Reputation: 757
Hi everyone I have such a problem, I have 2 asynchronous functions. I want only after the first is completely over, to run the second. This is what I tried to do:
run2functions = async () => {
await firstFunc();
await secondFunc();
};
firstFunc = () => {
console.log("first one");
//Api code for information from any server
}
secondFunc = () => {
console.log("second one");
//Api code for information from any server
}
run2functions();
But it does not always work, sometimes the code of the second function runs before the code of the first function, I'm not sure why, I used await to force the second to be only after the first one ends.
I only want the first function to end now to activate the second function.
Upvotes: 5
Views: 2602
Reputation: 1
async function rango(){
let amp = new Promise(function(resolve,reject){
setTimeout(function(){
console.log("First Promise");
},2000);
});
let volt = new Promise(function(resolve,reject){
setTimeout(function(){
console.log("Second Promise");
},3000);
});
let a = await amp;
let v = await volt;
return[a,v];
}
rango();
Upvotes: 0
Reputation: 206102
Make async
the functions that are await
able (return a Promise)
// DEMO ONLY Just a helper to wait some ms time and return a Promise
const wait = (t) => new Promise((r) => setTimeout(r, t));
const firstFunc = async () => {
await wait(1000); // Just to fake some wait time
console.log("first one");
}
const secondFunc = () => { // This does not need to be async
console.log("second one");
}
const run2functions = async() => {
await firstFunc(); // Await for this one
secondFunc(); // You don't need to await for this one
};
run2functions();
Will result in:
(waiting 1 sec....)
"first one"
"second one"
If you're waiting for both responses (i.e: one function takes 3sec to resolve, and the other one takes 2sec to resolve):
use Promise.all
// DEMO ONLY Just a helper to wait some ms time and return a Promise
const fakeFetch = (time, data) => new Promise((res, rej) => setTimeout(res, time, data));
// Functions that return a Promise (just like JS's fetch());
const one = () => fakeFetch( 3000, {message:"First!"} );
const two = () => fakeFetch( 2000, {message:"Second!"} );
Promise.all([one(), two()]).then((values) => {
// After 5 sec...
console.log(values); // In the exact order as the functions calls array
});
A real-world example of the above would be like:
const getJSON = (url) => fetch(url).then(res => res.json()); // Returns a Promise
Promise.all([getJSON("users.json"), getJSON("tasks.json")]).then((JSONs) => {
// After some unknown time... Both fetch Promises are resolved.
// Do some work with both JSON data:
console.log(JSONs); // In the exact order as the functions calls array
});
Upvotes: 5
Reputation: 26
function run2functions() {
firstFunc(secondFunc);
}
function firstFunc(cb) {
setTimeout(() => {
},1000);
cb();
console.log("first one");
}
function secondFunc() {
setTimeout(() => {
console.log("second one");
},1000);
}
run2functions();
Upvotes: -1
Reputation: 2494
If function one doesn't explicitly return a promise, and runs some async code, you can run into the situation you are describing.
You can solve this in two forms:
1 - Make firstFunc
async and make it only finish after all code has run
const firstFunc = async () => {
await getApiResponse();
...
}
2 - Make firstFunc return a Promise, that will make your main function properly await for it before moving on
const firstFunc = () => {
return getApiResponse();
}
Upvotes: 2
Reputation: 655
Async/await works only with fuctions which return Promise
. So your code should look something like that:
const run2functions = async () => {
await firstFunc();
await secondFunc();
};
const firstFunc = () => {
return new Promise((res) => {
// your async code here
console.log("first one");
resolve(res);
});
};
const secondFunc = () => {
return new Promise((res) => {
// your async code here
console.log("second one");
resolve(res);
});
};
await run2functions();
Additional resources
Upvotes: 2