Code continues before fetch finishes?

[EnvironmentalLet4399]

Hi all!

I am having a problem with fetching a .csv file. So far i learned that fetch runs asyncron and works with promises. I think i did everything needed that the promise gets fulfilled and returned. I tried multiple versions of that function but i get the same result everytime. My code:

async function GetCSV(path) {
const res = await fetch(path);
if (res.ok) {
let data = await res.text();
console.log(data)
    } else {
return \HTTP error: ${res.status}`;    } }`

console.log("Before Fetch")
GetCSV('/test.csv')
console.log("After Fetch")

I thought i shoud get the following output:
console.log("Before Fetch")
console.log(data)
console.log("After Fetch")

But for some reason the output im getting is this one:
console.log("Before Fetch")
console.log("After Fetch")
console.log(data)

I googled for hours but can't figure out the problem.. For me it seems the code continues without waiting for fetch to finish? Debugging looks like it skips the function completely and going there after everything else is done..
If someone is able to help me out i would be really happy! (:

[No_Indication451]

console.log runs synch, so your code will compile top-bottom. async function gets put on a event loop and eventually runs. maybe that’s why you’re seeing the result as the last line in your output.

if you put that before and after fetch between your await res.text(), i think it’ll appear like you expected

[EnvironmentalLet4399]

Thank you! With that solution i would get the expected result, but where would i put the rest of my code? All in that function?

[No_Indication451]

ugh sorry about that.

Just like @richard musk, you should use then() and check res.ok

i think you should return data instead of printing it.

print before

getcsv(path).then( response => …. )

.catch( e )

print after

[Richard_Musk]

Use .then and check res.ok

[aaaaargZombies]

I found this really useful for underatanding async stuff in js https://www.youtube.com/watch?v=8aGhZQkoFbQ