我是React.js的新手,对不起,如果这个问题听起来太愚蠢了。
当输入字段为空时,我正在尝试禁用按钮。 React中的最佳方法是什么?
我正在做以下事情:
<input ref="email"/>
<button disabled={!this.refs.email}>Let me in</button>
这是对的吗?
这不仅仅是动态属性的重复,因为我也很好奇将数据从一个元素传输/检查到另一个元素。
答案 0 :(得分:207)
您需要保持输入的当前值处于状态(或传递其值up to a parent via a callback function或sideways或&lt;您的应用程序&#39;这里的状态管理解决方案&gt; 使得它最终作为支柱传递回您的组件中,因此您可以为该按钮派生禁用的道具。
使用状态的示例:
<meta charset="UTF-8">
<script src="https://fb.me/react-0.13.3.js"></script>
<script src="https://fb.me/JSXTransformer-0.13.3.js"></script>
<div id="app"></div>
<script type="text/jsx;harmony=true">void function() { "use strict";
var App = React.createClass({
getInitialState() {
return {email: ''}
},
handleChange(e) {
this.setState({email: e.target.value})
},
render() {
return <div>
<input name="email" value={this.state.email} onChange={this.handleChange}/>
<button type="button" disabled={!this.state.email}>Button</button>
</div>
}
})
React.render(<App/>, document.getElementById('app'))
}()</script>
&#13;
答案 1 :(得分:3)
使用常量可以组合多个字段进行验证:
class LoginFrm extends React.Component {
constructor() {
super();
this.state = {
email: '',
password: '',
};
}
handleEmailChange = (evt) => {
this.setState({ email: evt.target.value });
}
handlePasswordChange = (evt) => {
this.setState({ password: evt.target.value });
}
handleSubmit = () => {
const { email, password } = this.state;
alert(`Welcome ${email} password: ${password}`);
}
render() {
const { email, password } = this.state;
const enabled =
email.length > 0 &&
password.length > 0;
return (
<form onSubmit={this.handleSubmit}>
<input
type="text"
placeholder="Email"
value={this.state.email}
onChange={this.handleEmailChange}
/>
<input
type="password"
placeholder="Password"
value={this.state.password}
onChange={this.handlePasswordChange}
/>
<button disabled={!enabled}>Login</button>
</form>
)
}
}
ReactDOM.render(<LoginFrm />, document.body);
&#13;
<script src="https://cdnjs.cloudflare.com/ajax/libs/react/15.1.0/react.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/react/15.1.0/react-dom.min.js"></script>
<body>
</body>
&#13;
答案 2 :(得分:1)
另一种检查方法是内联函数,以便在每个渲染(每个道具和状态改变)上都检查条件
const isDisabled = () =>
// condition check
这有效:
<button
type="button"
disabled={this.isDisabled()}
>
Let Me In
</button>
但这不起作用:
<button
type="button"
disabled={this.isDisabled}
>
Let Me In
</button>
答案 3 :(得分:0)
<button disabled={false}>button WORKS</button>
<button disabled={true}>button DOES NOT work</button>
现在只需使用 useState 或任何其他条件将 true/false 传递到按钮中,假设您使用的是 React。
答案 4 :(得分:0)
const Example = () => {
const [value, setValue] = React.useState("");
function handleChange(e) {
setValue(e.target.value);
}
return (
<input ref="email" value={value} onChange={handleChange}/>
<button disabled={!value}>Let me in</button>
);
}
export default Example;
答案 5 :(得分:-1)
这很简单,我们假设您通过扩展包含以下内容的Component构成了状态全类
class DisableButton extends Components
{
constructor()
{
super();
// now set the initial state of button enable and disable to be false
this.state = {isEnable: false }
}
// this function checks the length and make button to be enable by updating the state
handleButtonEnable(event)
{
const value = this.target.value;
if(value.length > 0 )
{
// set the state of isEnable to be true to make the button to be enable
this.setState({isEnable : true})
}
}
// in render you having button and input
render()
{
return (
<div>
<input
placeholder={"ANY_PLACEHOLDER"}
onChange={this.handleChangePassword}
/>
<button
onClick ={this.someFunction}
disabled = {this.state.isEnable}
/>
<div/>
)
}
}