How can you correctly write an asynchronous method on an object in typescript with a generic?

• 0
  At the moment I have come to such code, but knowing that there are different ways in JS, maybe there are other ways. Perhaps more compact and expressive?
  
  

  const ApiServiceModule = {
    get: async <T>(url: string): Promise<T> => await fetch(url)
      .then(response => response.json())
      .catch(err => console.error(err))
  }

  JavaScript Anonymous, Jul 30, 2019

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2 Answers
• 0
  What exactly do you want to shorten, it seems like there is nowhere to go: one type at the entrance - one at the exit?
  
  
  
  Well, you can remove async and await , they are not needed here:
  
  

  const ApiServiceModule = {
  
    get: <T>(url: string): Promise<T> => fetch(url)
  
      .then(response => response.json())
  
      .catch(err => console.error(err))
  
  }

  async / await is just sugar over Promise and in this case they do nothing, because fetch already returns Promise .
  
  
  
  P.S. Well, I wouldn't call it a method, it's a property with a function. The method has access to this .

  Anonymous

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• 0
  In addition to Aetae - generic is not needed here either
  
  Yes, parsing JSON returns any (for historical reasons) and it can be easily cast into any type with such a generic. But why use TypeScript at all if we always shut it up with our righteousness?
  
  Even if the data comes from our server, and we know their type for sure, they need to be checked, because there is no guarantee that no one was mistaken on the back, that we were not mistaken in the description of the type we are casting.
  
  Therefore, the type must be unknown for TypeScript to require checking everything before use:

  const ApiServiceModule = {
  
    get: (url: string): Promise<unknown> => fetch(url)
  
      .then(response => response.json())
  
      .catch(err => console.error(err))
  
  }

  Anonymous

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