use*_*448 8 javascript reactjs
我正在尝试通过另一个组件呈现一个按钮来引用和/或影响另一个组件的状态.
var Inputs = React.createClass({
getInitialState: function(){
return {count: 1};
},
add: function(){
this.setState({
count: this.state.count + 1
});
},
render: function(){
var items = [];
var inputs;
for (var i = 0; i < this.state.count; i++){
items.push(<input type="text" name={[i]} />);
items.push(<br />);
}
return (
<div className="col-md-9">
<form action="/" method="post" name="form1">
{items}
<input type="submit" className="btn btn-success" value="Submit Form" />
</form>
</div>
);
}
});
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我想编写一个能够访问Inputs中的add函数的新组件.我试着Inputs.add像这样直接引用它:
var Add = React.createClass({
render: function(){
return (
<input type="button" className="btn" value="Add an Input" onClick={Inputs.add} />
);
}
});
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但那没用.我如何通过另一个组件访问组件的功能,或通过另一个组件影响组件的状态?谢谢.
Mar*_*pse 10
您可以通过创建负责管理状态的父组件来完成此操作,然后将状态向下推送到子组件作为props.
/** @jsx React.DOM */
var Inputs = React.createClass({
render: function () {
var items = [];
var inputs;
for (var i = 0; i < this.props.count; i++) {
items.push( <input type="text" name={[i]} />);
items.push(<br />);
}
return (
<div className = "col-md-9">
<form action = "/" method = "post" name = "form1">
{items}
<input type="submit" className="btn btn-success" value = "Submit Form" />
</form>
</div>
);
}
});
var Add = React.createClass({
render: function () {
return (<input type = "button" className="btn" value="Add an Input" onClick={this.props.fnClick}/> );
}
});
var Parent = React.createClass({
getInitialState: function(){
return {count:1}
},
addInput: function(){
var newCount = this.state.count + 1;
this.setState({count: newCount});
},
render: function(){
return (
<div>
<Inputs count={this.state.count}></Inputs>
<Add fnClick={this.addInput}/>
</div>
);
}
});
React.renderComponent(<Parent></Parent> , document.body);
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