return a promise of an ongoing asynchronous call

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I'm trying to launch simultanuously many asynchrnous functions. Some of them make some http calls and I obviously don't want to make the same call multiple times. So, how can this be done using ES6 Promises and if not, what machanism would you recommend to deal with this kind of situations.

Example of the issue:

const func1 = () => async1();



const func2 = () => async2().then(async1);



return Promise.all([func1(), func2()]);

async1 and async2 are asynchronous functions returning Promises.
The issue is how to deal with three situations :

func1 and func2 launch async1 in the same time

func2 launches async1 while async1 is an ongoing asynchronous call in func2

async1 is launched by func2 after async1 has already been resolved by func1

The last situation is the only one I call deal with right now to prevent same calls from launching multiple times.

edited Nov 23 '18 at 14:21

asked Nov 23 '18 at 14:15

mouad

84110

@charlietfl async1 will be called twice in the sample code

– skyboyer
Nov 23 '18 at 14:26

Why you don't just call func2 without then(async1)?

– codtex
Nov 23 '18 at 14:27

@codtex beacuse func1 and func2 are simplified in the example above and in real life they are functions that can be called alone and within Promise.all. So I need I guess some kind of incapsulation of those async functions to not be called when they already have been called or wait if another function is waiting a response..

– mouad
Nov 23 '18 at 14:30

Situation is oversimplified to point we need to ask too many questions. In particular what would differ in async1() if called in func1 or func2? You can cache the promise but not sure if it needs complex cache or very simple one

– charlietfl
Nov 23 '18 at 14:31

1

@charlietfl is right, but I get your point anyway. In my opinion is better to rethink your logic design and avoid calling one async function from other functions multiple times. Otherwise something else I suggest is to write a simple wrapper class of Promise and set a running property (for example), when working with this object you can always check this property before executing. I can write u a simple example if this conception is not clear

– codtex
Nov 23 '18 at 14:34

add a comment |

Example of the issue:

const func1 = () => async1();



const func2 = () => async2().then(async1);



return Promise.all([func1(), func2()]);

async1 and async2 are asynchronous functions returning Promises.
The issue is how to deal with three situations :

func1 and func2 launch async1 in the same time

func2 launches async1 while async1 is an ongoing asynchronous call in func2

async1 is launched by func2 after async1 has already been resolved by func1

The last situation is the only one I call deal with right now to prevent same calls from launching multiple times.

edited Nov 23 '18 at 14:21

asked Nov 23 '18 at 14:15

mouad

84110

@charlietfl async1 will be called twice in the sample code

– skyboyer
Nov 23 '18 at 14:26

Why you don't just call func2 without then(async1)?

– codtex
Nov 23 '18 at 14:27

@codtex beacuse func1 and func2 are simplified in the example above and in real life they are functions that can be called alone and within Promise.all. So I need I guess some kind of incapsulation of those async functions to not be called when they already have been called or wait if another function is waiting a response..

– mouad
Nov 23 '18 at 14:30

Situation is oversimplified to point we need to ask too many questions. In particular what would differ in async1() if called in func1 or func2? You can cache the promise but not sure if it needs complex cache or very simple one

– charlietfl
Nov 23 '18 at 14:31

1

@charlietfl is right, but I get your point anyway. In my opinion is better to rethink your logic design and avoid calling one async function from other functions multiple times. Otherwise something else I suggest is to write a simple wrapper class of Promise and set a running property (for example), when working with this object you can always check this property before executing. I can write u a simple example if this conception is not clear

– codtex
Nov 23 '18 at 14:34

add a comment |

Example of the issue:

const func1 = () => async1();



const func2 = () => async2().then(async1);



return Promise.all([func1(), func2()]);

async1 and async2 are asynchronous functions returning Promises.
The issue is how to deal with three situations :

func1 and func2 launch async1 in the same time

func2 launches async1 while async1 is an ongoing asynchronous call in func2

async1 is launched by func2 after async1 has already been resolved by func1

The last situation is the only one I call deal with right now to prevent same calls from launching multiple times.

