Reputation: 3512
I am trying to create a sequence of functions that are chained together. The following does what I want, but I want to be able to call any of the functions alone to start the sequence. For example, if I called one() it would run one(), then two(), then three(), then complete(). If I called three() it would run three then complete(). Is there a way to 'link' each function to the one below it without calling it like one().then(two).then(three).then(complete);
but instead just one(), or just two(), etc?
function one() {
var d = $.Deferred();
console.log('one');
setTimeout(function() {
d.resolve();
}, 5000);
return d.promise();
}
function two() {
var d = $.Deferred();
console.log('two');
setTimeout(function() {
d.resolve();
}, 5000);
return d.promise();
}
function three() {
var d = $.Deferred();
console.log('three');
setTimeout(function() {
d.resolve();
}, 5000);
return d.promise();
}
function complete() {
console.log('done');
}
one().then(two).then(three).then(complete);
Upvotes: 0
Views: 141
Reputation: 171698
Assuming order is always the same just add a then()
inside each one and call the next in sequence there.
function one() {
var d = $.Deferred();
console.log('one');
setTimeout(d.resolve, 1000);
return d.promise().then(two);
}
function two() {
var d = $.Deferred();
console.log('two');
setTimeout(d.resolve, 1000);
return d.promise().then(three);
}
function three() {
var d = $.Deferred();
console.log('three');
setTimeout(d.resolve, 1000);
return d.promise().then(complete);
}
function complete() {
console.log('done');
}
one();
setTimeout(()=>{console.log('Start at three'); three()}, 5000)
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
Upvotes: 2