I'm trying to solve an issue. Currently, I'm opening a modal but I needed the combination of the 3 route observables for this. So, my code logic is something like this:
combineLatest([obs1, obs2, obs3])
.subscribe(([s1, s2, s3]: any) => {
openModal();
});
Since it will get executed 3 times, my modal will also be opened three times but I don't want it to do that but I know is expected behavior.
So, for this, the bad solution I implemented was using a flag like this:
let shouldOpen = true;
combineLatest([obs1, obs2, obs3])
.subscribe(([s1, s2, s3]: any) => {
if(shouldOpen) {
shouldOpen = false;
openModal();
}
});
But of course, that's not a good solution.
So, the question is, Is there a way for me to keep using combineLatest but getting it executed only once?
Here's a Stackblitz if you want to try with combineLatest.
Note: I cannot use
forkJoinorzip, I already tried those.
Since you can't use forkJoin (which is the right answer), skip the first two emissions from the combineLatest
combineLatest([obs1, obs2, obs3])
.pipe(skip(2))
.subscribe(([s1, s2, s3]: any) => {
openModal();
});
You could also use skipWhile to skip until all three are defined:
.pipe(skipWhile(([s1, s2, s3]) => s1===undefined || s2 === undefined || s3===undefined))