What is the best way to ensure the correct resolution order of several promises?

Question

What is the best way to ensure the correct resolution order of several promises?

Exploring the realm of modern JS and diving into ECMAScript 6 Promises. Experimenting with a simple test:

let slow = new Promise((resolve) => {
  setTimeout(function()
  {  
console.log('slow');
resolve();
  }, 2000, 'slow');
});

let instant = new Promise((resolve) => {
console.log('instant');
resolve();
});

let quick = new Promise((resolve) => {
  setTimeout(function()
  {  
console.log('quick');
resolve();
  }, 1000, 'quick');
});

Promise.all([slow, instant, quick]).then(function(results) {
  console.log('finished');
}, function(error) {
  console.log(error);
});

Desiring to ensure that "instant" is not logged before "slow" completes. Aiming for the console output to be: "quick", "slow", "instant", and "finished". They must all start asynchronously at the same time.

How can I make this happen?

javascript promise ecmascript-6 es6-promise

Answer 1

Answer №1

To clarify, your goal is to initiate all promises simultaneously and showcase the results of each promise in a specific order as they are received, right?

In that scenario, my approach would be:

let slow = new Promise((resolve) => {
  setTimeout(function()
  {
    // Instead of logging here, we resolve with the desired value
    resolve('slow');
  }, 2000, 'slow');
});

let instant = new Promise((resolve) => {
    resolve('instant');  
});

let quick = new Promise((resolve) => {
  setTimeout(function()
  {  
    resolve('quick');  
  }, 1000, 'quick');
});

// All Promises are now running. Let's display the results...

// First, wait for the result of `slow`...
slow.then((result) => {
  // Result received...
  console.log(result);
  
  // Now await the result of instant...
  instant.then((result) => {
    
    // Result obtained...
    console.log(result);
    
    // Now await the result of quick...
    quick.then((result) => {
      
      // Result acquired...
      console.log(result);
      
    }).then((result) => {
      // Completed
      console.log('finished');
    });
  });
});

Note that unlike cchamberlain's response, this method does not necessitate waiting for all promises to resolve prior to returning results. It returns the results as they arrive, while maintaining the specified order of results. (To confirm this, adjust the waiting time of quick to 2500ms, and observe that its result is displayed 500ms after instant.) Depending on your application, this may be beneficial.

The above code may seem cluttered, but fortunately, it can be significantly streamlined using the new async/await syntax in ES2017:

let slow = new Promise((resolve) => {
  setTimeout(function()
  {
    // Instead of logging here, we resolve with the desired value
    resolve('slow');
  }, 2000, 'slow');
});

let instant = new Promise((resolve) => {
    resolve('instant');  
});

let quick = new Promise((resolve) => {
  setTimeout(function()
  {  
    resolve('quick');  
  }, 1000, 'quick');
});

// All Promises are now running. Let's show the results...

async function logResults(...promises) {
  for (let promise of promises) {
    console.log(await promise);
  }
}

logResults(slow, instant, quick).then(() => console.log('finished'));

Test in Babel. Note: Presently, the above code won't work in modern browsers without Babel support (as of October 2016). In future browsers, it will.

Answer 2

To clarify, your goal is to initiate all promises simultaneously and showcase the results of each promise in a specific order as they are received, right?

In that scenario, my approach would be:

let slow = new Promise((resolve) => {
  setTimeout(function()
  {
    // Instead of logging here, we resolve with the desired value
    resolve('slow');
  }, 2000, 'slow');
});

let instant = new Promise((resolve) => {
    resolve('instant');  
});

let quick = new Promise((resolve) => {
  setTimeout(function()
  {  
    resolve('quick');  
  }, 1000, 'quick');
});

// All Promises are now running. Let's display the results...

// First, wait for the result of `slow`...
slow.then((result) => {
  // Result received...
  console.log(result);
  
  // Now await the result of instant...
  instant.then((result) => {
    
    // Result obtained...
    console.log(result);
    
    // Now await the result of quick...
    quick.then((result) => {
      
      // Result acquired...
      console.log(result);
      
    }).then((result) => {
      // Completed
      console.log('finished');
    });
  });
});

Note that unlike cchamberlain's response, this method does not necessitate waiting for all promises to resolve prior to returning results. It returns the results as they arrive, while maintaining the specified order of results. (To confirm this, adjust the waiting time of quick to 2500ms, and observe that its result is displayed 500ms after instant.) Depending on your application, this may be beneficial.

The above code may seem cluttered, but fortunately, it can be significantly streamlined using the new async/await syntax in ES2017:

let slow = new Promise((resolve) => {
  setTimeout(function()
  {
    // Instead of logging here, we resolve with the desired value
    resolve('slow');
  }, 2000, 'slow');
});

let instant = new Promise((resolve) => {
    resolve('instant');  
});

let quick = new Promise((resolve) => {
  setTimeout(function()
  {  
    resolve('quick');  
  }, 1000, 'quick');
});

// All Promises are now running. Let's show the results...

async function logResults(...promises) {
  for (let promise of promises) {
    console.log(await promise);
  }
}

logResults(slow, instant, quick).then(() => console.log('finished'));

Test in Babel. Note: Presently, the above code won't work in modern browsers without Babel support (as of October 2016). In future browsers, it will.

