logging errors in async functions - node.js

Question

I have this code:
async function getURL() {
try {
await fetch("http://www.blah.com");
return 0;
} catch (err) {
return err;
}
}
getURL().then( result => {
if (result === 0) console.log("success");
else console.log(result);
});
The fetch will fail and the error is logged to the console. How do I rework the code so it uses async and try/catch everywhere? That is, I'm looking to avoid doing getURL().then for the sake of consistency.
EDIT:
For those downvoting me, await getURL() won't work as it's invalid syntax.
EDIT2:
Tried this but it didn't catch the error:
async function getURL() {
return await fetch("http://www.blah.com");
}
let result = async function() {return await getURL();}
try {
result();
} catch (e) {
console.log(e);
}

Answer 1

You can wrap your whole code inside an instantly executed async function like this:
// service.js
async function getURL() {
return await fetch("http://www.blah.com");
}
// your.module.js
(async function() {
// do things...
try {
let result = await getURL();
} catch (e) {
console.log(e);
}
// do things...
res.send({});
});

Answer 2

Every time you need to catch an error from promise, either using new Promise, async-await or generator you need to use .then() or you can do something like this another async-await.
async function getURL() {
try {
await fetch("http://www.blah.com");
return 0; // EDIT: just returning value which is success
} catch (err) {
return err; // EDIT: returning value not rejecting a promise
}
}
async function main () {
try {
let result = await getURL();
if (result === 0) console.log("success");
console.log(result); // EDIT: error will be print.
}
catch (err) { // EDIT: getURL() never rejects so always success.
console.log(err);
}
});
main();
This situation doesn't really occurs as while our main function in server-side or client-side are async and handling this for us.
Like using express:
app.post('/api', async (req, res) => {
try {
let result = await getURL();
res.send(async);
}
catch(err) {
res.send(err);
}
});
EDIT: asyn-await doesn't reject or resolve a call, just return a value. thus must be used carefully.
function fetch(url) {
return new Promise( (resolve, reject) => {
let x = Math.floor(Math.random() * Math.floor(9) + 1);
// 50-50 resolve or reject
if(x%2===0) return resolve(false); //resolve with `false` statement
reject(true); // reject with `true` still a reject
});
}
async function getURL() {
try {
await fetch("http://www.blah.com");
return 0; // if fetch resolve
} catch (err) { //only if fetch reject
return err;
}
}
async function main () {
try {
let result = getURL();
if (result === 0) console.log("success"); //getURL never reject any call
console.log(result);
}
catch (err) { // getURL doesnt reject
console.log(err);
}
};
main();

Answer 3

I realize now async functions always return a promise. Even if you throw an error it still gets wrapped up into a promise. Therefore using try/catch won't help. This is how I ended up writing the code:
async function getURL() {
return await fetch("http://fake");
}
getURL().then( () => console.log("success")).catch( (e) => console.log(e));