I want to create a details page in Angular and want to pass an entry id with the RouterLink. Is there a way to get the parameter in the url?
The URL looks like: http://localhost:8080/details?date=13-09-2016
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2 Answers
Also RouteParams can be used to get the parameter. You can injected it to your component:
@Component(...)
class MyDetailsComponent {
final RouteParams _routeParams;
MyDetailsComponent(this._routeParams){
var date = _routeParams.get('date');
}
}
You can find more details here.
Comments
Implement OnActivate
@override
routerOnActivate(ComponentInstruction nextInstruction,
ComponentInstruction prevInstruction) {
String routeParamValue = nextInstruction.params['paramName'];
String queryParamValue = nextInstruction.urlParams['paramName'];
}
answered Sep 13, 2016 at 10:45
Günter Zöchbauer
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8 Comments
SepDev
Where I have to put this? Do I have to put this in the class part of the component?
Günter Zöchbauer
To the component added by the router. You need to change
params to queryParams for query parameters. param is for route parameters (path: '/some/:paramName/other')SepDev
It doesn't work. It is even not called I already changed it to queryParams but it isn't even called. My Routing looks like this:
@RouteConfig(const [ const Route(path: '/view', name: 'Home', component: SimpleKVComponent, useAsDefault: true), const Route(path: '/settings', name: 'Settings', component: SettingsComponent), const Route(path: '/details', name: 'Details', component: DetailsComponent) ])SepDev
And it is called by this line:
<td><a [routerLink]="['/Details', { date: history['link'] }]"><i class="glyphicon glyphicon-eye-open"></i></a></td>Günter Zöchbauer
Did you add
class DetailsComponent implements OnActivate {? |