Code won't execute in function containing return?

Question

I've run into some issues with my scoping, I'm new to javascript so my apologies for my inexperience. The issue i'm having is I don't understand why my console.log within the function groupA(thename) curly braces won't allow code to output. In visual studio's it greys the code out. I think it may be due to the return fetch(`${url}${thename}`), because the code won't grey out when I remove that return. Why does the return function do this?

My goal is to correctly chain each function together so the data pushed into birds_array can be returned via the groupA function. How would I go about correctly formatting these functions to allow this to be possible?

const birds_array = []

let groupAname = "Old World sparrow"
groupA(groupAname);

function groupA(thename) {
    return fetch(`${url}${thename}`)
    .then(response => response.text())
    .then(body => {
        const $ = cheerio.load(body);
        $('.llgymd').each( (index, element) => {
            const $element = $(element);
            const names = $element.text();
            groupA_array[index] = names;
        });


        groupA_loop()
        function groupA_loop(){
        for(j = 0; j < groupA_array.length; j++){
            if(!birds_array.includes(groupA_array[j])){
                birds_array.push(groupA_array[j]);
                }
                // code functions here
                console.log(birds_array);
            }
            // code functions here
            console.log(birds_array);
        }
        // code functions here
        console.log(birds_array);
    });
    // code does not function here
    console.log(birds_array);
}

console.log(birds_array); is dead code...this is async funtion. u have to use then to get value — xdeepakv, May 01 '20 at 08:10
Does this answer your question? [Understanding JavaScript promise object](https://stackoverflow.com/questions/39458201/understanding-javascript-promise-object) — xdeepakv, May 01 '20 at 08:12
I've looked into promises, async/await and callbacks. I don't understand how to implement it in this situation. Please show me an example. Thank you. — srb633, May 01 '20 at 08:13
It sounds like you already know the solution, as you see that inside the `then` callback the `console.log` statement works. — Bergi, May 01 '20 at 08:13
I've been learning javascript for about a week. Most of this code was through watching tutorials and trying to understand as I go. I learn by doing, so it would be really helpful if you can be more specific as I don't understand this syntax very well. I'd appreciate it but it's fine if you don't want to help. — srb633, May 01 '20 at 08:16
All the working console.logs are within your return statement so to say, if you follow the {} you'll see that the last console.log is after return. And that is "dead" code because the execution is left the function after the return finished. — Rijad Husic, May 01 '20 at 08:18

score 1 · Accepted Answer · answered May 01 '20 at 08:19

// Create a async function to await any async code. Else u get error.

async function main() {

      let groupAname = "Old World sparrow";
      // Await for the promise to be resolved. // Like pause call
      const birds_array = await groupA(groupAname);
      // Result done, print
      console.log(birds_array)
    }
main()

Sample:

const delayData = (num) => {
  return new Promise((r) => {
    setTimeout(() => {
      r(num);
    }, 1000);
  });
};
async function main() {
  const num = await delayData(1);

  console.log(num); // 1
  const num2 = await delayData(2);

  console.log(num2); // 2
}
main();

// Using then

delayData(1).then(num => {
  console.log(num) // 1
})
delayData(2).then(num => {
  console.log(num) // 2
})

Okay starting to understand. Thank you – srb633 May 01 '20 at 08:27 — srb633, May 01 '20 at 08:27

Code won't execute in function containing return?

1 Answers1