/TypeScript/Get url params in typescript

Get url params in typescript

import {Router, ActivatedRoute, Params} from '@angular/router';
import {OnInit, Component} from '@angular/core';

@Component({...})
export class MyComponent implements OnInit {

  constructor(private activatedRoute: ActivatedRoute) {}

  ngOnInit() {
    // Note: Below 'queryParams' can be replaced with 'params' depending on your requirements
    this.activatedRoute.queryParams.subscribe(params => {
        const userId = params['userId'];
        console.log(userId);
      });
  }

}

Source: stackoverflow.com

Similar codes

3 votes328 views1 answer
Label points in plot in r
TypeScript
0 vote328 views1 answer
'platformNativeScriptDynamic' is deprecated
TypeScript
0 vote330 views2 answers
When we dont have to show data of child but change in child should be displayed in parent automatically
TypeScript
0 vote336 views1 answer
Import xml elements in kotlin
TypeScript
0 vote338 views1 answer
'create' is deprecated
TypeScript
0 vote307 views1 answer
Render html contents from url in asp.net razor
TypeScript
0 vote341 views1 answer
Can we nested try statements in java
TypeScript
0 vote319 views2 answers
Moonspell (@moonspellofficial) • Instagram photos and videoswww.instagram.com › moonspellofficial 61.4k Followers, 619 Following, 2421 Posts - See Instagram photos and videos from Moonspell (@moonspellofficial)
TypeScript
0 vote312 views1 answer
Ionic scroll to item programmatically
TypeScript
0 vote353 views1 answer
- beyondcode/laravel-websockets is locked to version 1.12.0 and an update of this package was not requested.
TypeScript