How to parse external json request and put it into a variable?
1
I'm very new to coding, I'm hoping someone can help me. I have this little bit of code, I am trying to learn this process:
- Call Made to URL via GET request (json)
- Parse the response from the GET request
- Save the response into a variable that I will use later
Any and all help is appreciated!
const XMLHttpRequest = require("xmlhttprequest").XMLHttpRequest;
const awsURL = 'https://cors.io/?http://status.aws.amazon.com/data.json';
function Get(awsURL){
var request = new XMLHttpRequest();
request.open("GET", awsURL, false);
request.send(null);
return request.responseText;
}
var AWSJson = JSON.parse(Get(awsURL));
console.log("Archived Outages: "+AWSJson.service_name);javascript amazon-web-services
asked Nov 26 '18 at 20:16
Chamaco AranoChamaco Arano
62
add a comment |
1
I'm very new to coding, I'm hoping someone can help me. I have this little bit of code, I am trying to learn this process:
- Call Made to URL via GET request (json)
- Parse the response from the GET request
- Save the response into a variable that I will use later
Any and all help is appreciated!
const XMLHttpRequest = require("xmlhttprequest").XMLHttpRequest;
const awsURL = 'https://cors.io/?http://status.aws.amazon.com/data.json';
function Get(awsURL){
var request = new XMLHttpRequest();
request.open("GET", awsURL, false);
request.send(null);
return request.responseText;
}
var AWSJson = JSON.parse(Get(awsURL));
console.log("Archived Outages: "+AWSJson.service_name);javascript amazon-web-services
asked Nov 26 '18 at 20:16
Chamaco AranoChamaco Arano
62
Possible duplicate of How do I return the response from an asynchronous call?
– Patrick Hund
Nov 26 '18 at 20:20
So what error[s] are you getting with the above code when you run it?
– epascarello
Nov 26 '18 at 20:44
add a comment |
1
1
1
I'm very new to coding, I'm hoping someone can help me. I have this little bit of code, I am trying to learn this process:
- Call Made to URL via GET request (json)
- Parse the response from the GET request
- Save the response into a variable that I will use later
Any and all help is appreciated!
const XMLHttpRequest = require("xmlhttprequest").XMLHttpRequest;
const awsURL = 'https://cors.io/?http://status.aws.amazon.com/data.json';
function Get(awsURL){
var request = new XMLHttpRequest();
request.open("GET", awsURL, false);
request.send(null);
return request.responseText;
}
var AWSJson = JSON.parse(Get(awsURL));
console.log("Archived Outages: "+AWSJson.service_name);javascript amazon-web-services
asked Nov 26 '18 at 20:16
Chamaco AranoChamaco Arano
62
I'm very new to coding, I'm hoping someone can help me. I have this little bit of code, I am trying to learn this process:
- Call Made to URL via GET request (json)
- Parse the response from the GET request
- Save the response into a variable that I will use later
Any and all help is appreciated!
const XMLHttpRequest = require("xmlhttprequest").XMLHttpRequest;
const awsURL = 'https://cors.io/?http://status.aws.amazon.com/data.json';
function Get(awsURL){
var request = new XMLHttpRequest();
request.open("GET", awsURL, false);
request.send(null);
return request.responseText;
}
var AWSJson = JSON.parse(Get(awsURL));
console.log("Archived Outages: "+AWSJson.service_name);const XMLHttpRequest = require("xmlhttprequest").XMLHttpRequest;
const awsURL = 'https://cors.io/?http://status.aws.amazon.com/data.json';
function Get(awsURL){
var request = new XMLHttpRequest();
request.open("GET", awsURL, false);
request.send(null);
return request.responseText;
}
var AWSJson = JSON.parse(Get(awsURL));
console.log("Archived Outages: "+AWSJson.service_name);const XMLHttpRequest = require("xmlhttprequest").XMLHttpRequest;
const awsURL = 'https://cors.io/?http://status.aws.amazon.com/data.json';
function Get(awsURL){
var request = new XMLHttpRequest();
request.open("GET", awsURL, false);
request.send(null);
return request.responseText;
}
var AWSJson = JSON.parse(Get(awsURL));
console.log("Archived Outages: "+AWSJson.service_name);javascript amazon-web-services
javascript amazon-web-services
asked Nov 26 '18 at 20:16
Chamaco AranoChamaco Arano
62
asked Nov 26 '18 at 20:16
Chamaco AranoChamaco Arano
62
asked Nov 26 '18 at 20:16
Chamaco AranoChamaco Arano
62
asked Nov 26 '18 at 20:16
Chamaco AranoChamaco Arano
62
asked Nov 26 '18 at 20:16
Chamaco AranoChamaco Arano
62
62
Possible duplicate of How do I return the response from an asynchronous call?
