Async /await not working on making a fetch API call in javascript
1
I am making a fetch API call from my js file ,I have a doubt that when i am using Async/await still the code is executing in asynchronous manner.
I have also tried await at differnt place but its doesn't work.
let info
async function weather(){
// API call
let data=await fetch('http://api.openweathermap.org/data/2.5/weather?'+'&lat=20&lon=44'+'&units=metric'+'&APPID='+WEATHER_KEY)
console.log("inside API")
const res= await data.json();
console.log(res)
}
weather()
console.log("last part")
OUTPUT:
last part
inside API
"VALUE OF RES"
WHAT I EXPECT:
inside API
"VALUE OF RES"
last part
Any help will be highly appreciated..
javascript
asked Jan 3 at 15:47
Abhijeet kumarAbhijeet kumar
128
add a comment |
1
I am making a fetch API call from my js file ,I have a doubt that when i am using Async/await still the code is executing in asynchronous manner.
I have also tried await at differnt place but its doesn't work.
let info
async function weather(){
// API call
let data=await fetch('http://api.openweathermap.org/data/2.5/weather?'+'&lat=20&lon=44'+'&units=metric'+'&APPID='+WEATHER_KEY)
console.log("inside API")
const res= await data.json();
console.log(res)
}
weather()
console.log("last part")
OUTPUT:
last part
inside API
"VALUE OF RES"
WHAT I EXPECT:
inside API
"VALUE OF RES"
last part
Any help will be highly appreciated..
javascript
asked Jan 3 at 15:47
Abhijeet kumarAbhijeet kumar
128
3
JavaScript is asynchronous, the "last part" runs first because theawaithasn't returned a result yet.
– doublesharp
Jan 3 at 15:50
2
you'll need toawaitweather.
– Daniel A. White
Jan 3 at 15:50
And toawait weather()the whole things needs to be inside anotherasyncfunction.
– doublesharp
Jan 3 at 15:51
add await to weather function
– Sachin Kadam
Jan 3 at 15:52
As you wrote it yourself,weather()is asynchronous. Meaning non-blocking, meaningconsole.log("last part")won't wait for it to complete.
– Jeremy Thille
Jan 3 at 15:52
add a comment |
1
1
1
I am making a fetch API call from my js file ,I have a doubt that when i am using Async/await still the code is executing in asynchronous manner.
I have also tried await at differnt place but its doesn't work.
let info
async function weather(){
// API call
let data=await fetch('http://api.openweathermap.org/data/2.5/weather?'+'&lat=20&lon=44'+'&units=metric'+'&APPID='+WEATHER_KEY)
console.log("inside API")
const res= await data.json();
console.log(res)
}
weather()
console.log("last part")
OUTPUT:
last part
inside API
"VALUE OF RES"
WHAT I EXPECT:
inside API
"VALUE OF RES"
last part
Any help will be highly appreciated..
javascript
asked Jan 3 at 15:47
Abhijeet kumarAbhijeet kumar
128
I am making a fetch API call from my js file ,I have a doubt that when i am using Async/await still the code is executing in asynchronous manner.
I have also tried await at differnt place but its doesn't work.
let info
async function weather(){
// API call
let data=await fetch('http://api.openweathermap.org/data/2.5/weather?'+'&lat=20&lon=44'+'&units=metric'+'&APPID='+WEATHER_KEY)
console.log("inside API")
const res= await data.json();
console.log(res)
}
weather()
console.log("last part")
OUTPUT:
last part
inside API
"VALUE OF RES"
WHAT I EXPECT:
inside API
"VALUE OF RES"
last part
Any help will be highly appreciated..
javascript
javascript
asked Jan 3 at 15:47
Abhijeet kumarAbhijeet kumar
128
asked Jan 3 at 15:47
Abhijeet kumarAbhijeet kumar
128
asked Jan 3 at 15:47
Abhijeet kumarAbhijeet kumar
128
asked Jan 3 at 15:47
Abhijeet kumarAbhijeet kumar
128
asked Jan 3 at 15:47
Abhijeet kumarAbhijeet kumar
128
128
3
JavaScript is asynchronous, the "last part" runs first because theawaithasn't returned a result yet.
– doublesharp
Jan 3 at 15:50
2
you'll need toawaitweather.
– Daniel A. White
Jan 3 at 15:50
And toawait weather()the whole things needs to be inside anotherasyncfunction.
