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Return the Value of a function after aasy function ended
How do I return the response from an asynchronous call?Is there an “exists” function for jQuery?Which “href” value should I use for JavaScript links, “#” or “javascript:void(0)”?var functionName = function() vs function functionName() Set a default parameter value for a JavaScript functionHow can I get query string values in JavaScript?Sort array of objects by string property valueevent.preventDefault() vs. return falseWhy does ++[[]][+[]]+[+[]] return the string “10”?Copy array by valueHow do I return the response from an asynchronous call?
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I try to wait for a function to return a value of my mysql table, and use this as a return for my const ProjektNameIntentHandler. This is my code:
const ProjektNameIntentHandler =
canHandle(handlerInput)
return handlerInput.requestEnvelope.request.type === 'IntentRequest'
&& handlerInput.requestEnvelope.request.intent.name === 'ProjektNameIntent';
,
handle(handlerInput)
let getProjektName = queryDb()
getProjektName.then(function(result)
var projektName = result[0];
console.log(projektName.Test);
)
return handlerInput.responseBuilder
.speak(projektName.Test)
.withSimpleCard('Venture', projektName.Test)
.getResponse();
;
Now the problem is that get the result of ProjektNameIntentHandler before projektName got the result. First, I tried to put the second return into the scope of the function. But in this way, the result also belongs to the funtion and not as a return for my ProjektNameIntentHandler.
So all I try do archieve is that the second return for the handlerinput, waits for my getProjektName.then to finish. How can I do that?
javascript node.js json alexa alexa-skills-kit
edited Mar 24 at 14:23
Nino Filiu
3,66241632
asked Mar 24 at 14:15
OlliOlli
83
add a comment |
I try to wait for a function to return a value of my mysql table, and use this as a return for my const ProjektNameIntentHandler. This is my code:
const ProjektNameIntentHandler =
canHandle(handlerInput)
return handlerInput.requestEnvelope.request.type === 'IntentRequest'
&& handlerInput.requestEnvelope.request.intent.name === 'ProjektNameIntent';
,
handle(handlerInput)
let getProjektName = queryDb()
getProjektName.then(function(result)
var projektName = result[0];
console.log(projektName.Test);
)
return handlerInput.responseBuilder
.speak(projektName.Test)
.withSimpleCard('Venture', projektName.Test)
.getResponse();
;
Now the problem is that get the result of ProjektNameIntentHandler before projektName got the result. First, I tried to put the second return into the scope of the function. But in this way, the result also belongs to the funtion and not as a return for my ProjektNameIntentHandler.
So all I try do archieve is that the second return for the handlerinput, waits for my getProjektName.then to finish. How can I do that?
javascript node.js json alexa alexa-skills-kit
edited Mar 24 at 14:23
Nino Filiu
3,66241632
asked Mar 24 at 14:15
OlliOlli
83
add a comment |
I try to wait for a function to return a value of my mysql table, and use this as a return for my const ProjektNameIntentHandler. This is my code:
const ProjektNameIntentHandler =
canHandle(handlerInput)
return handlerInput.requestEnvelope.request.type === 'IntentRequest'
&& handlerInput.requestEnvelope.request.intent.name === 'ProjektNameIntent';
,
handle(handlerInput)
let getProjektName = queryDb()
getProjektName.then(function(result)
var projektName = result[0];
console.log(projektName.Test);
)
return handlerInput.responseBuilder
.speak(projektName.Test)
.withSimpleCard('Venture', projektName.Test)
.getResponse();
;
Now the problem is that get the result of ProjektNameIntentHandler before projektName got the result. First, I tried to put the second return into the scope of the function. But in this way, the result also belongs to the funtion and not as a return for my ProjektNameIntentHandler.
So all I try do archieve is that the second return for the handlerinput, waits for my getProjektName.then to finish. How can I do that?
javascript node.js json alexa alexa-skills-kit
edited Mar 24 at 14:23
Nino Filiu
3,66241632
asked Mar 24 at 14:15
OlliOlli
83
I try to wait for a function to return a value of my mysql table, and use this as a return for my const ProjektNameIntentHandler. This is my code:
const ProjektNameIntentHandler =
canHandle(handlerInput)
return handlerInput.requestEnvelope.request.type === 'IntentRequest'
&& handlerInput.requestEnvelope.request.intent.name === 'ProjektNameIntent';
,
handle(handlerInput)
let getProjektName = queryDb()
getProjektName.then(function(result)
var projektName = result[0];
console.log(projektName.Test);
)
return handlerInput.responseBuilder
.speak(projektName.Test)
.withSimpleCard('Venture', projektName.Test)
.getResponse();
;
Now the problem is that get the result of ProjektNameIntentHandler before projektName got the result. First, I tried to put the second return into the scope of the function. But in this way, the result also belongs to the funtion and not as a return for my ProjektNameIntentHandler.