– Patrick Hund
Nov 26 '18 at 20:20
So what error[s] are you getting with the above code when you run it?
– epascarello
Nov 26 '18 at 20:44
add a comment |
Possible duplicate of How do I return the response from an asynchronous call?
– Patrick Hund
Nov 26 '18 at 20:20
So what error[s] are you getting with the above code when you run it?
– epascarello
Nov 26 '18 at 20:44
Possible duplicate of How do I return the response from an asynchronous call?
– Patrick Hund
Nov 26 '18 at 20:20
Possible duplicate of How do I return the response from an asynchronous call?
– Patrick Hund
Nov 26 '18 at 20:20
So what error[s] are you getting with the above code when you run it?
– epascarello
Nov 26 '18 at 20:44
So what error[s] are you getting with the above code when you run it?
– epascarello
Nov 26 '18 at 20:44
add a comment |
1 Answer
1
active
oldest
votes
1
Return value in given address is not in correct json format. Enter the url in your browser and check the output !
answered Nov 26 '18 at 20:43
RaminRamin
1245
Also "awsURL" include more than one address.
– Ramin
Nov 26 '18 at 20:46
URL is using cors to read json file, that's why you see both URLS
– Chamaco Arano
Nov 26 '18 at 21:05
Anyway, it changes the result
– Ramin
Nov 26 '18 at 21:15
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
Return value in given address is not in correct json format. Enter the url in your browser and check the output !
answered Nov 26 '18 at 20:43
RaminRamin
1245
Also "awsURL" include more than one address.
– Ramin
Nov 26 '18 at 20:46
URL is using cors to read json file, that's why you see both URLS
– Chamaco Arano
Nov 26 '18 at 21:05
Anyway, it changes the result
– Ramin
Nov 26 '18 at 21:15
add a comment |
1
Return value in given address is not in correct json format. Enter the url in your browser and check the output !
answered Nov 26 '18 at 20:43
RaminRamin
1245
Also "awsURL" include more than one address.
– Ramin
Nov 26 '18 at 20:46
URL is using cors to read json file, that's why you see both URLS
– Chamaco Arano
Nov 26 '18 at 21:05
Anyway, it changes the result
– Ramin
Nov 26 '18 at 21:15
add a comment |
1
1
1
Return value in given address is not in correct json format. Enter the url in your browser and check the output !
answered Nov 26 '18 at 20:43
RaminRamin
1245
Return value in given address is not in correct json format. Enter the url in your browser and check the output !
answered Nov 26 '18 at 20:43
RaminRamin
1245
answered Nov 26 '18 at 20:43
RaminRamin
1245
answered Nov 26 '18 at 20:43
RaminRamin
1245
answered Nov 26 '18 at 20:43
RaminRamin
1245
1245
Also "awsURL" include more than one address.
– Ramin
Nov 26 '18 at 20:46
URL is using cors to read json file, that's why you see both URLS
– Chamaco Arano
Nov 26 '18 at 21:05
Anyway, it changes the result
– Ramin
Nov 26 '18 at 21:15
add a comment |
Also "awsURL" include more than one address.
– Ramin
Nov 26 '18 at 20:46
URL is using cors to read json file, that's why you see both URLS
– Chamaco Arano
Nov 26 '18 at 21:05
Anyway, it changes the result
– Ramin
Nov 26 '18 at 21:15
Also "awsURL" include more than one address.
– Ramin
Nov 26 '18 at 20:46
Also "awsURL" include more than one address.
– Ramin
Nov 26 '18 at 20:46
URL is using cors to read json file, that's why you see both URLS
– Chamaco Arano
Nov 26 '18 at 21:05
URL is using cors to read json file, that's why you see both URLS
– Chamaco Arano
Nov 26 '18 at 21:05
Anyway, it changes the result
– Ramin
Nov 26 '18 at 21:15
Anyway, it changes the result
– Ramin
Nov 26 '18 at 21:15
add a comment |
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Possible duplicate of How do I return the response from an asynchronous call?
– Patrick Hund
Nov 26 '18 at 20:20
So what error[s] are you getting with the above code when you run it?
– epascarello
Nov 26 '18 at 20:44