如何在reactjs中禁用链接？

Question

I have this in my reactjs app:

import Link from 'react-router/lib/Link'

Been trying to disable this link but this does not have the desired effect:

<Link disable={true}/>

It just renders it invisible. How can I disable( based on a condition) the reactjs Link?

Answer 1

Contain many issues on react-router , there is no support disabled attribute in Link Component, so you can try some with this issue :

1. onClick event

Use preventDefault() to handle onClick event.

/* YourComponent.js */
class YourComponent extends React.Component {
  render() {
    return (
      <Link onClick={e => e.preventDefault()} />
    );
  }
}

2. CSS's pointer-events attribute

/* YourComponent.js */
class YourComponent extends React.Component {
  render() {
    return (
      <Link className='disabled-link' />
    );
  }
}

/* css file */
.disable-link {
  pointer-events: none;
}

or you can use inline style

/* YourComponent.js */
class YourComponent extends React.Component {
  render() {
    return (
      <Link style={{ pointerEvents: 'none' }} />
    );
  }
}

What I used was method 2, it's more clearly for me on my project.

Answer 2

You could conditionally render something that looks like a disabled link based upon some state.

For instance in typescript:

export interface Location {
  pathname: string; 
  search: string;
  state: any;
  hash: string;
  key ?: string;
}
interface LinkProps {
  to: string | Location
  replace?:boolean
}
interface DisableLinkProps extends LinkProps {
  enabled: boolean
  linkText:string
}
export class DisableLink extends React.Component<DisableLinkProps, undefined> {
  render() {
    var element= this.props.enabled ? <span className="disableLinkDisabled">{this.props.linkText}</span> : <Link to={this.props.to} replace={this.props.replace}>{this.props.linkText}</Link>
    return element;
  }
}

interface DemoClassState {
  linkEnabled:boolean
}
export class DemoClass extends React.Component<undefined, DemoClassState> {
  constructor(props) {
    super(props);
    this.state = { linkEnabled:false }
  }
  toggleLinkEnabled = () => {
    this.setState((prevState) => {
        return {
            linkEnabled: !prevState.linkEnabled
        }
    });
  }
  render() {
    return <div>
        <DisableLink enabled={this.state.linkEnabled} to="/somewhere" linkText="Some link" />
        <button onClick={this.toggleLinkEnabled}>Toggle link enabled</button>
        </div>
  }
}

Answer 3

This is actually a bit tricky. And maybe even somewhat ill-advised.

https://css-tricks.com/how-to-disable-links/

The route (no pun intended) I took was to not render the link.

I composed a new component from react-router's Link .

import React from 'react';
import { Link } from 'react-router-dom';

export default function ToggleableLink(props)  {
    const { disabled, ...rest } = props;
    return disabled ? props.children : <Link {...rest}>{props.children}</Link>;
}

Usage:

<ToggleableLink disabled={!showTheLink}>Foobar</ToggleableLink>