– doublesharp
Jan 3 at 15:51
add await to weather function
– Sachin Kadam
Jan 3 at 15:52
As you wrote it yourself,weather()is asynchronous. Meaning non-blocking, meaningconsole.log("last part")won't wait for it to complete.
– Jeremy Thille
Jan 3 at 15:52
add a comment |
3
JavaScript is asynchronous, the "last part" runs first because theawaithasn't returned a result yet.
– doublesharp
Jan 3 at 15:50
2
you'll need toawaitweather.
– Daniel A. White
Jan 3 at 15:50
And toawait weather()the whole things needs to be inside anotherasyncfunction.
– doublesharp
Jan 3 at 15:51
add await to weather function
– Sachin Kadam
Jan 3 at 15:52
As you wrote it yourself,weather()is asynchronous. Meaning non-blocking, meaningconsole.log("last part")won't wait for it to complete.
– Jeremy Thille
Jan 3 at 15:52
3
3
JavaScript is asynchronous, the "last part" runs first because the
await hasn't returned a result yet.– doublesharp
Jan 3 at 15:50
JavaScript is asynchronous, the "last part" runs first because the
await hasn't returned a result yet.– doublesharp
Jan 3 at 15:50
2
2
you'll need to
await weather.– Daniel A. White
Jan 3 at 15:50
you'll need to
await weather.– Daniel A. White
Jan 3 at 15:50
And to
await weather() the whole things needs to be inside another async function.– doublesharp
Jan 3 at 15:51
And to
await weather() the whole things needs to be inside another async function.– doublesharp
Jan 3 at 15:51
add await to weather function
– Sachin Kadam
Jan 3 at 15:52
add await to weather function
– Sachin Kadam
Jan 3 at 15:52
As you wrote it yourself,
weather() is asynchronous. Meaning non-blocking, meaning console.log("last part") won't wait for it to complete.– Jeremy Thille
Jan 3 at 15:52
As you wrote it yourself,
weather() is asynchronous. Meaning non-blocking, meaning console.log("last part") won't wait for it to complete.– Jeremy Thille
Jan 3 at 15:52
add a comment |
2 Answers
2
active
oldest
votes
1
The easiest way is to wrap it all in another async function so that you can await weather().
// this function fetches the weather then logs the result
async function weather() {
// API call
let data = await fetch('http://api.openweathermap.org/data/2.5/weather?'+'&lat=20&lon=44'+'&units=metric'+'&APPID='+WEATHER_KEY);
console.log('inside API');
const res = await data.json();
console.log(res);
}
// this async function awaits for weather() to return
async function run() {
await weather();
console.log('last part');
}
// this runs first
run();
answered Jan 3 at 15:54
doublesharpdoublesharp
19.7k33961
add a comment |
0
// async function
async function fetchAsync () {
// await response of fetch call
let response = await fetch('http://api.openweathermap.org/data/2.5/weather?'+'&lat=20&lon=44'+'&units=metric'+'&APPID='+WEATHER_KEY);
// only proceed once promise is resolved
let data = await response.json();
// only proceed once second promise is resolved
return data;
}
// trigger async function
// log response or catch error of fetch promise
fetchAsync()
.then(data => console.log(data))
.catch(err => console.log(err))
answered Jan 3 at 15:53
vinoth svinoth s
795
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
The easiest way is to wrap it all in another async function so that you can await weather().
// this function fetches the weather then logs the result
async function weather() {
// API call
let data = await fetch('http://api.openweathermap.org/data/2.5/weather?'+'&lat=20&lon=44'+'&units=metric'+'&APPID='+WEATHER_KEY);
console.log('inside API');
const res = await data.json();
console.log(res);
}
// this async function awaits for weather() to return
async function run() {
await weather();
console.log('last part');
}
// this runs first
run();
answered Jan 3 at 15:54
doublesharpdoublesharp
19.7k33961
add a comment |
1
The easiest way is to wrap it all in another async function so that you can await weather().
// this function fetches the weather then logs the result
async function weather() {
// API call
let data = await fetch('http://api.openweathermap.org/data/2.5/weather?'+'&lat=20&lon=44'+'&units=metric'+'&APPID='+WEATHER_KEY);
console.log('inside API');
const res = await data.json();
console.log(res);
}
// this async function awaits for weather() to return
async function run() {
await weather();
console.log('last part');
}
// this runs first
run();
answered Jan 3 at 15:54
doublesharpdoublesharp
19.7k33961
add a comment |
1
1
1
The easiest way is to wrap it all in another async function so that you can await weather().
