Unable to print property of returned JSON object from get request

Question

<script>

        var apikey = "someKey";
        var SEid = "0137454dfgsdfgsdfgzxv4435099:pnqp_fvtmsy";
        var query = "Flower";
        var returnedData;

        $.get("https://www.googleapis.com/customsearch/v1?key=" + apikey + "&cx=" + SEid + "&q=" + query, function(data){

        console.log(data);
        returnedData = data;
        });

        console.log(returnedData.items[1].formattedUrl);
        </script>

I keep getting back the error Uncaught TypeError: Cannot read property 'items' of undefined

but I can't figure out why the variable returned Data is undefined. I'm not trying to even reference it until after I've called the get request.

Please don't flag this question as invalid, I'm really trying to figure this out and have tried for the past 2 days various methods but I just can't figure it out.

@brandenM `$.get` makes an asynchronous call which mean we never know when the response will be back & `js` will not wait for it.So `console.log(returnedData.items[1].formattedUrl);` will be executed instead of waiting for the result.When it is executing `returnedData` is undefined because till now no response has arrived from `$.get` call. Please check the duplicate.The problem & solution is clearly defined there.Hope it will be useful — brk, Mar 21 '18 at 04:44

score 1 · Answer 1 · answered Mar 21 '18 at 04:42

1

The $.get is running asynchronously - once the script reaches that line, it simply sends out the request, it doesn't wait for the request to complete before continuing.

You can't return a value that's generated asynchronously. You could, however, return a promise that, when resolved, resolves with the desired value.

answered Mar 21 '18 at 04:42

CertainPerformance

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run `console.log(returnedData.items[1].formattedUrl);` right after you define it, within the curly brackets, and I bet you'll see it. – Dan Oswalt Mar 21 '18 at 04:43
Yea but then how do I access it for later and or put it into a global variable? – brandenM Mar 21 '18 at 14:55
@brandenM like I said, return a Promise that resolves with the desired value. – CertainPerformance Mar 21 '18 at 18:11

score 0 · Answer 2 · answered Mar 21 '18 at 04:40

0

Using returnedData.toJSON().items or just returnedData.get(items).get(0) directly.

answered Mar 21 '18 at 04:40

Hearen

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Unable to print property of returned JSON object from get request

2 Answers2