Promise.all.then Not all code paths return a value
I have the following function that I want to return Promise< number >
async fetchCommentLines(commitDict: CommitDict): Promise < number > {
if (GitCommentService.isLoggedIn()) {
const commentLines = Object.values(commitDict).map(async commit => {
// ...
// do the job and return number
// ...
return lineNums;
});
Promise.all(commentLines)
.then(commitLines => {
return Array.prototype.concat.apply(, commitLines);
});
} else {
return as number;
}
}
Firstly I got "function lacks ending return statement and return type does not include 'undefined'"
And then I added undefined (so return type becomes Promise< number | undefined >)
But I'm getting "not all code paths return a value" this time.
It seems I'm not considering a possible code path with below code
Promise.all(...)
.then(val => {return ...})
What I'm missing?
I also tried this
Promise.all(...)
.then(val => {return ...})
.catch(e => {return ...})
But it was not helpful
Note: My main purpose is to return Promise< number >, not Promise< number | undefined >
javascript typescript promise async-await es6-promise
asked Dec 29 '18 at 13:26
user3790180user3790180
10819
add a comment |
I have the following function that I want to return Promise< number >
async fetchCommentLines(commitDict: CommitDict): Promise < number > {
if (GitCommentService.isLoggedIn()) {
const commentLines = Object.values(commitDict).map(async commit => {
// ...
// do the job and return number
// ...
return lineNums;
});
Promise.all(commentLines)
.then(commitLines => {
return Array.prototype.concat.apply(, commitLines);
});
} else {
return as number;
}
}
Firstly I got "function lacks ending return statement and return type does not include 'undefined'"
And then I added undefined (so return type becomes Promise< number | undefined >)
But I'm getting "not all code paths return a value" this time.
It seems I'm not considering a possible code path with below code
Promise.all(...)
.then(val => {return ...})
What I'm missing?
I also tried this
Promise.all(...)
.then(val => {return ...})
.catch(e => {return ...})
But it was not helpful
Note: My main purpose is to return Promise< number >, not Promise< number | undefined >
javascript typescript promise async-await es6-promise
asked Dec 29 '18 at 13:26
user3790180user3790180
10819
add a comment |
I have the following function that I want to return Promise< number >
async fetchCommentLines(commitDict: CommitDict): Promise < number > {
if (GitCommentService.isLoggedIn()) {
const commentLines = Object.values(commitDict).map(async commit => {
// ...
// do the job and return number
// ...
return lineNums;
});
Promise.all(commentLines)
.then(commitLines => {
return Array.prototype.concat.apply(, commitLines);
});
} else {
return as number;
}
}
Firstly I got "function lacks ending return statement and return type does not include 'undefined'"
And then I added undefined (so return type becomes Promise< number | undefined >)
But I'm getting "not all code paths return a value" this time.
It seems I'm not considering a possible code path with below code
Promise.all(...)
.then(val => {return ...})
What I'm missing?
I also tried this
Promise.all(...)
.then(val => {return ...})
.catch(e => {return ...})
But it was not helpful
Note: My main purpose is to return Promise< number >, not Promise< number | undefined >
javascript typescript promise async-await es6-promise
asked Dec 29 '18 at 13:26
user3790180user3790180
10819
I have the following function that I want to return Promise< number >
async fetchCommentLines(commitDict: CommitDict): Promise < number > {
if (GitCommentService.isLoggedIn()) {
const commentLines = Object.values(commitDict).map(async commit => {
// ...
// do the job and return number
// ...
return lineNums;
});
Promise.all(commentLines)
.then(commitLines => {
return Array.prototype.concat.apply(, commitLines);
});
} else {
return as number;
}
}
Firstly I got "function lacks ending return statement and return type does not include 'undefined'"
And then I added undefined (so return type becomes Promise< number | undefined >)
But I'm getting "not all code paths return a value" this time.
It seems I'm not considering a possible code path with below code
Promise.all(...)
.then(val => {return ...})
What I'm missing?
I also tried this
Promise.all(...)
