How to VueJS router-link active style
https://router.vuejs.org/en/api/router-link.html add attribute active-class="active" eg:
<ul class="nav navbar-nav">
<router-link tag="li" active-class="active" to="/" exact><a>Home</a></router-link>
<router-link tag="li" active-class="active" to="/about"><a>About</a></router-link>
<router-link tag="li" active-class="active" to="/permission-list"><a>Permisison</a></router-link>
</ul>
The :active pseudo-class is not the same as adding a class to style the element.
The :active CSS pseudo-class represents an element (such as a button) that is being activated by the user. When using a mouse, "activation" typically starts when the mouse button is pressed down and ends when it is released.
What we are looking for is a class, such as .active
, which we can use to style the navigation item.
For a clearer example of the difference between :active
and .active
see the following snippet:
li:active {
background-color: #35495E;
}
li.active {
background-color: #41B883;
}
<ul>
<li>:active (pseudo-class) - Click me!</li>
<li class="active">.active (class)</li>
</ul>
Vue-Router
vue-router
automatically applies two active classes, .router-link-active
and .router-link-exact-active
, to the <router-link>
component.
router-link-active
This class is applied automatically to the <router-link>
component when its target route is matched.
The way this works is by using an inclusive match behavior. For example, <router-link to="/foo">
will get this class applied as long as the current path starts with /foo/
or is /foo
.
So, if we had <router-link to="/foo">
and <router-link to="/foo/bar">
, both components would get the router-link-active
class when the path is /foo/bar
.
router-link-exact-active
This class is applied automatically to the <router-link>
component when its target route is an exact match. Take into consideration that both classes, router-link-active
and router-link-exact-active
, will be applied to the component in this case.
Using the same example, if we had <router-link to="/foo">
and <router-link to="/foo/bar">
, the router-link-exact-active
class would only be applied to <router-link to="/foo/bar">
when the path is /foo/bar
.
The exact prop
Lets say we have <router-link to="/">
, what will happen is that this component will be active for every route. This may not be something that we want, so we can use the exact
prop like so: <router-link to="/" exact>
. Now the component will only get the active class applied when it is an exact match at /
.
CSS
We can use these classes to style our element, like so:
nav li:hover,
nav li.router-link-active,
nav li.router-link-exact-active {
background-color: indianred;
cursor: pointer;
}
The <router-link>
tag was changed using the tag
prop, <router-link tag="li" />
.
Change default classes globally
If we wish to change the default classes provided by vue-router
globally, we can do so by passing some options to the vue-router
instance like so:
const router = new VueRouter({
routes,
linkActiveClass: "active",
linkExactActiveClass: "exact-active",
})
Change default classes per component instance (<router-link>
)
If instead we want to change the default classes per <router-link>
and not globally, we can do so by using the active-class
and exact-active-class
attributes like so:
<router-link to="/foo" active-class="active">foo</router-link>
<router-link to="/bar" exact-active-class="exact-active">bar</router-link>
v-slot API
Vue Router 3.1.0+ offers low level customization through a scoped slot. This comes handy when we wish to style the wrapper element, like a list element <li>
, but still keep the navigation logic in the anchor element <a>
.
<router-link
to="/foo"
v-slot="{ href, route, navigate, isActive, isExactActive }"
>
<li
:class="[isActive && 'router-link-active', isExactActive && 'router-link-exact-active']"
>
<a :href="href" @click="navigate">{{ route.fullPath }}</a>
</li>
</router-link>