react-router：如何禁用<link> ，如果它是活跃的？

Question

How can I disable a <Link> in react-router, if its URL already active? Eg if my URL wouldn't change on a click on <Link> I want to prevent clicking at all or render a <span> instead of a <Link> .

The only solution which comes to my mind is using activeClassName (or activeStyle ) and setting pointer-events: none; , but I'd rather like to use a solution which works in IE9 and IE10.

Answer 1

You can use CSS's pointer-events attribute. This will work with most of the browsers. For example your JS code:

class Foo extends React.Component {
  render() {
    return (
      <Link to='/bar' className='disabled-link'>Bar</Link>
    );
  }
}

and CSS:

.disabled-link {
  pointer-events: none;
}

Links:

• pointer-events CSS property
• How to disable HTML links

The How to disable HTML links answer attached suggested using both disabled and pointer-events: none for maximum browser-support.

a[disabled] {
    pointer-events: none;
}

Link to source: How to disable Link

Answer 2

I'm not going to ask why you would want this behavior, but I guess you can wrap <Link /> in your own custom link component.

<MyLink to="/foo/bar" linktext="Maybe a link maybe a span" route={this.props.route} />

class MyLink extends Component {
    render () {
        if(this.props.route === this.props.to){
            return <span>{this.props.linktext}</span>
        }
        return <Link to={this.props.to}>{this.props.linktext}</Link>
    }
}

(ES6, but you probably get the general idea...)

Answer 3

这对我有用：

<Link to={isActive ? '/link-to-route' : '#'} />

Answer 4

Another possibility is to disable the click event if clicking already on the same path. Here is a solution that works with react-router v4 .

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

class SafeLink extends Component {
    onClick(event){
        if(this.props.to === this.props.history.location.pathname){
            event.preventDefault();
        }

        // Ensure that if we passed another onClick method as props, it will be called too
        if(this.props.onClick){
            this.props.onClick();
        }
    }

    render() {
        const { children, onClick, ...other } = this.props;
        return <Link onClick={this.onClick.bind(this)} {...other}>{children}</Link>
    }
}

export default withRouter(SafeLink);

You can then use your link as (any extra props from Link would work):

<SafeLink className="some_class" to="/some_route">Link text</SafeLink>

Answer 5

All the goodness of React Router NavLink with the disable ability.

import React from "react"; // v16.3.2
import { withRouter, NavLink } from "react-router-dom"; // v4.2.2

export const Link = withRouter(function Link(props) {
  const { children, history, to, staticContext, ...rest } = props;
  return <>
    {history.location.pathname === to ?
      <span>{children}</span>
      :
      <NavLink {...{to, ...rest}}>{children}</NavLink>
    }
  </>
});

Answer 6

React Router's Route component has three different ways to render content based on the current route. While component is most typically used to show a component only during a match, the children component takes in a ({match}) => {return <stuff/>} callback that can render things cased on match even when the routes don't match .

I've created a NavLink class that replaces a Link with a span and adds a class when its to route is active.

class NavLink extends Component {
  render() {
    var { className, activeClassName, to, exact, ...rest } = this.props;
    return(
      <Route
        path={to}
        exact={exact}
        children={({ match }) => {
          if (match) {
            return <span className={className + " " + activeClassName}>{this.props.children}</span>;
          } else {
            return <Link className={className} to={to} {...rest}/>;
          }
        }}
      />
    );
  }
}

Then create a navlink like so

<NavLink to="/dashboard" className="navlink" activeClassName="active">

React Router's NavLink does something similar, but that still allows the user to click into the link and push history.

Answer 7

Based on nbeuchat's answer and component - I've created an own improved version of component that overrides react router's Link component for my project.

In my case I had to allow passing an object to to prop (as native react-router-dom link does), also I've added a checking of search query and hash along with the pathname

import PropTypes from 'prop-types';
import React, { Component } from 'react';
import { Link as ReactLink } from 'react-router-dom';
import { withRouter } from "react-router";

const propTypes = {
  to: PropTypes.oneOfType([PropTypes.string, PropTypes.func, PropTypes.object]),
  location: PropTypes.object,
  children: PropTypes.node,
  onClick: PropTypes.func,
  disabled: PropTypes.bool,
  staticContext: PropTypes.object
};

class Link extends Component {
  handleClick = (event) => {
    if (this.props.disabled) {
      event.preventDefault();
    }

    if (typeof this.props.to === 'object') {
      let {
        pathname,
        search = '',
        hash = ''
      } = this.props.to;
      let { location } = this.props;

      // Prepend with ? to match props.location.search
      if (search[0] !== '?') {
        search = '?' + search;
      }

      if (
        pathname === location.pathname
        && search === location.search
        && hash === location.hash
      ) {
        event.preventDefault();
      }
    } else {
      let { to, location } = this.props;

      if (to === location.pathname + location.search + location.hash) {
        event.preventDefault();
      }
    }

    // Ensure that if we passed another onClick method as props, it will be called too
    if (this.props.onClick) {
      this.props.onClick(event);
    }
  };

  render() {
    let { onClick, children, staticContext, ...restProps } = this.props;
    return (
      <ReactLink
        onClick={ this.handleClick }
        { ...restProps }
      >
        { children }
      </ReactLink>
    );
  }
}

Link.propTypes = propTypes;

export default withRouter(Link);

Answer 8

Another option to solve this problem would be to use a ConditionalWrapper component which renders the <Link> tag based on a condition.

This is the ConditionalWrapper component which I used based on this blog here https://blog.hackages.io/conditionally-wrap-an-element-in-react-a8b9a47fab2 :

const ConditionalWrapper = ({ condition, wrapper, children }) =>
    condition ? wrapper(children) : children;

export default ConditionalWrapper

This is how we have used it:

const SearchButton = () => {
    const {
        searchData,
    } = useContext(SearchContext)

    const isValid = () => searchData?.search.length > 2

    return (<ConditionalWrapper condition={isValid()}
                                wrapper={children => <Link href={buildUrl(searchData)}>{children}</Link>}>
            <a
                className={`ml-auto bg-${isValid()
                    ? 'primary'
                    : 'secondary'} text-white font-filosofia italic text-lg md:text-2xl px-4 md:px-8 pb-1.5`}>{t(
                    'search')}</a>
        </ConditionalWrapper>
    )
}

This solution does not render the Link element and avoids also code duplication.

Answer 9

... const [isActive, setIsActive] = useState(true); ...

you can try this, this worked for me.

Answer 10

I think you should you atrribute to=null to set disable a link.

See an example at here https://stackoverflow.com/a/44709182/4787879

Answer 11

如果它适合你的设计，在它上面放一个 div，然后操作 z-index。