You can add the type as JSON. there is a type in typescript for that.
sample code
import { Component, OnInit } from '@angular/core';
@Component({
selector: 'app-array-json-dates',
templateUrl: './array-json-dates.component.html',
styleUrls: ['./array-json-dates.component.scss']
})
export class ArrayJsonDatesComponent implements OnInit {
jsonObject: JSON;
arrayObj: any = [
{
id: 1,
name: "john",
lastModified: '2011-01-01 02:00:00'
},
{
id: 2,
name: "Franc",
lastModified: '2001-01-01 02:00:00'
},
{
id: 3,
name: "Andrew",
lastModified: '2021-01-01 02:00:00'
},
{
id: 11,
name: "Mark",
lastModified: '2020-01-01 02:00:00'
},
{
id: 12,
name: "Eric",
lastModified: '2020-02-01 02:00:00'
},
{
id: 8,
name: "Tony",
lastModified: '1990-01-01 02:00:00'
}
]
constructor() {
this.jsonObject = <JSON>this.arrayObj;
}
ngOnInit(): void {
}
In your case
getArticoliByDesc = (descrizione : string) => {
return this.httpClient.get<JSON>(`http://${this.server}:${this.port}/api/articoli/cerca/descrizione/${descrizione}`) //ALT + 0096 | ALT GR + '
}
But still creating a Model class which matched your api response would be the best way to do this. Even though api returns json, Angular http will convert it in to an object. Which you can then generalize by adding your model
Ex:
export class ApiModel{
property1: string;
property2: string;
property3: string;
}
getArticoliByDesc = (descrizione : string) => {
return this.httpClient.get<ApiModel>(`http://${this.server}:${this.port}/api/articoli/cerca/descrizione/${descrizione}`) //ALT + 0096 | ALT GR + '
}