Nested (or child) synced-cron jobs

Is it possible to spawn another cron job derived from this parent cron job as in this scenario as I do not see any examples in doc.

name: ‘Cron job HTTP get from external API’,
schedule: function(parser) {
return parser.text(‘every 15 seconds’);
job: function(intendedAt) {‘getServiceA’, ‘userName’, function (err, respJson) {
if (err) {
console.log("Reason of error is : ", err.reason);
} else {
console.log("respJson: ", respJson);
console.log('job should be running at:');
console.log('intendedAt :' ,intendedAt);


Meteor.startup(() => {

    getServiceA: function (userName) {
    const nextOneDay = moment().add(1, 'days').unix();
    const url = '", EJSON.stringify(nextOneDay )";
    var result ='GET', url, {}, function (err, resp) {

   // How to start a list of one time future child SyncedCron jobs, 

‘postServiceB’ with time information extracted from getServiceA result like
// i.e
postServiceB 20 mins from extracted time 2016-04-13 10:10:00
postServiceB 60 mins from extracted time 2016-04-15 10:50:00
postServiceB 80 mins from extracted time 2016-04-16 20:10:00

Or do you do another SyncedCron.start() in parent cron job (getServiceA) to start child cron job?

Here shows how to fire one time job, but not sure how to structure this as a child job

var newCronJobId = Meteor.uuid();
name: newCronJobId,
schedule: function (parser) {
return parser.text(‘after 20 secs’);
job: function () {
// do stuff

I don’t quite follow your example, but you can call any server side code you want with SyncedCron. I’d suggest just calling some server code that then triggers whatever parent/child jobs you want to have run in order. So have something like a CronManager object with a startAll function, that is called from SyncedCron, and starts all server processes you want to have run/monitor together.

Thanks Huge, basically a recurring parent cron job that extracts time information from one URL, and using that time information to fire off a list of one-time child cron jobs to another URL service…

Is this a too simple or too difficult question? Hence no replies? Thanks