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I'm trying to wrap my mind around promises in Javascript. I was in the illusion that once a Promise was resolved it could never go back to rejected. To test that I wrote a little script. I see that the first messages that come back are the resolve messages "1 resolve 2" etc. I did expect the first message to be "0 reject 1".
for (let i = 0; i < 10; i++)
let p = new Promise((resolve, reject) =>
let a = 1 + (i % 2)
if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)
)
p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
)
at the console:
[Log] 1 resolve 2
[Log] 3 resolve 2
[Log] 5 resolve 2
[Log] 7 resolve 2
[Log] 9 resolve 2
[Log] 0 reject 1
[Log] 2 reject 1
[Log] 4 reject 1
[Log] 6 reject 1
[Log] 8 reject 1
< Promise status: "pending"
thanks for you help....
After reading
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Promise/then
I got to this code. The catch is removed.
for (let i = 0; i < 10; i++)
let p = new Promise((resolve, reject) =>
let a = 1 + (i % 2)
if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)
)
p.then((message) =>
console.log(message)
, failed =>
console.log(failed)
)
at the console:
[Log] 0 reject 1
[Log] 1 resolve 2
[Log] 2 reject 1
[Log] 3 resolve 2
[Log] 4 reject 1
[Log] 5 resolve 2
[Log] 6 reject 1
[Log] 7 resolve 2
[Log] 8 reject 1
[Log] 9 resolve 2
< Promise status: "pending"
javascript es6-promise
|
show 1 more comment
I'm trying to wrap my mind around promises in Javascript. I was in the illusion that once a Promise was resolved it could never go back to rejected. To test that I wrote a little script. I see that the first messages that come back are the resolve messages "1 resolve 2" etc. I did expect the first message to be "0 reject 1".
for (let i = 0; i < 10; i++)
let p = new Promise((resolve, reject) =>
let a = 1 + (i % 2)
if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)
)
p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
)
at the console:
[Log] 1 resolve 2
[Log] 3 resolve 2
[Log] 5 resolve 2
[Log] 7 resolve 2
[Log] 9 resolve 2
[Log] 0 reject 1
[Log] 2 reject 1
[Log] 4 reject 1
[Log] 6 reject 1
[Log] 8 reject 1
< Promise status: "pending"
thanks for you help....
After reading
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Promise/then
I got to this code. The catch is removed.
for (let i = 0; i < 10; i++)
let p = new Promise((resolve, reject) =>
let a = 1 + (i % 2)
if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)
)
p.then((message) =>
console.log(message)
, failed =>
console.log(failed)
)
at the console:
[Log] 0 reject 1
[Log] 1 resolve 2
[Log] 2 reject 1
[Log] 3 resolve 2
[Log] 4 reject 1
[Log] 5 resolve 2
[Log] 6 reject 1
[Log] 7 resolve 2
[Log] 8 reject 1
[Log] 9 resolve 2
< Promise status: "pending"
javascript es6-promise
3
You are creating new promise on every iteration
– brk
1 hour ago
You have 10 promises, they don't "go back to rejected" like you said.
– Kev
1 hour ago
3
Is your question why you see the resolved promises before the rejected? I.e. why is the output [1, 3, 5, 7, 9, 0, 2, 4, 6, 8] instead of [0, 1, 2, 3, 4, ...]?
– junvar
1 hour ago
2
Yes Junvar. That is my question.
– Edwin
1 hour ago
3
99% sure it's because the.thenand.catcheach take a tick on the event loop. So the rejections are all a single tick behind your resolves.
– jhpratt
1 hour ago
|
show 1 more comment
I'm trying to wrap my mind around promises in Javascript. I was in the illusion that once a Promise was resolved it could never go back to rejected. To test that I wrote a little script. I see that the first messages that come back are the resolve messages "1 resolve 2" etc. I did expect the first message to be "0 reject 1".
for (let i = 0; i < 10; i++)
let p = new Promise((resolve, reject) =>
let a = 1 + (i % 2)
if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)
)
p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
)
at the console:
[Log] 1 resolve 2
[Log] 3 resolve 2
[Log] 5 resolve 2
[Log] 7 resolve 2
[Log] 9 resolve 2
[Log] 0 reject 1
[Log] 2 reject 1
[Log] 4 reject 1
[Log] 6 reject 1
[Log] 8 reject 1
< Promise status: "pending"
thanks for you help....
