Question

我目前尝试在我的应用程序中获得不错的路由，但在我当前的实现中，我必须获取并检查路由订阅中的所有参数。有没有什么方法可以从路线中获得employees或departments的“静态”参数，我可以提前做出改变，而不是检查所有参数值的每种可能性？
如果我的任何路线都不合逻辑或只是错误的话，我也愿意接受改进。

export const EmployeeManagementRoutes: RouterConfig = [
{
path: 'employee-management',
component: EmployeeManagementComponent,
children: [
  //display all employees
  { 
    path: '', 
    //is there any way to make this redirict relative to the component the router is for?
    redirectTo: '/employee-management/employees/department/all',
    pathMatch: 'full'
  },
  //display all employees from a specific department
  { 
    path: 'employees//department/:department',  
    component: EmployeeManagementTableComponent 
  },
  //open dialog for deleting or creating a new employee or    
  if option is 'show' and id is not undefined show a specific employee 
  { 
    path: 'employees/action/:option/:id',   
    component: EmployeeManagementTableComponent 
  },
  //display all departments
  { 
    path: 'departments',  
    component: EmployeeManagementTableComponent 
  },
  //open dialog for deleting or creating a new department
  { 
    path: 'departments/action/:option',  
    component: EmployeeManagementTableComponent 
  },

]
}
];

Answer 1

您可以使用ActivatedRoute。这不仅仅是为了获得参数。类似的东西：

ngOnInit() {
     let path = this.route.url.value.map(val => val.path);
    }

会将url的部分作为字符串数组提供给你。例如["employees","department","management"].

Answer 2

解析你的网址：

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

export class TopNavigationComponent implements OnInit {
  constructor(
    private router: Router
  ) {}
  ngOnInit() {
    let parsedUrl = this.router.parseUrl(this.router.url);
    console.log(parsedUrl);  
  }
}

在收到的对象中，解析的URL位于：根 - ＆GT; CHILDREN-＆GT;伯 - ＆GT;段

您可以阅读更多here

Angular2 Hot从url获取所有参数？

2 个答案: