Interface type check with Typescript

Question

This question is the direct analogon to Class type check with TypeScript

I need to find out at runtime if a variable of type any implements an interface. Here\'s my

Answer 1

You can validate a TypeScript type at runtime using ts-validate-type, like so (does require a Babel plugin though):

const user = validateType<{ name: string }>(data);

Answer 2

I found an example from @progress/kendo-data-query in file filter-descriptor.interface.d.ts

Checker

declare const isCompositeFilterDescriptor: (source: FilterDescriptor | CompositeFilterDescriptor) => source is CompositeFilterDescriptor;

Example usage

const filters: Array<FilterDescriptor | CompositeFilterDescriptor> = filter.filters;

filters.forEach((element: FilterDescriptor | CompositeFilterDescriptor) => {
    if (isCompositeFilterDescriptor(element)) {
        // element type is CompositeFilterDescriptor
    } else {
        // element type is FilterDescriptor
    }
});

Answer 3

I would like to point out that TypeScript does not provide a direct mechanism for dynamically testing whether an object implements a particular interface.

Instead, TypeScript code can use the JavaScript technique of checking whether an appropriate set of members are present on the object. For example:

var obj : any = new Foo();

if (obj.someInterfaceMethod) {
    ...
}

Answer 4

Working with string literals is difficult because if you want to refactor you method or interface names then it could be possible that your IDE don't refactor these string literals. I provide you mine solution which works if there is at least one method in the interface

export class SomeObject implements interfaceA {
  public methodFromA() {}
}

export interface interfaceA {
  methodFromA();
}

Check if object is of type interface:

const obj = new SomeObject();
const objAsAny = obj as any;
const objAsInterfaceA = objAsAny as interfaceA;
const isObjOfTypeInterfaceA = objAsInterfaceA.methodFromA != null;
console.log(isObjOfTypeInterfaceA)

Note: We will get true even if we remove 'implements interfaceA' because the method still exists in the SomeObject class

Answer 5

Here's the solution I came up with using classes and lodash: (it works!)

// TypeChecks.ts
import _ from 'lodash';

export class BakedChecker {
    private map: Map<string, string>;

    public constructor(keys: string[], types: string[]) {
        this.map = new Map<string, string>(keys.map((k, i) => {
            return [k, types[i]];
        }));
        if (this.map.has('__optional'))
            this.map.delete('__optional');
    }

    getBakedKeys() : string[] {
        return Array.from(this.map.keys());
    }

    getBakedType(key: string) : string {
        return this.map.has(key) ? this.map.get(key) : "notfound";
    }
}

export interface ICheckerTemplate {
    __optional?: any;
    [propName: string]: any;
}

export function bakeChecker(template : ICheckerTemplate) : BakedChecker {
    let keys = _.keysIn(template);
    if ('__optional' in template) {
        keys = keys.concat(_.keysIn(template.__optional).map(k => '?' + k));
    }
    return new BakedChecker(keys, keys.map(k => {
        const path = k.startsWith('?') ? '__optional.' + k.substr(1) : k;
        const val = _.get(template, path);
        if (typeof val === 'object') return val;
        return typeof val;
    }));
}

export default function checkType<T>(obj: any, template: BakedChecker) : obj is T {
    const o_keys = _.keysIn(obj);
    const t_keys = _.difference(template.getBakedKeys(), ['__optional']);
    return t_keys.every(tk => {
        if (tk.startsWith('?')) {
            const ak = tk.substr(1);
            if (o_keys.includes(ak)) {
                const tt = template.getBakedType(tk);
                if (typeof tt === 'string')
                    return typeof _.get(obj, ak) === tt;
                else {
                    return checkType<any>(_.get(obj, ak), tt);
                }
            }
            return true;
        }
        else {
            if (o_keys.includes(tk)) {
                const tt = template.getBakedType(tk);
                if (typeof tt === 'string')
                    return typeof _.get(obj, tk) === tt;
                else {
                    return checkType<any>(_.get(obj, tk), tt);
                }
            }
            return false;
        }
    });
}

custom classes:

// MyClasses.ts

import checkType, { bakeChecker } from './TypeChecks';

class Foo {
    a?: string;
    b: boolean;
    c: number;

    public static _checker = bakeChecker({
        __optional: {
            a: ""
        },
        b: false,
        c: 0
    });
}

class Bar {
    my_string?: string;
    another_string: string;
    foo?: Foo;

    public static _checker = bakeChecker({
        __optional: {
            my_string: "",
            foo: Foo._checker
        },
        another_string: ""
    });
}

to check the type at runtime:

if (checkType<Bar>(foreign_object, Bar._checker)) { ... }