After reading
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Promise/then
I got to this code. The catch is removed.
for (let i = 0; i < 10; i++)
let p = new Promise((resolve, reject) =>
let a = 1 + (i % 2)
if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)
)
p.then((message) =>
console.log(message)
, failed =>
console.log(failed)
)
at the console:
[Log] 0 reject 1
[Log] 1 resolve 2
[Log] 2 reject 1
[Log] 3 resolve 2
[Log] 4 reject 1
[Log] 5 resolve 2
[Log] 6 reject 1
[Log] 7 resolve 2
[Log] 8 reject 1
[Log] 9 resolve 2
< Promise status: "pending"
javascript es6-promise
I'm trying to wrap my mind around promises in Javascript. I was in the illusion that once a Promise was resolved it could never go back to rejected. To test that I wrote a little script. I see that the first messages that come back are the resolve messages "1 resolve 2" etc. I did expect the first message to be "0 reject 1".
for (let i = 0; i < 10; i++)
let p = new Promise((resolve, reject) =>
let a = 1 + (i % 2)
if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)
)
p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
)
at the console:
[Log] 1 resolve 2
[Log] 3 resolve 2
[Log] 5 resolve 2
[Log] 7 resolve 2
[Log] 9 resolve 2
[Log] 0 reject 1
[Log] 2 reject 1
[Log] 4 reject 1
[Log] 6 reject 1
[Log] 8 reject 1
< Promise status: "pending"
thanks for you help....
After reading
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Promise/then
I got to this code. The catch is removed.
for (let i = 0; i < 10; i++)
let p = new Promise((resolve, reject) =>
let a = 1 + (i % 2)
if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)
)
p.then((message) =>
console.log(message)
, failed =>
console.log(failed)
)
at the console:
[Log] 0 reject 1
[Log] 1 resolve 2
[Log] 2 reject 1
[Log] 3 resolve 2
[Log] 4 reject 1
[Log] 5 resolve 2
[Log] 6 reject 1
[Log] 7 resolve 2
[Log] 8 reject 1
[Log] 9 resolve 2
< Promise status: "pending"
javascript es6-promise
javascript es6-promise
edited 38 mins ago
Edwin
asked 1 hour ago
EdwinEdwin
513
513
3
You are creating new promise on every iteration
– brk
1 hour ago
You have 10 promises, they don't "go back to rejected" like you said.
– Kev
1 hour ago
3
Is your question why you see the resolved promises before the rejected? I.e. why is the output [1, 3, 5, 7, 9, 0, 2, 4, 6, 8] instead of [0, 1, 2, 3, 4, ...]?
– junvar
1 hour ago
2
Yes Junvar. That is my question.
– Edwin
1 hour ago
3
99% sure it's because the.thenand.catcheach take a tick on the event loop. So the rejections are all a single tick behind your resolves.
– jhpratt
1 hour ago
|
show 1 more comment
3
You are creating new promise on every iteration
– brk
1 hour ago
You have 10 promises, they don't "go back to rejected" like you said.
– Kev
1 hour ago
3
Is your question why you see the resolved promises before the rejected? I.e. why is the output [1, 3, 5, 7, 9, 0, 2, 4, 6, 8] instead of [0, 1, 2, 3, 4, ...]?
– junvar
1 hour ago
2
Yes Junvar. That is my question.
– Edwin
1 hour ago
3
99% sure it's because the.thenand.catcheach take a tick on the event loop. So the rejections are all a single tick behind your resolves.
– jhpratt
1 hour ago
3
3
You are creating new promise on every iteration
– brk
1 hour ago
You are creating new promise on every iteration
– brk
1 hour ago
You have 10 promises, they don't "go back to rejected" like you said.
– Kev
1 hour ago
You have 10 promises, they don't "go back to rejected" like you said.
– Kev
1 hour ago
3
3
Is your question why you see the resolved promises before the rejected? I.e. why is the output [1, 3, 5, 7, 9, 0, 2, 4, 6, 8] instead of [0, 1, 2, 3, 4, ...]?
– junvar
1 hour ago
Is your question why you see the resolved promises before the rejected? I.e. why is the output [1, 3, 5, 7, 9, 0, 2, 4, 6, 8] instead of [0, 1, 2, 3, 4, ...]?