So all I try do archieve is that the second return for the handlerinput, waits for my getProjektName.then to finish. How can I do that?
javascript node.js json alexa alexa-skills-kit
javascript node.js json alexa alexa-skills-kit
edited Mar 24 at 14:23
Nino Filiu
3,66241632
asked Mar 24 at 14:15
OlliOlli
83
edited Mar 24 at 14:23
Nino Filiu
3,66241632
asked Mar 24 at 14:15
OlliOlli
83
edited Mar 24 at 14:23
Nino Filiu
3,66241632
edited Mar 24 at 14:23
Nino Filiu
3,66241632
edited Mar 24 at 14:23
Nino Filiu
3,66241632
3,66241632
asked Mar 24 at 14:15
OlliOlli
83
asked Mar 24 at 14:15
OlliOlli
83
asked Mar 24 at 14:15
OlliOlli
83
83
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
As you indeed guessed, as of now, you return undefined because the asynchronous function getProjektName has not yet resolved. The thing is, inside a synchronous function, you can't wait for a function to finish executing - you can only do that in asynchronous functions... However, you could make handle asynchronous! If that fits your requirements, you can modify your code as such:
const ProjektNameIntentHandler =
// ...
async handle(handlerInput) // 'async' makes the function asynchronous
let result = await queryDb(); // wait for the promise to resolve
let projektName = result[0];
console.log(projektName.Test);
return handlerInput.responseBuilder
.speak(projektName.Test)
.withSimpleCard('Venture', projektName.Test)
.getResponse();
;
I'll skip the lengthly explanations of how to embrace the asynchronicity of JavaScript - there's already a similar question with a high-quality answer that does just that!
answered Mar 24 at 14:37
Nino FiliuNino Filiu
3,66241632
OMG IT FINALLY WORKS! Thank you so so much! I was brachiate through soooo many different methods and nothing worked. I cant beliefe it finaly came to an end!
– Olli
Mar 24 at 14:45
Great! I'm happy to hear that! By the way - it can feel weird for new users, but thank-you comments should be avoided, instead, simply upvote and/or accept the question - here's why. Happy coding @0lli1234!
– Nino Filiu
Mar 24 at 14:50
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
As you indeed guessed, as of now, you return undefined because the asynchronous function getProjektName has not yet resolved. The thing is, inside a synchronous function, you can't wait for a function to finish executing - you can only do that in asynchronous functions... However, you could make handle asynchronous! If that fits your requirements, you can modify your code as such:
const ProjektNameIntentHandler =
// ...
async handle(handlerInput) // 'async' makes the function asynchronous
let result = await queryDb(); // wait for the promise to resolve
let projektName = result[0];
console.log(projektName.Test);
return handlerInput.responseBuilder
.speak(projektName.Test)
.withSimpleCard('Venture', projektName.Test)
.getResponse();
;
I'll skip the lengthly explanations of how to embrace the asynchronicity of JavaScript - there's already a similar question with a high-quality answer that does just that!
answered Mar 24 at 14:37
Nino FiliuNino Filiu
3,66241632
OMG IT FINALLY WORKS! Thank you so so much! I was brachiate through soooo many different methods and nothing worked. I cant beliefe it finaly came to an end!
– Olli
Mar 24 at 14:45
Great! I'm happy to hear that! By the way - it can feel weird for new users, but thank-you comments should be avoided, instead, simply upvote and/or accept the question - here's why. Happy coding @0lli1234!
– Nino Filiu
Mar 24 at 14:50
add a comment |
As you indeed guessed, as of now, you return undefined because the asynchronous function getProjektName has not yet resolved. The thing is, inside a synchronous function, you can't wait for a function to finish executing - you can only do that in asynchronous functions... However, you could make handle asynchronous! If that fits your requirements, you can modify your code as such:
const ProjektNameIntentHandler =
// ...
async handle(handlerInput) // 'async' makes the function asynchronous
let result = await queryDb(); // wait for the promise to resolve
let projektName = result[0];
console.log(projektName.Test);
return handlerInput.responseBuilder
.speak(projektName.Test)
.withSimpleCard('Venture', projektName.Test)
.getResponse();
;
I'll skip the lengthly explanations of how to embrace the asynchronicity of JavaScript - there's already a similar question with a high-quality answer that does just that!
answered Mar 24 at 14:37
Nino FiliuNino Filiu
3,66241632
OMG IT FINALLY WORKS! Thank you so so much! I was brachiate through soooo many different methods and nothing worked. I cant beliefe it finaly came to an end!