.then(val => {return ...})
.catch(e => {return ...})
But it was not helpful
Note: My main purpose is to return Promise< number >, not Promise< number | undefined >
javascript typescript promise async-await es6-promise
javascript typescript promise async-await es6-promise
asked Dec 29 '18 at 13:26
user3790180user3790180
10819
asked Dec 29 '18 at 13:26
user3790180user3790180
10819
edited Dec 29 '18 at 13:33
user3790180
asked Dec 29 '18 at 13:26
user3790180user3790180
10819
asked Dec 29 '18 at 13:26
user3790180user3790180
10819
asked Dec 29 '18 at 13:26
user3790180user3790180
10819
10819
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Your branch with Promise.all never issues a return (value) statement. The then callback does, but not the code outside the then callback.
You probably want to return the result of Promise.all().then().
return Promise.all(commentLines)
// ^^^^^^
.then(commitLines => {
return Array.prototype.concat.apply(, commitLines);
});
answered Dec 29 '18 at 13:29
T.J. CrowderT.J. Crowder
682k12112081305
1
My purpose with it was to flatten anumberintonumberSo[[1,2],[3,4]]into[1,2,3,4]I'll try the other ways you mentioned
– user3790180
Dec 29 '18 at 13:46
@user3790180 - Ah! In that case, ignore me. :-) (I've removed that part of the answer.) You might look intoArray.prototype.flat. It's at Stage 3 and already supported by Chrome and Firefox, easily polyfilled for others... :-) That would bereturn Promise.all(commentLines).then(commitLines => commitLines.flat());
– T.J. Crowder
Dec 29 '18 at 13:55
add a comment |
you should return the Promise
return Promise.all(commentLines).then(...)
//....
or wait for the promise and return the result object
let lines = await Promise.all(commentLines)
return .concat(...lines)
edited Dec 29 '18 at 13:41
Bergi
367k58546873
answered Dec 29 '18 at 13:29
apple appleapple apple
2,5401723
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
Your branch with Promise.all never issues a return (value) statement. The then callback does, but not the code outside the then callback.
You probably want to return the result of Promise.all().then().
return Promise.all(commentLines)
// ^^^^^^
.then(commitLines => {
return Array.prototype.concat.apply(, commitLines);
});
answered Dec 29 '18 at 13:29
T.J. CrowderT.J. Crowder
682k12112081305
1
My purpose with it was to flatten anumberintonumberSo[[1,2],[3,4]]into[1,2,3,4]I'll try the other ways you mentioned
– user3790180
Dec 29 '18 at 13:46
@user3790180 - Ah! In that case, ignore me. :-) (I've removed that part of the answer.) You might look intoArray.prototype.flat. It's at Stage 3 and already supported by Chrome and Firefox, easily polyfilled for others... :-) That would bereturn Promise.all(commentLines).then(commitLines => commitLines.flat());
– T.J. Crowder
Dec 29 '18 at 13:55
add a comment |
Your branch with Promise.all never issues a return (value) statement. The then callback does, but not the code outside the then callback.
You probably want to return the result of Promise.all().then().
return Promise.all(commentLines)
// ^^^^^^
.then(commitLines => {
return Array.prototype.concat.apply(, commitLines);
});
answered Dec 29 '18 at 13:29
T.J. CrowderT.J. Crowder
682k12112081305
1
My purpose with it was to flatten anumberintonumberSo[[1,2],[3,4]]into[1,2,3,4]I'll try the other ways you mentioned
– user3790180
Dec 29 '18 at 13:46
@user3790180 - Ah! In that case, ignore me. :-) (I've removed that part of the answer.) You might look intoArray.prototype.flat. It's at Stage 3 and already supported by Chrome and Firefox, easily polyfilled for others... :-) That would bereturn Promise.all(commentLines).then(commitLines => commitLines.flat());
– T.J. Crowder
Dec 29 '18 at 13:55
add a comment |
Your branch with Promise.all never issues a return (value) statement. The then callback does, but not the code outside the then callback.
You probably want to return the result of Promise.all().then().
return Promise.all(commentLines)
// ^^^^^^
.then(commitLines => {
return Array.prototype.concat.apply(, commitLines);
});
answered Dec 29 '18 at 13:29
T.J. CrowderT.J. Crowder
682k12112081305
Your branch with Promise.all never issues a return (value) statement. The then callback does, but not the code outside the then callback.
