如何在Angular 6中获取特定的网址模式

Question

I would like to get a specific URL pattern. The sample is given below:

localhost/reports/orderproducts?filter=1&from=01.10.2018&to=31.10.2018&admins=45%2C32%2C43&product=0&categories=5%2C2

Right now I am getting:

localhost/reports/orderproducts?filter=1&from=01.10.2018&to=31.10.2018&admins=45&admins=32&admins=43&product=0&categories=5&categories=2

    submitFiltration(form: NgForm) : void {
      let params : Object = MiscUtils.filterObject(this.formData);
      params['filter'] = 1;
      this.router.navigate([], {queryParams: params});
      this.submitted.emit(this.formData);
    }

param object

Any help is appreciated. The best help would be to help me to understand the approach. Thanks in advance

Answer 1

You are trying to show as URL Encoded string fo example in you admins=45%2C32%2 actualy is admins=45,32,43 if you want to show it like this you must add the array with a join(',') to you get split it by coma.

Example:

let stringParam = adminsArray.join(',');