function pro(){
return new Promise(resolve=>{
resolve()
})
}
function a(){
return pro().then(res=>{
setTimeout(()=>{
console.log(1)
},2000)
})
}
function b(){
return pro().then(res=>{
console.log(2)
})
}
function c(){
return pro().then(res=>{
console.log(3)
})
}
a().then(b).then(c);
//2 3 1
Why isn"t the final result 1 / 2 / 3?
is it correct to return Promise in this way?
because your a method just creates a timer, it doesn't mean that the timer callback is executed.
function a() {
return pro().then(res => {
return new Promise((reslove) => {
setTimeout(() => {
console.log(1)
reslove();
}, 2000)
})
})
}there is nothing wrong with grammar, but I think what you are trying to say is:
function pro() {
return new Promise(resolve => {
resolve()
})
}
function a() {
return pro().then(res => {
return new Promise((resolve, reject) => {
setTimeout(() => {
console.log(1);
resolve()
},
2000)
})
})
}
function b() {
return pro().then(res => {
console.log(2)
})
}
function c() {
return pro().then(res => {
console.log(3)
})
}
a().then(b).then(c);your code, a, immediately returns a promise of resolved, so it goes straight to b without waiting. To output sequentially, you must resolve
in setTimeout The order ofis no problem, except that setTimeout is a macro task and others are micro tasks. For more information, please see https://codeshelper.com/a/11...
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in Promise, no matter where the resolve is, is it always executed last? Who can talk about the operating mechanism, or share a link. Thank you! the code is as follows: actions.getForbiddenFuncList = function(context){ return new Promise((resolve, re...