You probably want to return the result of Promise.all().then().
return Promise.all(commentLines)
// ^^^^^^
.then(commitLines => {
return Array.prototype.concat.apply(, commitLines);
});
answered Dec 29 '18 at 13:29
T.J. CrowderT.J. Crowder
682k12112081305
edited Dec 29 '18 at 13:56
answered Dec 29 '18 at 13:29
T.J. CrowderT.J. Crowder
682k12112081305
answered Dec 29 '18 at 13:29
T.J. CrowderT.J. Crowder
682k12112081305
answered Dec 29 '18 at 13:29
T.J. CrowderT.J. Crowder
682k12112081305
682k12112081305
1
My purpose with it was to flatten anumberintonumberSo[[1,2],[3,4]]into[1,2,3,4]I'll try the other ways you mentioned
– user3790180
Dec 29 '18 at 13:46
@user3790180 - Ah! In that case, ignore me. :-) (I've removed that part of the answer.) You might look intoArray.prototype.flat. It's at Stage 3 and already supported by Chrome and Firefox, easily polyfilled for others... :-) That would bereturn Promise.all(commentLines).then(commitLines => commitLines.flat());
– T.J. Crowder
Dec 29 '18 at 13:55
add a comment |
1
My purpose with it was to flatten anumberintonumberSo[[1,2],[3,4]]into[1,2,3,4]I'll try the other ways you mentioned
– user3790180
Dec 29 '18 at 13:46
@user3790180 - Ah! In that case, ignore me. :-) (I've removed that part of the answer.) You might look intoArray.prototype.flat. It's at Stage 3 and already supported by Chrome and Firefox, easily polyfilled for others... :-) That would bereturn Promise.all(commentLines).then(commitLines => commitLines.flat());
– T.J. Crowder
Dec 29 '18 at 13:55
1
1
My purpose with it was to flatten a
number into number So [[1,2],[3,4]] into [1,2,3,4] I'll try the other ways you mentioned– user3790180
Dec 29 '18 at 13:46
My purpose with it was to flatten a
number into number So [[1,2],[3,4]] into [1,2,3,4] I'll try the other ways you mentioned– user3790180
Dec 29 '18 at 13:46
@user3790180 - Ah! In that case, ignore me. :-) (I've removed that part of the answer.) You might look into
Array.prototype.flat. It's at Stage 3 and already supported by Chrome and Firefox, easily polyfilled for others... :-) That would be return Promise.all(commentLines).then(commitLines => commitLines.flat());– T.J. Crowder
Dec 29 '18 at 13:55
@user3790180 - Ah! In that case, ignore me. :-) (I've removed that part of the answer.) You might look into
Array.prototype.flat. It's at Stage 3 and already supported by Chrome and Firefox, easily polyfilled for others... :-) That would be return Promise.all(commentLines).then(commitLines => commitLines.flat());– T.J. Crowder
Dec 29 '18 at 13:55
add a comment |
you should return the Promise
return Promise.all(commentLines).then(...)
//....
or wait for the promise and return the result object
let lines = await Promise.all(commentLines)
return .concat(...lines)
edited Dec 29 '18 at 13:41
Bergi
367k58546873
answered Dec 29 '18 at 13:29
apple appleapple apple
2,5401723
add a comment |
you should return the Promise
return Promise.all(commentLines).then(...)
//....
or wait for the promise and return the result object
let lines = await Promise.all(commentLines)
return .concat(...lines)
edited Dec 29 '18 at 13:41
Bergi
367k58546873
answered Dec 29 '18 at 13:29
apple appleapple apple
2,5401723
add a comment |
you should return the Promise
return Promise.all(commentLines).then(...)
//....
or wait for the promise and return the result object
let lines = await Promise.all(commentLines)
return .concat(...lines)
edited Dec 29 '18 at 13:41
Bergi
367k58546873
answered Dec 29 '18 at 13:29
apple appleapple apple
2,5401723
you should return the Promise
return Promise.all(commentLines).then(...)
//....
or wait for the promise and return the result object
let lines = await Promise.all(commentLines)
return .concat(...lines)
edited Dec 29 '18 at 13:41
Bergi
367k58546873
answered Dec 29 '18 at 13:29
apple appleapple apple
2,5401723
edited Dec 29 '18 at 13:41
Bergi
367k58546873
edited Dec 29 '18 at 13:41
Bergi
367k58546873
edited Dec 29 '18 at 13:41
Bergi
367k58546873
367k58546873
answered Dec 29 '18 at 13:29
apple appleapple apple
2,5401723
answered Dec 29 '18 at 13:29
apple appleapple apple
2,5401723
answered Dec 29 '18 at 13:29
apple appleapple apple
2,5401723
2,5401723
add a comment |
add a comment |
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