Snippet for Avoiding the async/await hell medium article

correctuse-example-1.js

async function selectPizza() {
  const pizzaData = await getPizzaData()    // async call
  const chosenPizza = choosePizza()    // sync call
  await addPizzaToCart(chosenPizza)    // async call
}

async function selectDrink() {
  const drinkData = await getDrinkData()    // async call
  const chosenDrink = chooseDrink()    // sync call
  await addDrinkToCart(chosenDrink)    // async call
}

(async () => {
  const pizzaPromise = selectPizza()
  const drinkPromise = selectDrink()
  await pizzaPromise
  await drinkPromise
  orderItems()    // async call
})()

// Although I prefer it this way 

Promise.all([selectPizza(), selectDrink()]).then(orderItems)   // async call

Missing closing parenthesis here. Should be:
Promise.all([selectPizza(), selectDrink()]).then(orderItems) // async call
<3

Thanks @anton-kl fixed it now

Why not just use

(async () =>{
  await Promise.all([selectPizza(), selectDrink()])
  await orderItems() // can also be return or even without anything, if we want to fire-and-forget
})()

why mix async/await with .then? any special reason?

L24 should be started with await

I have updated the gist.

why mix async/await with .then? any special reason?

@ATGardner if we use await then the code following the Promise.all statement will have to wait for the function calls. Why do it when you have a much shorter syntax and which doesn't have a side effect.

I believe we should use .then() first and if we intentionally want to wait in the entire function then use async/await

The last statement should be
return orderItems();
if, say it returns a order number in Promise?

Even it does not explicitly return a value, there is a Promise created can be used for the return of outer async ().