– junvar
1 hour ago
2
2
Yes Junvar. That is my question.
– Edwin
1 hour ago
Yes Junvar. That is my question.
– Edwin
1 hour ago
3
3
99% sure it's because the
.then and .catch each take a tick on the event loop. So the rejections are all a single tick behind your resolves.– jhpratt
1 hour ago
99% sure it's because the
.then and .catch each take a tick on the event loop. So the rejections are all a single tick behind your resolves.– jhpratt
1 hour ago
|
show 1 more comment
4 Answers
4
active
oldest
votes
You can see what's going on under the hood by using your the console of your browser's dev tools and, possibly, setting break points (this articles might be helpful if u're using Chrome or Firefox):

As you can see, all your 10 promises are created before any of them are executed (resolve/reject).
Interestingly, in your code the resolved promises are handled first.
If you define the handlers in two separate definitions, you'll get the expected results:
p.then((message) =>
console.log(message)
)
p.catch((message) =>
console.log(message)
)
Output:

i am so surprize by the conclusion of your answer. I never imagined than chain vs multiple instruction can have this huge impact. Really good sharing
– Yanis-git
41 mins ago
add a comment |
The point is, Promises are made to be used as Asynchronous calls, so when you execute your loop and for each iteration you creates a new promise, you are creating new instances, and each one of those can be executed in their own time.
But what this even means? The explanation is, when you create 10 new Promises in a loop, each promise will be executed in his own time and probablly will mess up with your promise solving order.
What you can do to solve it? You can use await command to wait each promise to solve, like the code bellow:
for (let i = 0; i < 10; i++)
let p = new Promise((resolve, reject) =>
let a = 1 + (i % 2)
if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)
)
await p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
)
Or you could try use Promise.all() which will basically solve the order for you, see the official docs here
1
well, you can remove the.thenif you're usingawait
– pushkin
50 mins ago
1
Yes I could use async-await; and do prefer that. It's that I want to get to understand the promise.
– Edwin
47 mins ago
add a comment |
Because JavaScript are mono thread :
- promise
- eventListener
- setTimeout
- setInterval
previous listed method are not part of javascript enterpreter (V8 Engine for example), it delegate to the event loop which are part of browser or nodejs. more information here
Basically this code are delegate to 3th party (node, browser) which will decide himself when and on which order this collection of microtasks will be executed and return to the main thread.
is Why Following code :
let p = new Promise((resolve, reject) =>
resolve('micro task thread');
);
p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
);
console.log('main thread');will print "main thread" then "micro task thread" but you don't have any obvious delayed code. Is because all main thread function call will be execute before doing task on eventLoop.
Most of the time, event loop will execute the collection as first come first rendered.
@Edwin i have not understand what you try to say. Ludovico post interesting answer. And my answer contain, i think, some tips to understand how async traitment work under the hood.
– Yanis-git
39 mins ago
add a comment |
I did found a solution here:
MDN promise then
for (let i = 0; i < 10; i++)
let p = new Promise((resolve, reject) =>
let a = 1 + (i % 2)
if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)
)
p.then((message) =>
console.log(message)
, failed =>
console.log(failed)
)
I will note that though this is absolutely valid, most developers will never wrote code like that (and some would probably be confused as to it's behavior).
– jhpratt
30 mins ago
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can see what's going on under the hood by using your the console of your browser's dev tools and, possibly, setting break points (this articles might be helpful if u're using Chrome or Firefox):

As you can see, all your 10 promises are created before any of them are executed (resolve/reject).
Interestingly, in your code the resolved promises are handled first.
If you define the handlers in two separate definitions, you'll get the expected results:
p.then((message) =>
console.log(message)
)
p.catch((message) =>
console.log(message)
)
Output:

i am so surprize by the conclusion of your answer. I never imagined than chain vs multiple instruction can have this huge impact. Really good sharing
– Yanis-git
41 mins ago
add a comment |
You can see what's going on under the hood by using your the console of your browser's dev tools and, possibly, setting break points (this articles might be helpful if u're using Chrome or Firefox):

As you can see, all your 10 promises are created before any of them are executed (resolve/reject).