Answer 3

Answer №2

The problem arises because the logging occurs within the setTimeout function, rather than directly after resolving the promise.

const sluggish = new Promise((resolve) => {
  setTimeout(() => {
    console.log('sluggish - from setTimeout');
    resolve('sluggish - from resolve');
  }, 2000, 'sluggish');
});

const immediate = new Promise((resolve) => {
  console.log('immediate - from setTimeout');
  resolve('immediate - from resolve');
});

const speedy = new Promise((resolve) => {
    setTimeout(() => {
    console.log('speedy - from setTimeout');
    resolve('speedy -from resolve');
  }, 1000, 'speedy');
});

Promise.all([sluggish, immediate, speedy]).then((output) => {
  console.log('output -', output);
  console.log('done');
}, (error) => {
  console.log(error);
});

Supplying a value to the resolve method will aggregate all results in the Promise.all. The responses return as an array from each promise, allowing for iteration upon completion of all promises.

Answer 4

The problem arises because the logging occurs within the setTimeout function, rather than directly after resolving the promise.

const sluggish = new Promise((resolve) => {
  setTimeout(() => {
    console.log('sluggish - from setTimeout');
    resolve('sluggish - from resolve');
  }, 2000, 'sluggish');
});

const immediate = new Promise((resolve) => {
  console.log('immediate - from setTimeout');
  resolve('immediate - from resolve');
});

const speedy = new Promise((resolve) => {
    setTimeout(() => {
    console.log('speedy - from setTimeout');
    resolve('speedy -from resolve');
  }, 1000, 'speedy');
});

Promise.all([sluggish, immediate, speedy]).then((output) => {
  console.log('output -', output);
  console.log('done');
}, (error) => {
  console.log(error);
});

Supplying a value to the resolve method will aggregate all results in the Promise.all. The responses return as an array from each promise, allowing for iteration upon completion of all promises.

Answer 5

Answer №3

UPDATED

In order to ensure that the results are logged in the correct order, you cannot simply kick them all off at the same time with your current code. One approach could be to schedule each promise and then print the results upon completion:

let slow = new Promise((resolve) => {
  setTimeout(() => resolve('slow'), 2000)
})

let instant = new Promise((resolve) => {
  // Consider using setImmediate for async execution in future updates
  setTimeout(() => resolve('instant'), 0)
})

let quick = new Promise((resolve) => {
  setTimeout(() => resolve('quick'), 1000)
})

Promise.all([slow, instant, quick]).then(function(results) {
  for(let result of results) {
    console.log(result)
  }
  console.log('finished')
}, function(err) {
  console.error(err)
})

This scheduling approach ensures that the results are printed in the expected order.

Answer 6

UPDATED

In order to ensure that the results are logged in the correct order, you cannot simply kick them all off at the same time with your current code. One approach could be to schedule each promise and then print the results upon completion:

let slow = new Promise((resolve) => {
  setTimeout(() => resolve('slow'), 2000)
})

let instant = new Promise((resolve) => {
  // Consider using setImmediate for async execution in future updates
  setTimeout(() => resolve('instant'), 0)
})

let quick = new Promise((resolve) => {
  setTimeout(() => resolve('quick'), 1000)
})

Promise.all([slow, instant, quick]).then(function(results) {
  for(let result of results) {
    console.log(result)
  }
  console.log('finished')
}, function(err) {
  console.error(err)
})

This scheduling approach ensures that the results are printed in the expected order.

Answer 7

Answer №4

If you need to address the requirements mentioned in Question:

"instant" should not echo/log before "slow"

and

all elements must start at the same time asynchronously.

You can simply rearrange the elements within the iterable object passed to Promise.all(), or modify the resulting array within the .then() chain of Promise.all() before using console.log() on each element of the resulting array.

If the requirement is:

"How do I wait for another promise?"

or

"How do I ensure the order of resolution for multiple promises?"

refer to this specific Answer.

Promise.all collects an array of values from all promises within the iterable object it was provided. The resultant array maintains the order of the original iterable object, not based on the sequence of promise resolutions. If any item in the iterable array is not a promise, it is treated as one by Promise.resolve.

let slow = new Promise((resolve) => {
  setTimeout(function(value) {  
resolve(value);
  }, 2000, "slow");
});

let instant = new Promise((resolve) => {
resolve("instant");
});

let quick = new Promise((resolve) => {
  setTimeout(function(value) {  
resolve(value);
  }, 1000, "quick");
});

Promise.all([slow, instant, quick]).then(function(results) {
  console.log("finished");
  console.log(results.join("\n"))
}, function(error) {
  console.log(error);
});

Answer 8

If you need to address the requirements mentioned in Question:

"instant" should not echo/log before "slow"

and

all elements must start at the same time asynchronously.

You can simply rearrange the elements within the iterable object passed to Promise.all(), or modify the resulting array within the .then() chain of Promise.all() before using console.log() on each element of the resulting array.