edited Nov 23 '18 at 14:21

asked Nov 23 '18 at 14:15

mouad

84110

Example of the issue:

const func1 = () => async1();



const func2 = () => async2().then(async1);



return Promise.all([func1(), func2()]);

async1 and async2 are asynchronous functions returning Promises.
The issue is how to deal with three situations :

func1 and func2 launch async1 in the same time

func2 launches async1 while async1 is an ongoing asynchronous call in func2

async1 is launched by func2 after async1 has already been resolved by func1

The last situation is the only one I call deal with right now to prevent same calls from launching multiple times.

javascript node.js asynchronous es6-promise

edited Nov 23 '18 at 14:21

asked Nov 23 '18 at 14:15

mouad

84110

edited Nov 23 '18 at 14:21

asked Nov 23 '18 at 14:15

mouad

84110

edited Nov 23 '18 at 14:21

asked Nov 23 '18 at 14:15

mouad

84110

asked Nov 23 '18 at 14:15

mouad

84110

asked Nov 23 '18 at 14:15

mouad

84110

@charlietfl async1 will be called twice in the sample code

– skyboyer
Nov 23 '18 at 14:26

Why you don't just call func2 without then(async1)?

– codtex
Nov 23 '18 at 14:27

@codtex beacuse func1 and func2 are simplified in the example above and in real life they are functions that can be called alone and within Promise.all. So I need I guess some kind of incapsulation of those async functions to not be called when they already have been called or wait if another function is waiting a response..

– mouad
Nov 23 '18 at 14:30

Situation is oversimplified to point we need to ask too many questions. In particular what would differ in async1() if called in func1 or func2? You can cache the promise but not sure if it needs complex cache or very simple one

– charlietfl
Nov 23 '18 at 14:31

1

@charlietfl is right, but I get your point anyway. In my opinion is better to rethink your logic design and avoid calling one async function from other functions multiple times. Otherwise something else I suggest is to write a simple wrapper class of Promise and set a running property (for example), when working with this object you can always check this property before executing. I can write u a simple example if this conception is not clear

– codtex
Nov 23 '18 at 14:34

add a comment |

@charlietfl async1 will be called twice in the sample code

– skyboyer
Nov 23 '18 at 14:26

Why you don't just call func2 without then(async1)?

– codtex
Nov 23 '18 at 14:27

@codtex beacuse func1 and func2 are simplified in the example above and in real life they are functions that can be called alone and within Promise.all. So I need I guess some kind of incapsulation of those async functions to not be called when they already have been called or wait if another function is waiting a response..

– mouad
Nov 23 '18 at 14:30

Situation is oversimplified to point we need to ask too many questions. In particular what would differ in async1() if called in func1 or func2? You can cache the promise but not sure if it needs complex cache or very simple one

– charlietfl
Nov 23 '18 at 14:31

1

@charlietfl is right, but I get your point anyway. In my opinion is better to rethink your logic design and avoid calling one async function from other functions multiple times. Otherwise something else I suggest is to write a simple wrapper class of Promise and set a running property (for example), when working with this object you can always check this property before executing. I can write u a simple example if this conception is not clear

– codtex
Nov 23 '18 at 14:34

@charlietfl async1 will be called twice in the sample code

– skyboyer
Nov 23 '18 at 14:26

Why you don't just call func2 without then(async1)?

– codtex
Nov 23 '18 at 14:27

@codtex beacuse func1 and func2 are simplified in the example above and in real life they are functions that can be called alone and within Promise.all. So I need I guess some kind of incapsulation of those async functions to not be called when they already have been called or wait if another function is waiting a response..

