I'm learning how Asynchronous JavaScript works, and I'm trying to print to the console the numbers 1 and 2 (in this order). The function that logs 1 has a setTimeout, and as such, the order is always inverted. I know why this happens. I just don't know how to make it work as I would like to. I've tried this:
function first(){
setTimeout(
()=>console.log(1),
1000
)
return Promise.resolve(true)
}
function second(){
console.log(2)
}
function a(){
first()
.then(second())
}
console.log("running...")
a()
and also this:
async function first(){
setTimeout(
()=>console.log(1),
2000
)
return true
}
function second(){
console.log(2)
}
async function a(){
await first()
second()
}
console.log("running...")
a()
Both of which print
running...
2
1
The desired output would be
running...
1
2
What am I doing wrong?
Would callbacks not be acceptable?
function first(callback) {
setTimeout(() => {
console.log(1);
callback();
}, 2000);
}
function second() {
console.log(2)
}
// When first is completed, second will run.
first(() => second());
console.log("This can run at any time, while we're waiting for our callback.");Read about Js promises (developer.mozilla.org) here his your code working
function first(){
return new Promise((resolve)=>{
console.log('waiting')
setTimeout( ()=>{console.log(1); resolve('waiting finished')}, 2000 )
}
);
}
function second(){
console.log(2)
}
async function a(){
await first().then(res => console.log(res))
second()
}
console.log("running...")
a()