Javascript function returns "un defined"

Question

1.00/5 (1 vote)

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JavaScript

function Sti(addr, tag, id, tt) {
  let the;
  (async () => {
      let response = await fetch('http://funapi.eu5.org/our_api.php?addr='+addr+'&id='+id+'&tag='+tag+'&tt='+tt);

      let text = await response.text();

      the = text;
  })()
  return the;
}

let then = Sti("https://stackoverflow.com", "div", "left-sidebar", "html");

document.getElementById('h3').innerHTML = then;

Tryit Editor v3.6[^]
after running this code the

HTML

<p id=h3></p>

block just shows "undefined" why?? I normally use var instead of let recently started using let I noticed some subtle differences is this because of that??

What I have tried:

I tried using var to declare the variable inside the async function!!
https://i.imgur.com/Oa4xKVe.png
this is what console shows.

Posted 12-Jun-21 4:33am

Ɗααɳιടԋ Sყҽԃ

Updated 14-Jun-21 4:31am

v2

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4 solutions

Solution 2

If you build the URL based on the above, you get http://funapi.eu5.org/our_api.php?addr=https://stackoverflow.com&id=left-sidebar&tag=div&tt=html[^]. Paste that into your browser to check the result.

Posted 13-Jun-21 0:48am

Richard MacCutchan

Comments

Ɗααɳιടԋ Sყҽԃ 13-Jun-21 7:09am

that is what i want

Richard MacCutchan 13-Jun-21 7:13am

What is?

Ɗααɳιടԋ Sყҽԃ 13-Jun-21 7:15am

i wanted to get this https://imgur.com/ke5agRB

Richard MacCutchan 13-Jun-21 7:16am

Well you got it, so now what do you want to do?

Ɗααɳιടԋ Sყҽԃ 13-Jun-21 11:30am

I want to get it in the JavaScript code, My question clearly says that the function returns "un defined" i want it to return what the page "http://funapi.eu5.org/our_api.php?addr=https://stackoverflow.com&id=left-sidebar&tag=div&tt=html" consists of!!

Richard MacCutchan 13-Jun-21 11:45am

I have tried a few variations on your script but it fails every time. I have a suspicion it is something to do with the async calls and the fetch method. Sadly I am not a Javascript expert so am out of other ideas. You need to do some more debugging to find out why it is not returning any data.

Solution 3

Its seems your code is clear but the issue causing when you assigning the response from the url to variable.

let text = await response.text();

Solution:
1. check the generated URL by your code whether working or not and retuning proper response

2. Pre-check the response before assigning to var or let

3. If URL response contains any property with your HTML or TEXT value then you have to assign the property to your variable not a full response.

Posted 14-Jun-21 1:58am

Prabhu Ravichandran

Updated 14-Jun-21 2:14am

v2

Comments

Richard MacCutchan 14-Jun-21 8:15am

Please do not use bold text for all the answer. People here are capable of reading normal weight fonts.

Ɗααɳιടԋ Sყҽԃ 14-Jun-21 10:32am

no that is working without the Sti() function

Solution 1

You should try using this:

JavaScript

var the;

instead of

JavaScript

let the;

Posted 12-Jun-21 9:27am

A-Games

Comments

Ɗααɳιടԋ Sყҽԃ 12-Jun-21 16:15pm

I've tried that it still doesn't work

A-Games 12-Jun-21 18:16pm

Oh. Well, i experemented with ur code some more and i dont know :\

A-Games 12-Jun-21 18:16pm

srry

Ɗααɳιടԋ Sყҽԃ 13-Jun-21 7:05am

no prob

A-Games 25-Jun-21 22:34pm

im not good with PHP

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Richard Deeming · Accepted Answer · 2021-06-14T04:31:00

Your Sti function returns before the inner async function has completed.

You can't write a sync-over-async wrapper in Javascript. You need to make the Sti function async, and change the calling code to await the result.

JavaScript

(async function(){
    async function Sti(addr, tag, id, tt) {
        let response = await fetch(...);
        let text = await response.text();
        return text;
    }
    
    let then = await Sti("https://stackoverflow.com", "div", "left-sidebar", "html");
    document.getElementById('h3').innerHTML = then;
})();

You'll have two other issues running that code on TryItEditor:

You're trying to load an http: URL from an https: site. Your browser will probably block that request.
The URL you're trying to load doesn't allow cross-origin requests.