If the requirement is:

"How do I wait for another promise?"

or

"How do I ensure the order of resolution for multiple promises?"

refer to this specific Answer.

Promise.all collects an array of values from all promises within the iterable object it was provided. The resultant array maintains the order of the original iterable object, not based on the sequence of promise resolutions. If any item in the iterable array is not a promise, it is treated as one by Promise.resolve.

let slow = new Promise((resolve) => {
  setTimeout(function(value) {  
resolve(value);
  }, 2000, "slow");
});

let instant = new Promise((resolve) => {
resolve("instant");
});

let quick = new Promise((resolve) => {
  setTimeout(function(value) {  
resolve(value);
  }, 1000, "quick");
});

Promise.all([slow, instant, quick]).then(function(results) {
  console.log("finished");
  console.log(results.join("\n"))
}, function(error) {
  console.log(error);
});

Answer 9

Answer №5

At the outset, your code initiates all 3 promises simultaneously. The logging of "finished" is also done correctly. Based on my understanding of the question, you aim to process the results of the promises in a sequential manner while executing them concurrently.

let slow = new Promise((resolve) => {
  setTimeout(function()
  {  
    resolve();
  }, 2000);
});

let instant = new Promise((resolve) => {
    resolve();
});

let quick = new Promise((resolve) => {
  setTimeout(function()
  {  
    resolve();
  }, 1000);
});

instant.then(function(results) {
  console.log("instant");
}).then(function(){return quick;}).then(function(results) {
  console.log("quick");
}).then(function(){return slow;}).then(function(results) {
  console.log("slow");
}).then(function(){ return Promise.all([slow, instant, quick]);}).then(function(results) {
  console.log('finished');
}).catch(function(error) {
  console.log(error);   
});

This approach ensures that you handle the resolutions in a specific order.

Note: In your given example, the use of setTimeout guarantees sequential calling based on time order. Therefore, your current code will already log "instant", "quick", "slow", and then "finished". However, the code I've shared secures this order for any assortment of promises with varying resolution times.

Answer 10

At the outset, your code initiates all 3 promises simultaneously. The logging of "finished" is also done correctly. Based on my understanding of the question, you aim to process the results of the promises in a sequential manner while executing them concurrently.

let slow = new Promise((resolve) => {
  setTimeout(function()
  {  
    resolve();
  }, 2000);
});

let instant = new Promise((resolve) => {
    resolve();
});

let quick = new Promise((resolve) => {
  setTimeout(function()
  {  
    resolve();
  }, 1000);
});

instant.then(function(results) {
  console.log("instant");
}).then(function(){return quick;}).then(function(results) {
  console.log("quick");
}).then(function(){return slow;}).then(function(results) {
  console.log("slow");
}).then(function(){ return Promise.all([slow, instant, quick]);}).then(function(results) {
  console.log('finished');
}).catch(function(error) {
  console.log(error);   
});

This approach ensures that you handle the resolutions in a specific order.

Note: In your given example, the use of setTimeout guarantees sequential calling based on time order. Therefore, your current code will already log "instant", "quick", "slow", and then "finished". However, the code I've shared secures this order for any assortment of promises with varying resolution times.

Answer 11

Answer №6

Promise.all() is not needed for achieving the desired outcome.

You can accomplish the task by making function calls to return the Promise constructor, passing the necessary value to be logged through resolve() or reject(). Store the values in an array. Use .then() to handle the returned Promise value, add it to the array, and return the array to the callback function parameter in the subsequent .then() of the chain. Retrieve the array of chained Promise values in the final .then() of the chain.

let results = [];

let pace = value => {console.log(value); results.push(value); return results};

let slow = () => new Promise((resolve) => {
  setTimeout((value) => {  
    resolve(value);  
  }, 2000, "slow");
});

let instant = () => new Promise((resolve) => {
    resolve("instant");  
});

let quick = () => new Promise((resolve) => {
  setTimeout((value) => {  
    resolve(value);  
  }, 1000, "quick");
});

slow().then(pace)
.then(instant).then(pace)
.then(quick).then(pace)
.then(res => console.log("All done, results:", res))
.catch(error => console.log(error));

Answer 12

Promise.all() is not needed for achieving the desired outcome.

You can accomplish the task by making function calls to return the Promise constructor, passing the necessary value to be logged through resolve() or reject(). Store the values in an array. Use .then() to handle the returned Promise value, add it to the array, and return the array to the callback function parameter in the subsequent .then() of the chain. Retrieve the array of chained Promise values in the final .then() of the chain.

let results = [];

let pace = value => {console.log(value); results.push(value); return results};

let slow = () => new Promise((resolve) => {
  setTimeout((value) => {  
    resolve(value);  
  }, 2000, "slow");
});

let instant = () => new Promise((resolve) => {
    resolve("instant");  
});

let quick = () => new Promise((resolve) => {
  setTimeout((value) => {  
    resolve(value);  
  }, 1000, "quick");
});

slow().then(pace)
.then(instant).then(pace)
.then(quick).then(pace)
.then(res => console.log("All done, results:", res))
.catch(error => console.log(error));

What is the best way to ensure the correct resolution order of several promises?

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

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