– mouad
Nov 23 '18 at 14:30

Situation is oversimplified to point we need to ask too many questions. In particular what would differ in async1() if called in func1 or func2? You can cache the promise but not sure if it needs complex cache or very simple one

– charlietfl
Nov 23 '18 at 14:31

@charlietfl is right, but I get your point anyway. In my opinion is better to rethink your logic design and avoid calling one async function from other functions multiple times. Otherwise something else I suggest is to write a simple wrapper class of Promise and set a running property (for example), when working with this object you can always check this property before executing. I can write u a simple example if this conception is not clear

– codtex
Nov 23 '18 at 14:34

add a comment |

2 Answers
2

active

oldest

votes

You can reuse a promise object any time so if you store a request promise the first time it gets called you can return the stored promise on subsequent calls and avoid making multiple requests to same resource

const async1 = (() => {

  let promise;



  return (arg) => {

    console.log(arg, ' = ', promise ? 'Cached request' : 'New request');

    promise = promise || new Promise((res) => setTimeout(res, 500));

    return promise.then(()=>arg);

  }



})()



Promise.all([async1(1),async1(2)]).then(res=>console.log(res))

edited Nov 23 '18 at 15:18

answered Nov 23 '18 at 15:10

charlietfl

141k1391125

thanks ! this is exactly what I was looking for

– mouad
Nov 23 '18 at 15:32

add a comment |

If I understand it right you want to ensure async1 in your example will not be called twice.

The easy and (quite low)-level of achieving that is memoization.

Say with lodash's _.memoize() it would be

const async1 = _.memoize(async1, () => 1)

// wrapping func1 and func2 is not actually required in this case

const func1 = _.memoize(() => async1()); 

const func2 = _.memoize(() => async2().then(async1));

return Promise.all([func1(), func2()]);

Beware since you are unable to transparently switch from memoized to _un_memoized implementations on fly.

[UPD] since memoization relies on arguments passed you may need to pass resolver callback. Say in your case async1 may either got zero arguments or something coming from previous promise(when used in .then). So arguments are different but since we know all arguments don't matter we can pass resolver that returns constant as a key(say, '1')

edited Nov 23 '18 at 14:40

answered Nov 23 '18 at 14:34

skyboyer

4,23811333

add a comment |

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2 Answers
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2 Answers
2

active

oldest

votes

const async1 = (() => {

  let promise;



  return (arg) => {

    console.log(arg, ' = ', promise ? 'Cached request' : 'New request');

    promise = promise || new Promise((res) => setTimeout(res, 500));

    return promise.then(()=>arg);

  }



})()



Promise.all([async1(1),async1(2)]).then(res=>console.log(res))

edited Nov 23 '18 at 15:18

answered Nov 23 '18 at 15:10

charlietfl

141k1391125

thanks ! this is exactly what I was looking for

– mouad
Nov 23 '18 at 15:32

add a comment |

const async1 = (() => {

  let promise;



  return (arg) => {

    console.log(arg, ' = ', promise ? 'Cached request' : 'New request');

    promise = promise || new Promise((res) => setTimeout(res, 500));

    return promise.then(()=>arg);

  }



})()



Promise.all([async1(1),async1(2)]).then(res=>console.log(res))

edited Nov 23 '18 at 15:18

answered Nov 23 '18 at 15:10

charlietfl

141k1391125

thanks ! this is exactly what I was looking for

– mouad
Nov 23 '18 at 15:32

add a comment |

const async1 = (() => {

  let promise;



  return (arg) => {

    console.log(arg, ' = ', promise ? 'Cached request' : 'New request');

    promise = promise || new Promise((res) => setTimeout(res, 500));

    return promise.then(()=>arg);

  }



})()



Promise.all([async1(1),async1(2)]).then(res=>console.log(res))

edited Nov 23 '18 at 15:18

answered Nov 23 '18 at 15:10

charlietfl

141k1391125

const async1 = (() => {

  let promise;



  return (arg) => {

    console.log(arg, ' = ', promise ? 'Cached request' : 'New request');

    promise = promise || new Promise((res) => setTimeout(res, 500));

    return promise.then(()=>arg);

  }



})()



Promise.all([async1(1),async1(2)]).then(res=>console.log(res))

const async1 = (() => {

  let promise;



  return (arg) => {

    console.log(arg, ' = ', promise ? 'Cached request' : 'New request');

    promise = promise || new Promise((res) => setTimeout(res, 500));

    return promise.then(()=>arg);