// this function fetches the weather then logs the result
async function weather() {
// API call
let data = await fetch('http://api.openweathermap.org/data/2.5/weather?'+'&lat=20&lon=44'+'&units=metric'+'&APPID='+WEATHER_KEY);
console.log('inside API');
const res = await data.json();
console.log(res);
}
// this async function awaits for weather() to return
async function run() {
await weather();
console.log('last part');
}
// this runs first
run();
answered Jan 3 at 15:54
doublesharpdoublesharp
19.7k33961
The easiest way is to wrap it all in another async function so that you can await weather().
// this function fetches the weather then logs the result
async function weather() {
// API call
let data = await fetch('http://api.openweathermap.org/data/2.5/weather?'+'&lat=20&lon=44'+'&units=metric'+'&APPID='+WEATHER_KEY);
console.log('inside API');
const res = await data.json();
console.log(res);
}
// this async function awaits for weather() to return
async function run() {
await weather();
console.log('last part');
}
// this runs first
run();
answered Jan 3 at 15:54
doublesharpdoublesharp
19.7k33961
edited Jan 3 at 16:08
answered Jan 3 at 15:54
doublesharpdoublesharp
19.7k33961
answered Jan 3 at 15:54
doublesharpdoublesharp
19.7k33961
answered Jan 3 at 15:54
doublesharpdoublesharp
19.7k33961
19.7k33961
add a comment |
add a comment |
0
// async function
async function fetchAsync () {
// await response of fetch call
let response = await fetch('http://api.openweathermap.org/data/2.5/weather?'+'&lat=20&lon=44'+'&units=metric'+'&APPID='+WEATHER_KEY);
// only proceed once promise is resolved
let data = await response.json();
// only proceed once second promise is resolved
return data;
}
// trigger async function
// log response or catch error of fetch promise
fetchAsync()
.then(data => console.log(data))
.catch(err => console.log(err))
answered Jan 3 at 15:53
vinoth svinoth s
795
add a comment |
0
// async function
async function fetchAsync () {
// await response of fetch call
let response = await fetch('http://api.openweathermap.org/data/2.5/weather?'+'&lat=20&lon=44'+'&units=metric'+'&APPID='+WEATHER_KEY);
// only proceed once promise is resolved
let data = await response.json();
// only proceed once second promise is resolved
return data;
}
// trigger async function
// log response or catch error of fetch promise
fetchAsync()
.then(data => console.log(data))
.catch(err => console.log(err))
answered Jan 3 at 15:53
vinoth svinoth s
795
add a comment |
0
0
0
// async function
async function fetchAsync () {
// await response of fetch call
let response = await fetch('http://api.openweathermap.org/data/2.5/weather?'+'&lat=20&lon=44'+'&units=metric'+'&APPID='+WEATHER_KEY);
// only proceed once promise is resolved
let data = await response.json();
// only proceed once second promise is resolved
return data;
}
// trigger async function
// log response or catch error of fetch promise
fetchAsync()
.then(data => console.log(data))
.catch(err => console.log(err))
answered Jan 3 at 15:53
vinoth svinoth s
795
// async function
async function fetchAsync () {
// await response of fetch call
let response = await fetch('http://api.openweathermap.org/data/2.5/weather?'+'&lat=20&lon=44'+'&units=metric'+'&APPID='+WEATHER_KEY);
// only proceed once promise is resolved
let data = await response.json();
// only proceed once second promise is resolved
return data;
}
// trigger async function
// log response or catch error of fetch promise
fetchAsync()
.then(data => console.log(data))
.catch(err => console.log(err))
answered Jan 3 at 15:53
vinoth svinoth s
795
answered Jan 3 at 15:53
vinoth svinoth s
795
answered Jan 3 at 15:53
vinoth svinoth s
795
answered Jan 3 at 15:53
vinoth svinoth s
795
795
add a comment |
add a comment |
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3
JavaScript is asynchronous, the "last part" runs first because the
awaithasn't returned a result yet.– doublesharp
Jan 3 at 15:50
2
you'll need to
awaitweather.– Daniel A. White
Jan 3 at 15:50
And to
await weather()the whole things needs to be inside anotherasyncfunction.– doublesharp
Jan 3 at 15:51
add await to weather function
– Sachin Kadam
Jan 3 at 15:52
As you wrote it yourself,
weather()is asynchronous. Meaning non-blocking, meaningconsole.log("last part")won't wait for it to complete.– Jeremy Thille
Jan 3 at 15:52