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我試圖讓該瀏覽器進入一個網頁被按下時後退按鈕的網址登陸網址,我在組件中添加該代碼角4沒有得到來自onpopstate
import {Component, OnInit} from '@angular/core';
import {LocalDataSource} from 'ng2-smart-table';
import {TableComponent} from '../../../table/table.component';
import {PagerService} from '../../../services/pager/pager.service';
import {ActivatedRoute, Params, Router} from '@angular/router';
import {Subscription} from 'rxjs/Subscription';
import {PlatformLocation} from '@angular/common';
@Component({
selector: 'app-test',
templateUrl: './test.component.html',
styleUrls: ['./test.component.css', '../../../app.component.css']
})
export class TestComponent implements OnInit {
source: LocalDataSource;
currentPage: number;
totalCount: number;
pager: any = {};
paramsSubscription: Subscription;
urlParams: Params;
constructor(private router: Router, private location: PlatformLocation, private route: ActivatedRoute) {
this.location.onPopState(() => {
console.log('Back Pressed');
this.getURLParams();
// Giving the URL parameters of the page from where back button is clicked
console.log(this.urlParams['page']);
});
}
getURLParams() {
this.paramsSubscription = this.route.params.subscribe(
(params: Params) => {
this.urlParams = params;
}
);
}
ngOnInit() {
// Methods to populate data
}
}
這是檢測後退按鈕並在控制檯中顯示參數。但是,控制檯正在顯示按下後退按鈕的頁面,而不是其着陸的頁面。我想根據着陸頁動態更新數據。
有沒有辦法通過任何方法檢測登陸頁面參數? 請讓我知道如果我失去了一些東西。
謝謝!
這是什麼方法做:'this.getURLParams()'我沒有看到它在所提供的代碼示例。 – DeborahK
錯過了那個,現在更新了代碼!它只是獲取URL參數,基於此我更新了網頁的數據部分。 –