– Olli
Mar 24 at 14:45
Great! I'm happy to hear that! By the way - it can feel weird for new users, but thank-you comments should be avoided, instead, simply upvote and/or accept the question - here's why. Happy coding @0lli1234!
– Nino Filiu
Mar 24 at 14:50
add a comment |
As you indeed guessed, as of now, you return undefined because the asynchronous function getProjektName has not yet resolved. The thing is, inside a synchronous function, you can't wait for a function to finish executing - you can only do that in asynchronous functions... However, you could make handle asynchronous! If that fits your requirements, you can modify your code as such:
const ProjektNameIntentHandler =
// ...
async handle(handlerInput) // 'async' makes the function asynchronous
let result = await queryDb(); // wait for the promise to resolve
let projektName = result[0];
console.log(projektName.Test);
return handlerInput.responseBuilder
.speak(projektName.Test)
.withSimpleCard('Venture', projektName.Test)
.getResponse();
;
I'll skip the lengthly explanations of how to embrace the asynchronicity of JavaScript - there's already a similar question with a high-quality answer that does just that!
answered Mar 24 at 14:37
Nino FiliuNino Filiu
3,66241632
As you indeed guessed, as of now, you return undefined because the asynchronous function getProjektName has not yet resolved. The thing is, inside a synchronous function, you can't wait for a function to finish executing - you can only do that in asynchronous functions... However, you could make handle asynchronous! If that fits your requirements, you can modify your code as such:
const ProjektNameIntentHandler =
// ...
async handle(handlerInput) // 'async' makes the function asynchronous
let result = await queryDb(); // wait for the promise to resolve
let projektName = result[0];
console.log(projektName.Test);
return handlerInput.responseBuilder
.speak(projektName.Test)
.withSimpleCard('Venture', projektName.Test)
.getResponse();
;
I'll skip the lengthly explanations of how to embrace the asynchronicity of JavaScript - there's already a similar question with a high-quality answer that does just that!
answered Mar 24 at 14:37
Nino FiliuNino Filiu
3,66241632
answered Mar 24 at 14:37
Nino FiliuNino Filiu
3,66241632
answered Mar 24 at 14:37
Nino FiliuNino Filiu
3,66241632
answered Mar 24 at 14:37
Nino FiliuNino Filiu
3,66241632
3,66241632
OMG IT FINALLY WORKS! Thank you so so much! I was brachiate through soooo many different methods and nothing worked. I cant beliefe it finaly came to an end!
– Olli
Mar 24 at 14:45
Great! I'm happy to hear that! By the way - it can feel weird for new users, but thank-you comments should be avoided, instead, simply upvote and/or accept the question - here's why. Happy coding @0lli1234!
– Nino Filiu
Mar 24 at 14:50
add a comment |
OMG IT FINALLY WORKS! Thank you so so much! I was brachiate through soooo many different methods and nothing worked. I cant beliefe it finaly came to an end!
– Olli
Mar 24 at 14:45
Great! I'm happy to hear that! By the way - it can feel weird for new users, but thank-you comments should be avoided, instead, simply upvote and/or accept the question - here's why. Happy coding @0lli1234!
– Nino Filiu
Mar 24 at 14:50
OMG IT FINALLY WORKS! Thank you so so much! I was brachiate through soooo many different methods and nothing worked. I cant beliefe it finaly came to an end!
– Olli
Mar 24 at 14:45
OMG IT FINALLY WORKS! Thank you so so much! I was brachiate through soooo many different methods and nothing worked. I cant beliefe it finaly came to an end!
– Olli
Mar 24 at 14:45
Great! I'm happy to hear that! By the way - it can feel weird for new users, but thank-you comments should be avoided, instead, simply upvote and/or accept the question - here's why. Happy coding @0lli1234!
– Nino Filiu
Mar 24 at 14:50
Great! I'm happy to hear that! By the way - it can feel weird for new users, but thank-you comments should be avoided, instead, simply upvote and/or accept the question - here's why. Happy coding @0lli1234!
– Nino Filiu
Mar 24 at 14:50
add a comment |
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