NOTE
This problem somehow happens only with a specific server-side api. Therefore it addresses the wrong problem. I'll not delete it since it have answers and comments.
I'm trying to execute a few ajax requests, do some stuff after each is done and some other stuff after all of them are done, for that I'm using the code below:
let
myarr = [],
myfunc = arg => myarr.push(arg);
$.when(
$.post(myparams).done(myfunc),
$.post(otherparams).done(myfunc),
$.post(yetanother).done(myfunc)
// it comes out with only one arg
).then(e => console.log(myarr));
But when it comes to execute the then block it usually has only executed the done of the first operation, how could I fix that?
I'm sorry if it's a duplicate but honestly I didn't even knew what to search for :/
Comment
I also tried to create my own deferreds where I would execute the ajax and resolve them inside the done block, but that yielded the same results.
Using only done or only then, same.
Per jQuery's documentation on $.when():
Each argument [of
.then()] is an array with the following structure:[ data, statusText, jqXHR ]
Meaning you could do something like this...
$.when(
$.post(myparams),
$.post(otherparams),
$.post(yetanother)
).then((res1, res2, res3) => { //Arg for each result
myfunc(res1[0]); //Call myfunc for result 1's data
myfunc(res2[0]); //Call myfunc for result 2's data
myfunc(res3[0]); //Call myfunc for result 3's data
});
Though perhaps a cleaner version might be something like this instead...
let
myarr = [],
myfunc = arg => myarr.push(arg);
$.when(
$.get('https://jsonplaceholder.typicode.com/todos/1'),
$.get('https://jsonplaceholder.typicode.com/todos/2'),
$.get('https://jsonplaceholder.typicode.com/todos/3')
).then((...results) => { //Get all results as an array
results.map(r=>r[0]).forEach(myfunc); //Call myfunc for each result's data
console.log(myarr);
});<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>