How can I disable a <Link> in react-router, if its URL already active? E.g. 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.
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:
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
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>
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>
}
</>
});
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.
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);
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.
const [isActive, setIsActive] = useState(true);
<Link to={isActive ? '/link-to-route' : null} />
you can try this, this worked for me.