"点击按钮后删除所有其他"活动"类。
请看这里:https://codepen.io/azat-io/pen/RWjyZX
var Tags = React.createClass({
setFilter: function(filter) {
this.props.onChangeFilter(filter);
},
render: function() {
return <div className="tags">
<button className="btn active" onClick={this.setFilter.bind(this, '')}>All</button>
<button className="btn" onClick={this.setFilter.bind(this, 'male')}>male</button>
<button className="btn" onClick={this.setFilter.bind(this, 'female')}>female</button>
<button className="btn" onClick={this.setFilter.bind(this, 'child')}>child</button>
<button className="btn" onClick={this.setFilter.bind(this, 'blonde')}>blonde</button>
</div>
}
});
var Kid = React.createClass({
render: function() {
return <ul>
<li>{this.props.name}</li>
</ul>
}
});
var List = React.createClass({
getInitialState: function() {
return {
filter: ''
};
},
changeFilter: function(filter) {
this.setState({
filter: filter
});
},
render: function() {
var list = this.props.Data;
if (this.state.filter !== '') {
list = list.filter((i)=> i.tags.indexOf(this.state.filter) !== -1);
console.log(list);
}
list = list.map(function(Props){
return <Kid {...Props} />
});
return <div>
<h2>Kids Finder</h2>
<Tags onChangeFilter={this.changeFilter}/>
{list}
</div>
}
});
var options = {
Data: [{
name: 'Eric Cartman',
tags: ['male', 'child']
},{
name: 'Wendy Testaburger',
tags: ['female', 'child']
},{
name: 'Randy Marsh',
tags: ['male']
},{
name: 'Butters Stotch',
tags: ['male', 'blonde', 'child']
},{
name: 'Bebe Stevens',
tags: ['female', 'blonde', 'child']
}]
};
var element = React.createElement(List, options);
React.render(element, document.body);
我如何在这里做得更好?
答案 0 :(得分:38)
很简单。 看看这个
https://codepen.io/anon/pen/mepogj?editors=001
基本上你想要处理组件的状态,以便检查当前活动的组件。你需要包括
getInitialState: function(){}
//and
isActive: function(){}
查看链接上的代码
答案 1 :(得分:17)
这非常有用:
https://github.com/JedWatson/classnames
你可以做像
这样的事情classNames('foo', 'bar'); // => 'foo bar'
classNames('foo', { bar: true }); // => 'foo bar'
classNames({ 'foo-bar': true }); // => 'foo-bar'
classNames({ 'foo-bar': false }); // => ''
classNames({ foo: true }, { bar: true }); // => 'foo bar'
classNames({ foo: true, bar: true }); // => 'foo bar'
// lots of arguments of various types
classNames('foo', { bar: true, duck: false }, 'baz', { quux: true }); // => 'foo bar baz quux'
// other falsy values are just ignored
classNames(null, false, 'bar', undefined, 0, 1, { baz: null }, ''); // => 'bar 1'
或像这样使用
var btnClass = classNames('btn', this.props.className, {
'btn-pressed': this.state.isPressed,
'btn-over': !this.state.isPressed && this.state.isHovered
});
答案 2 :(得分:4)
由于您已经有<Tags>
组件在其父级上调用一个函数,因此您不需要其他状态:只需将过滤器作为道具传递给<Tags>
组件,并在渲染按钮时使用它。像这样:
将<Tags>
组件中的渲染功能更改为:
render: function() {
return <div className = "tags">
<button className = {this._checkActiveBtn('')} onClick = {this.setFilter.bind(this, '')}>All</button>
<button className = {this._checkActiveBtn('male')} onClick = {this.setFilter.bind(this, 'male')}>male</button>
<button className = {this._checkActiveBtn('female')} onClick = {this.setFilter.bind(this, 'female')}>female</button>
<button className = {this._checkActiveBtn('blonde')} onClick = {this.setFilter.bind(this, 'blonde')}>blonde</button>
</div>
},
在<Tags>
内添加一个函数:
_checkActiveBtn: function(filterName) {
return (filterName == this.props.activeFilter) ? "btn active" : "btn";
}
在您的<List>
组件中,将过滤器状态作为道具传递给<tags>
组件:
return <div>
<h2>Kids Finder</h2>
<Tags filter = {this.state.filter} onChangeFilter = {this.changeFilter} />
{list}
</div>
然后它应该按预期工作。 Codepen here(希望链接有效)
答案 3 :(得分:1)
来自他们的网站。
render() {
let className = 'menu';
if (this.props.isActive) {
className += ' menu-active';
}
return <span className={className}>Menu</span>
}
答案 4 :(得分:0)
您也可以使用纯js来完成此操作,就像使用jQuery的旧方法一样
如果您想要简单的方法,请尝试
Traceback (most recent call last): File "script.py", line 45, in <module>
edytta.sell_artwork(girl_with_mandolin, '$6M (USD)')
File "script.py", line 30, in sell_artwork
veneer.add_listing(listed_artwork)
TypeError: add_listing() missing 1 required positional argument: 'new_listing'