CODEWITHSUNDEEP
javascriptnode.jsfirebasegoogle-cloud-functions
KVNA
KVNA

Reputation: 907

How to pass a Promise.all return to another function Cloud Functions for Firebase

I am trying to set a few variables from Firebase and then pass those into a anotherfunction. Currently, the Promise.all is properly setting foo and bar, however, I can't tell if foo and bar are being passed into then, and also Firebase is not properly scoped.

The Promise.all block is based on a tutorial found here: https://www.youtube.com/watch?v=NgZIb6Uwpjc&t=305s

exports.someFunction = functions.database.ref(`/data`).onWrite(event => {
  const condition = event.data.val()
  if (condition) {
    // Get the data at this location only once, returns a promise, to ensure retrieval of foo and bar
    const foo = event.data.adminRef.child('foo').once('value')
    const bar = event.data.adminRef.child('bar').once('value')

    return Promise.all([foo, bar]).then(results => {
      const foo = results[0].val()
      const bar = results[1].val()
      // Properly sets foo and bar
      // As far as I can tell foo and bar are not passed into 'then'
    }).then([foo, bar] => {
      return someModule.anotherFunction({
        "foo": foo,
        "bar": bar
      })
    })
  } else {
    console.log('Fail');
  }
});

How can I pass foo and bar into anotherFunction and set the response of that function to Firebase?

Upvotes: 3

Views: 3178

Answers (2)

Jaromanda X
Jaromanda X

Reputation: 1

Here's where you've gone wrong - see the comments in the code

return Promise.all([foo, bar]).then(results => {
  const foo = results[0].val()
  const bar = results[1].val()
  // you dont' return anything so, the following .then gets undefined argument
}).then([foo, bar] => {
  //    ^^^^^^^^^^^^^ invalid syntax, you need .then(([foo, bar]) =>
  return someModule.anotherFunction({
    "foo": foo,
    "bar": bar
  })

To simplify your code, just remove }).then([foo, bar] => { altogether!!

return Promise.all([foo, bar])
.then(results => {
    const foo = results[0].val()
    const bar = results[1].val()
    return someModule.anotherFunction({
        "foo": foo,
        "bar": bar
    }))
.then ...

But, if there's more to the actual code than you show, you can do

return Promise.all([foo, bar])
.then(results => results.map(result => result.val()))
.then(([foo, bar]) => someModule.anotherFunction({
    "foo": foo,
    "bar": bar
}))
.then ...

or

return Promise.all([foo, bar])
.then(([foo, bar]) => ([foo.val(), bar.val()]))
.then(([foo, bar]) => someModule.anotherFunction({
    "foo": foo,
    "bar": bar
}))
.then ...

Upvotes: 6

f.khantsis
f.khantsis

Reputation: 3540

in the first then add 

return Promise.resolve([foo,bar]);

or (as per @Jaromanda X)

return [foo, bar];

Upvotes: 4

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