Interestingly, in your code the resolved promises are handled first.
If you define the handlers in two separate definitions, you'll get the expected results:
p.then((message) =>
console.log(message)
)
p.catch((message) =>
console.log(message)
)
Output:

i am so surprize by the conclusion of your answer. I never imagined than chain vs multiple instruction can have this huge impact. Really good sharing
– Yanis-git
41 mins ago
add a comment |
You can see what's going on under the hood by using your the console of your browser's dev tools and, possibly, setting break points (this articles might be helpful if u're using Chrome or Firefox):

As you can see, all your 10 promises are created before any of them are executed (resolve/reject).
Interestingly, in your code the resolved promises are handled first.
If you define the handlers in two separate definitions, you'll get the expected results:
p.then((message) =>
console.log(message)
)
p.catch((message) =>
console.log(message)
)
Output:

You can see what's going on under the hood by using your the console of your browser's dev tools and, possibly, setting break points (this articles might be helpful if u're using Chrome or Firefox):

As you can see, all your 10 promises are created before any of them are executed (resolve/reject).
Interestingly, in your code the resolved promises are handled first.
If you define the handlers in two separate definitions, you'll get the expected results:
p.then((message) =>
console.log(message)
)
p.catch((message) =>
console.log(message)
)
Output:

answered 1 hour ago
ludovicoludovico
775
775
i am so surprize by the conclusion of your answer. I never imagined than chain vs multiple instruction can have this huge impact. Really good sharing
– Yanis-git
41 mins ago
add a comment |
i am so surprize by the conclusion of your answer. I never imagined than chain vs multiple instruction can have this huge impact. Really good sharing
– Yanis-git
41 mins ago
i am so surprize by the conclusion of your answer. I never imagined than chain vs multiple instruction can have this huge impact. Really good sharing
– Yanis-git
41 mins ago
i am so surprize by the conclusion of your answer. I never imagined than chain vs multiple instruction can have this huge impact. Really good sharing
– Yanis-git
41 mins ago
add a comment |
The point is, Promises are made to be used as Asynchronous calls, so when you execute your loop and for each iteration you creates a new promise, you are creating new instances, and each one of those can be executed in their own time.
But what this even means? The explanation is, when you create 10 new Promises in a loop, each promise will be executed in his own time and probablly will mess up with your promise solving order.
What you can do to solve it? You can use await command to wait each promise to solve, like the code bellow:
for (let i = 0; i < 10; i++)
let p = new Promise((resolve, reject) =>
let a = 1 + (i % 2)
if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)
)
await p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
)
Or you could try use Promise.all() which will basically solve the order for you, see the official docs here
1
well, you can remove the.thenif you're usingawait
– pushkin
50 mins ago
1
Yes I could use async-await; and do prefer that. It's that I want to get to understand the promise.
– Edwin
47 mins ago
add a comment |
The point is, Promises are made to be used as Asynchronous calls, so when you execute your loop and for each iteration you creates a new promise, you are creating new instances, and each one of those can be executed in their own time.
But what this even means? The explanation is, when you create 10 new Promises in a loop, each promise will be executed in his own time and probablly will mess up with your promise solving order.
What you can do to solve it? You can use await command to wait each promise to solve, like the code bellow:
for (let i = 0; i < 10; i++)
let p = new Promise((resolve, reject) =>
let a = 1 + (i % 2)
if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)
)
await p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
)
Or you could try use Promise.all() which will basically solve the order for you, see the official docs here
1
well, you can remove the.thenif you're usingawait
– pushkin
50 mins ago
1
Yes I could use async-await; and do prefer that. It's that I want to get to understand the promise.
– Edwin
47 mins ago
add a comment |
The point is, Promises are made to be used as Asynchronous calls, so when you execute your loop and for each iteration you creates a new promise, you are creating new instances, and each one of those can be executed in their own time.
But what this even means? The explanation is, when you create 10 new Promises in a loop, each promise will be executed in his own time and probablly will mess up with your promise solving order.
What you can do to solve it? You can use await command to wait each promise to solve, like the code bellow:
for (let i = 0; i < 10; i++)
let p = new Promise((resolve, reject) =>
let a = 1 + (i % 2)
if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)
)
await p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
)
Or you could try use Promise.all() which will basically solve the order for you, see the official docs here
The point is, Promises are made to be used as Asynchronous calls, so when you execute your loop and for each iteration you creates a new promise, you are creating new instances, and each one of those can be executed in their own time.
