Comments (4)
My gut instinct is that this doesn't belong in UISrefActive. It seems trivial enough to do it "vanilla" style.
how can I do so that the class is always active irrespective of the parameters given as a default, so in essence be active as long as the state is in the dynamicView path
I might put the default parameter values on the dynamic state(s), then link to the state without any explicit values.
const state = {
name: 'dyn1',
params: {
page: 0,
filterType: { }
}
}
<UISref to={dynamicView}>
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so so far, I just handled it vanilla
import { UISref, UIView, UISrefActive, StateService } from '@uirouter/react';
....
const stateService = this.props.transition.router.stateService as StateService;
const currentRoute = stateService.$current.name;
<UISref to={dynamicView} params={{ bucketId, page, viewtype }}>
<li role="presentation" key={index++} className={currentRoute == dynamicView && "active" }>
...
</li>
</UISref>
I still be interested to learn other ways to handle it with UISrefActive, if any
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I understand your needs, there currently isn't a way to override the isActive
logic, but shouldn't be too much work to implement it as function prop in the component.
An example could be:
<UISrefActive class="active" key={index} isActive={() =>{}}>
<UISref to={dynamicView} params={{ page, filterType }}><a>link</a></UISref>
</UISrefActive>
And the isActive
prop would a function called with the router
instance as argument, so one can implement whatever logic needed and return a bool
.
Would like to know what @christopherthielen thinks about it.
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