That is not true. Try the following code:
Konstantin Rouda
101

You are completely right, when the article was written I believe my thoughts were of a scenario like:

function foo () {
return new Promise(function(resolve, reject) {
setTimeout(resolve(‘resolved value’), 1000);
});
};
foo()
.then(function () {
new Promise(function(resolve, reject) {
setTimeout(resolve(‘resolved value expected on the next then’), 1000);
});
})
.then(function (val) {
console.log(val)
})

Which, like you stated, will return and log undefined instead of “resolved value expected on the next then”. That can be fixed by doing a return of new Promise(function(resolve, reject))… on the first .then


Regarding the reject consider the following:

function foo () {
return new Promise(function(resolve, reject) {
setTimeout(resolve(‘resolved value’), 1000);
});
};
foo()
.then(function () {
error.onPurpose
})
.then(function () {
console.log(`won’t execute`)
})
.then(function () {
console.log(`won’t execute also`)
})
.catch(function(error) {
console.log(‘error of not defined’, error)
})
.then(function (val) {
console.log(‘will execute regardless of error or not’);
});

What I meant was that an error is thrown on the first .then which makes it “skip” the second and third .then and be caught by the .catch , regardless of an error being caught the last .then , and any other that follows, will execute all the same (contrary to what happens with the “skipped” ones that are before the .catch)


The article will be edited accordingly so not to mislead anyone else, thank you for the feedback.