But what this even means? The explanation is, when you create 10 new Promises in a loop, each promise will be executed in his own time and probablly will mess up with your promise solving order.
What you can do to solve it? You can use await command to wait each promise to solve, like the code bellow:
for (let i = 0; i < 10; i++)
let p = new Promise((resolve, reject) =>
let a = 1 + (i % 2)
if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)
)
await p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
)
Or you could try use Promise.all() which will basically solve the order for you, see the official docs here
answered 1 hour ago
Esdras XavierEsdras Xavier
45017
45017
1
well, you can remove the.thenif you're usingawait
– pushkin
50 mins ago
1
Yes I could use async-await; and do prefer that. It's that I want to get to understand the promise.
– Edwin
47 mins ago
add a comment |
1
well, you can remove the.thenif you're usingawait
– pushkin
50 mins ago
1
Yes I could use async-await; and do prefer that. It's that I want to get to understand the promise.
– Edwin
47 mins ago
1
1
well, you can remove the
.then if you're using await– pushkin
50 mins ago
well, you can remove the
.then if you're using await– pushkin
50 mins ago
1
1
Yes I could use async-await; and do prefer that. It's that I want to get to understand the promise.
– Edwin
47 mins ago
Yes I could use async-await; and do prefer that. It's that I want to get to understand the promise.
– Edwin
47 mins ago
add a comment |
Because JavaScript are mono thread :
- promise
- eventListener
- setTimeout
- setInterval
previous listed method are not part of javascript enterpreter (V8 Engine for example), it delegate to the event loop which are part of browser or nodejs. more information here
Basically this code are delegate to 3th party (node, browser) which will decide himself when and on which order this collection of microtasks will be executed and return to the main thread.
is Why Following code :
let p = new Promise((resolve, reject) =>
resolve('micro task thread');
);
p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
);
console.log('main thread');will print "main thread" then "micro task thread" but you don't have any obvious delayed code. Is because all main thread function call will be execute before doing task on eventLoop.
Most of the time, event loop will execute the collection as first come first rendered.
@Edwin i have not understand what you try to say. Ludovico post interesting answer. And my answer contain, i think, some tips to understand how async traitment work under the hood.
– Yanis-git
39 mins ago
add a comment |
Because JavaScript are mono thread :
- promise
- eventListener
- setTimeout
- setInterval
previous listed method are not part of javascript enterpreter (V8 Engine for example), it delegate to the event loop which are part of browser or nodejs. more information here
Basically this code are delegate to 3th party (node, browser) which will decide himself when and on which order this collection of microtasks will be executed and return to the main thread.
is Why Following code :
let p = new Promise((resolve, reject) =>
resolve('micro task thread');
);
p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
);
console.log('main thread');will print "main thread" then "micro task thread" but you don't have any obvious delayed code. Is because all main thread function call will be execute before doing task on eventLoop.
Most of the time, event loop will execute the collection as first come first rendered.
@Edwin i have not understand what you try to say. Ludovico post interesting answer. And my answer contain, i think, some tips to understand how async traitment work under the hood.
– Yanis-git
39 mins ago
add a comment |
Because JavaScript are mono thread :
- promise
- eventListener
- setTimeout
- setInterval
previous listed method are not part of javascript enterpreter (V8 Engine for example), it delegate to the event loop which are part of browser or nodejs. more information here
Basically this code are delegate to 3th party (node, browser) which will decide himself when and on which order this collection of microtasks will be executed and return to the main thread.
is Why Following code :
let p = new Promise((resolve, reject) =>
resolve('micro task thread');
);
p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
);
console.log('main thread');will print "main thread" then "micro task thread" but you don't have any obvious delayed code. Is because all main thread function call will be execute before doing task on eventLoop.
Most of the time, event loop will execute the collection as first come first rendered.
