Node.JS API call returns UNDEFINED

Question

I am trying to make my first API call with NodeJS by using node-fetch, however I am getting UNDEFINED as error.

The call works great on my browser link, however with the code below, it returns undefined:

import fetch from 'node-fetch';

const latLongAirports = [{
        "name": "Madrid",
        "iata": "MAD",
        "lat": 40.49565434242003,
        "long": -3.574541319609411,
    },{
        "name": "Los Angeles",
        "iata": "LAX",
        "lat": 33.93771087455066,
        "long": -118.4007447751959,
    },{
        "name": "Mexico City",
        "iata": "MEX",
        "lat": 19.437281814699613,
        "long": -99.06588831304731,}
]

export function getTemperature(iata){
    let data = latLongAirports.find(el => el.iata === iata);
    var url = "http://www.7timer.info/bin/api.pl?lon=" + data.long + "&lat=" + data.lat + "&product=astro&output=json"
    console.log(url);
    async () =>{
        const response = fetch(url);
        const data2 = await response.json();
        console.log(data2);
    }
}

console.log(getTemperature('MEX'));

Any ideas why I am not getting the data?

Does this answer your question? Why calling react setState method doesn't mutate the state immediately? — poPaTheGuru
– poPaTheGuru, Commented Oct 28, 2021 at 9:01
getTemperature doesn't even have a return statement. Also you have constructed a async ()=>{} but never invoke it. — Ricky Mo
– Ricky Mo, Commented Oct 28, 2021 at 9:03
@poPaTheGuru fetch(url) and await response.json() don't run at all. console.log is run at the same time with your await response.json() no, there is no multi-threading. so will always be undefined no, it is undefined because there is no return, but you could e.g. return a Promise. — t.niese
– t.niese, Commented Oct 28, 2021 at 9:04
@jabaa - Certainly answers part of it, but there's a lot more going on above than just that. :-) — T.J. Crowder
– T.J. Crowder, Commented Oct 28, 2021 at 14:11

Ricky Mo · Accepted Answer · 2021-10-28 09:08:08Z

2

getTemperature() missing return statement
fetch() not awaited
you have constructed an async ()=>{} but never invoke it

Simply make getTemperature() become async and await to it, also await to fetch and return the result.

export async function getTemperature(iata){
    let data = latLongAirports.find(el => el.iata === iata);
    var url = "http://www.7timer.info/bin/api.pl?lon=" + data.long + "&lat=" + data.lat + "&product=astro&output=json"
    console.log(url);
    
    const response = await fetch(url);
    const data2 = await response.json();
    console.log(data2);
    return data2;
}

console.log(await getTemperature('MEX'));

answered Oct 28, 2021 at 9:08

Ricky Mo

7,7682 gold badges22 silver badges41 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

T.J. Crowder Over a year ago

Nice one! Minor nit pick: Should check response.ok before calling response.json, details in my blog post here.

turbopasi · Accepted Answer · 2021-10-28 09:08:51Z

1

I suggest making the whole getTemperature function an async function. Also use a try ... catch block when await something.

export async function getTemperature(iata) {

  /* ... */
  
  try {
  
    const response = await fetch(url);
    const data2 = await response.json();
    return data2;
  
  } catch (ex) {
  
    throw ex;
  
  }

}

Now your getTemperature function will return a promise as well


// then/catch

getTemperature("MEX").then((result) => {
  console.log(result);
})
.catch((error) => {
  console.log(error.message);
});


// or async/await

( async function () {
  
  try {
    const result = await getTemperature("MEX");
    console.log(result);
  } catch (ex) {
    console.log(ex.message);
  }

})();

answered Oct 28, 2021 at 9:08

turbopasi

3,6152 gold badges20 silver badges49 bronze badges

3 Comments

T.J. Crowder Over a year ago

"Also use a try ... catch block when await something." I strongly recommend not doing that. Instead, let the error (rejection) propagate to the caller, except in top-level entry points (event handlers, mostly) where there's nothing useful the error can propagate to. Also, try { ... } catch (ex) { throw ex; } is completely pointless anyway.

turbopasi Over a year ago

I actually believed without the try ... catch block inside getTemperature the exception will be unhandled and the app crashes - is this not the case ?? Then I learned something 😀 thanks !

T.J. Crowder Over a year ago

Right, that's not the case. :-) You're handling the exception (catch), but then throwing it again (throw), which is effectively the same as not handling it (since you're not doing anything else in the catch block, like cleanup). Either way, the function throws an exception (rejects its promise). Allowing a function to throw/reject is fine, you just handle exceptions at a higher level.

Collectives™ on Stack Overflow

Node.JS API call returns UNDEFINED

2 Answers 2

1 Comment

3 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

3 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related