  }



})()



Promise.all([async1(1),async1(2)]).then(res=>console.log(res))

const async1 = (() => {

  let promise;



  return (arg) => {

    console.log(arg, ' = ', promise ? 'Cached request' : 'New request');

    promise = promise || new Promise((res) => setTimeout(res, 500));

    return promise.then(()=>arg);

  }



})()



Promise.all([async1(1),async1(2)]).then(res=>console.log(res))

edited Nov 23 '18 at 15:18

answered Nov 23 '18 at 15:10

charlietfl

141k1391125

edited Nov 23 '18 at 15:18

answered Nov 23 '18 at 15:10

charlietfl

141k1391125

answered Nov 23 '18 at 15:10

charlietfl

141k1391125

answered Nov 23 '18 at 15:10

charlietfl

141k1391125

thanks ! this is exactly what I was looking for

– mouad
Nov 23 '18 at 15:32

add a comment |

thanks ! this is exactly what I was looking for

– mouad
Nov 23 '18 at 15:32

thanks ! this is exactly what I was looking for

– mouad
Nov 23 '18 at 15:32

add a comment |

If I understand it right you want to ensure async1 in your example will not be called twice.

The easy and (quite low)-level of achieving that is memoization.

Say with lodash's _.memoize() it would be

const async1 = _.memoize(async1, () => 1)

// wrapping func1 and func2 is not actually required in this case

const func1 = _.memoize(() => async1()); 

const func2 = _.memoize(() => async2().then(async1));

return Promise.all([func1(), func2()]);

Beware since you are unable to transparently switch from memoized to _un_memoized implementations on fly.

edited Nov 23 '18 at 14:40

answered Nov 23 '18 at 14:34

skyboyer

4,23811333

add a comment |

If I understand it right you want to ensure async1 in your example will not be called twice.

The easy and (quite low)-level of achieving that is memoization.

Say with lodash's _.memoize() it would be

const async1 = _.memoize(async1, () => 1)

// wrapping func1 and func2 is not actually required in this case

const func1 = _.memoize(() => async1()); 

const func2 = _.memoize(() => async2().then(async1));

return Promise.all([func1(), func2()]);

Beware since you are unable to transparently switch from memoized to _un_memoized implementations on fly.

edited Nov 23 '18 at 14:40

answered Nov 23 '18 at 14:34

skyboyer

4,23811333

add a comment |

If I understand it right you want to ensure async1 in your example will not be called twice.

The easy and (quite low)-level of achieving that is memoization.

Say with lodash's _.memoize() it would be

const async1 = _.memoize(async1, () => 1)

// wrapping func1 and func2 is not actually required in this case

const func1 = _.memoize(() => async1()); 

const func2 = _.memoize(() => async2().then(async1));

return Promise.all([func1(), func2()]);

Beware since you are unable to transparently switch from memoized to _un_memoized implementations on fly.

edited Nov 23 '18 at 14:40

answered Nov 23 '18 at 14:34

skyboyer

4,23811333

If I understand it right you want to ensure async1 in your example will not be called twice.

The easy and (quite low)-level of achieving that is memoization.

Say with lodash's _.memoize() it would be

const async1 = _.memoize(async1, () => 1)

// wrapping func1 and func2 is not actually required in this case

const func1 = _.memoize(() => async1()); 

const func2 = _.memoize(() => async2().then(async1));

return Promise.all([func1(), func2()]);

Beware since you are unable to transparently switch from memoized to _un_memoized implementations on fly.

edited Nov 23 '18 at 14:40

answered Nov 23 '18 at 14:34

skyboyer

4,23811333

edited Nov 23 '18 at 14:40

answered Nov 23 '18 at 14:34

skyboyer

4,23811333

answered Nov 23 '18 at 14:34

skyboyer

4,23811333

answered Nov 23 '18 at 14:34

skyboyer

4,23811333

add a comment |

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