Because JavaScript are mono thread :
- promise
- eventListener
- setTimeout
- setInterval
previous listed method are not part of javascript enterpreter (V8 Engine for example), it delegate to the event loop which are part of browser or nodejs. more information here
Basically this code are delegate to 3th party (node, browser) which will decide himself when and on which order this collection of microtasks will be executed and return to the main thread.
is Why Following code :
let p = new Promise((resolve, reject) =>
resolve('micro task thread');
);
p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
);
console.log('main thread');will print "main thread" then "micro task thread" but you don't have any obvious delayed code. Is because all main thread function call will be execute before doing task on eventLoop.
Most of the time, event loop will execute the collection as first come first rendered.
let p = new Promise((resolve, reject) =>
resolve('micro task thread');
);
p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
);
console.log('main thread'); let p = new Promise((resolve, reject) =>
resolve('micro task thread');
);
p.then((message) =>
console.log(message)
).catch((message) =>
console.log(message)
);
console.log('main thread');answered 49 mins ago
Yanis-gitYanis-git
2,6291725
2,6291725
@Edwin i have not understand what you try to say. Ludovico post interesting answer. And my answer contain, i think, some tips to understand how async traitment work under the hood.
– Yanis-git
39 mins ago
add a comment |
@Edwin i have not understand what you try to say. Ludovico post interesting answer. And my answer contain, i think, some tips to understand how async traitment work under the hood.
– Yanis-git
39 mins ago
@Edwin i have not understand what you try to say. Ludovico post interesting answer. And my answer contain, i think, some tips to understand how async traitment work under the hood.
– Yanis-git
39 mins ago
@Edwin i have not understand what you try to say. Ludovico post interesting answer. And my answer contain, i think, some tips to understand how async traitment work under the hood.
– Yanis-git
39 mins ago
add a comment |
I did found a solution here:
MDN promise then
for (let i = 0; i < 10; i++)
let p = new Promise((resolve, reject) =>
let a = 1 + (i % 2)
if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)
)
p.then((message) =>
console.log(message)
, failed =>
console.log(failed)
)
I will note that though this is absolutely valid, most developers will never wrote code like that (and some would probably be confused as to it's behavior).
– jhpratt
30 mins ago
add a comment |
I did found a solution here:
MDN promise then
for (let i = 0; i < 10; i++)
let p = new Promise((resolve, reject) =>
let a = 1 + (i % 2)
if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)
)
p.then((message) =>
console.log(message)
, failed =>
console.log(failed)
)
I will note that though this is absolutely valid, most developers will never wrote code like that (and some would probably be confused as to it's behavior).
– jhpratt
30 mins ago
add a comment |
I did found a solution here:
MDN promise then
for (let i = 0; i < 10; i++)
let p = new Promise((resolve, reject) =>
let a = 1 + (i % 2)
if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)
)
p.then((message) =>
console.log(message)
, failed =>
console.log(failed)
)
I did found a solution here:
MDN promise then
for (let i = 0; i < 10; i++)
let p = new Promise((resolve, reject) =>
let a = 1 + (i % 2)
if (a === 2)
resolve(i + ' resolve ' + a)
else
reject(i + ' reject ' + a)
)
p.then((message) =>
console.log(message)
, failed =>
console.log(failed)
)
answered 33 mins ago
EdwinEdwin
513
513
I will note that though this is absolutely valid, most developers will never wrote code like that (and some would probably be confused as to it's behavior).
– jhpratt
30 mins ago
add a comment |
I will note that though this is absolutely valid, most developers will never wrote code like that (and some would probably be confused as to it's behavior).
– jhpratt
30 mins ago
I will note that though this is absolutely valid, most developers will never wrote code like that (and some would probably be confused as to it's behavior).
– jhpratt
30 mins ago
I will note that though this is absolutely valid, most developers will never wrote code like that (and some would probably be confused as to it's behavior).
– jhpratt
30 mins ago
add a comment |
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3
You are creating new promise on every iteration
– brk
1 hour ago
You have 10 promises, they don't "go back to rejected" like you said.
– Kev
1 hour ago
3
Is your question why you see the resolved promises before the rejected? I.e. why is the output [1, 3, 5, 7, 9, 0, 2, 4, 6, 8] instead of [0, 1, 2, 3, 4, ...]?
– junvar
1 hour ago
2
Yes Junvar. That is my question.
– Edwin
1 hour ago
3
99% sure it's because the
.thenand.catcheach take a tick on the event loop. So the rejections are all a single tick behind your resolves.– jhpratt
1 hour ago