I have a method -
export default getLink= () => {
const url = 'http://....'
return url;
}
I want to use it as -
import getLink from './link';
const linkUrl= `${getLink}/food/new`;
But it seems the getLink returns me -
isfunction () {return'http://....';}/food/new
How can I return just the http link
You are returning a function there, since you can grab the value by invoking this function and assigning it a variable.
import getLink from './link';
const url = getLink();
const linkUrl= `${url}/food/new`;
Also, with this code actually you can't export your function like that for arrow functions. If you use default then you shouldn't use a name in your function declaration to export your function.
Instead use:
export default () => {
const url = 'http://www.foo.com'
return url;
};
or first assign it to a variable then use default:
const getLink = () => {
const url = 'http://www.foo.com'
return url;
};
export default getLink;
One other alternative is using a named export instead of default.
export const getLink = () => {
const url = 'http://www.foo.com'
return url;
}
then import it like:
import { getLink } from "/.link";
You are not calling the function, you are concatenating the variable pointing to the function to the other string.
You should call the function to get its output instead:
const linkUrl= `${getLink()}/food/new`;
Since you're using the variable pointing to the function in a template string, the output you get is the string representation of that function (i.e. its code), obtained by calling the .toString method of the function. You can see this yourself if your try:
console.log(getLink.toString());