Question

因此，当前我正在路由到页面以从策略中获取文档列表。我还将引入所有策略，因此您可以浏览选择表单，并使用传递的策略编号将其重新路由到该表单。有没有一种方法可以选择使用新网址进行浏览的选项。

 <mat-form-field name="selectPolicy">
      <mat-select placeholder="Change Policy">
        <mat-option *ngFor="let policy of policies" [routerLink]="['../', policy.policyNbr]">{{policy.policyTypeDesc}}
          #{{policy.policyNbr}}
        </mat-option>
      </mat-select>
    </mat-form-field>

截至目前，这会更改我的网址，但更改后确实会提交到新页面。

Answer 1

from collections import defaultdict

record_keys = ['name', 'rating', 'price', 'cuisine']


def load(file):
    with open(file) as file:
        data = file.read()

    restaurants = []
    # chop up input on each blank line (2 newlines in a row) 
    for record in data.split("\n\n"):
        fields = record.split("\n")

        # build a dictionary by zipping together the fixed set
        # of field names and the values from this particular record
        restaurant = dict(zip(record_keys, fields))

        # split chops apart the type cuisine on comma, then _.strip()
        # removes any leading/trailing whitespace on each type of cuisine 
        restaurant['cuisine'] = [_.strip() for _ in restaurant['cuisine'].split(",")]
        restaurants.append(restaurant)

    return restaurants


def build_index(database, key, value):
    index = defaultdict(set)
    for record in database:
        for v in record.get(key, []):
            # defaultdict will create a set if one is not present or add to it if one does
            index[v].add(record[value])

    return index


restaurant_db = load('/var/tmp/r')
print(restaurant_db)

by_type = build_index(restaurant_db, 'cuisine', 'name')
print(by_type)

//assuming you have a route set up to take the value as a param
navigateTo(value){
  console.log(value);
  this.router.navigate(['../',value]);

}

Answer 2

在该类中，如果您有一个，可以在ngOnInit之后

使用Material Select和Angular 6的routerLink路由页面

2 个答案: