Add mongoose query (promise) to array and do parallel operations

Question

I'm using mongoose to do some operation in MongoDB. I'd like to seach all tasks created by user, then set value isPerforming to false and save it. Documentation says that mongoose query (with .exec() function) is promise. There's many tasks so I think to push them to array and do parallel save operations using Q.all([...]). The problem is when I try to do .push(). My code stops after first .push() operation. Is there any other way to do it?

 function stopUserTasks(userid) {

    var deferred = Q.defer();
    var query = Task.find({'_creator': userid}).exec();

    query.then(function(data, err) {
        console.log('found: ');
        console.log(data);
        if (err) {
          deferred.reject(err);
        }
        return data;
      })
      .then(function(data, err) {

        var len = data.length;
        var saveTasksPromises = [];

        for(var i = len; i--; ) {
          console.log(data[i]._id);
          saveTasksPromises.push(Task.save({'_id': data[i]._id, 'isPerforming': false}).exec() );
        }
        return saveTasksPromises;

      })
      .then(function(data, err) {
        console.log(data);

        deferred.resolve();
      });

      return deferred.promise;

  }// #stopUserTasks

Please don't use the explicit promise creation antipattern. If you have to convert one type of promises (mongoose) to another (q), use Promise.resolve(otherPromise) or Q(otherPromise). — Tamas Hegedus
– Tamas Hegedus, Commented Nov 16, 2016 at 16:45

Basim Hennawi · Accepted Answer · 2016-11-16 16:31:02Z

1

Model.save() returns promise actually, so no need to write Model.save().exec() so your line need to be adjusted as follow:

saveTasksPromises.push(Task.save({'_id': element._id, 'isPerforming': false}))

Also using Q.all, the whole function could be as follow:

function stopUserTasks(userid) {
  var deferred = Q.defer();
  var promise = Task.find({'_creator': userid}).exec();

  promise
    .then(function(data) {
      var len = data.length;
      var saveTasksPromises = [];

      for(var i = len; i--; ) {
        saveTasksPromises.push(Task.save({'_id': data[i]._id, 'isPerforming': false}));
      }
      return Q.all(saveTasksPromises);
    })
    .then(function(data) {
      deferred.resolve(data);
    });
    .catch(function(err) {
      deferred.reject(err);
    });
    return deferred.promise;
}

edited Nov 16, 2016 at 16:31

answered Nov 16, 2016 at 16:11

Basim Hennawi

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5 Comments

miuosh Over a year ago

It's not working. deferred promise isn't resolved or rejected. In console appear only first data._id. Of course element._id should be replaced by data[i]._id.

Basim Hennawi Over a year ago

Check the edited answer above, after putting Q.all in action.

miuosh Over a year ago

It almost work.. I figure out by myself how to do it thanks to yours suggestions. It works when in loop we change saveTasksPromises.push(Task.save({'_id': data[i]._id, 'isPerforming': false})); to var task = data[i]; task.isPerforming = false; saveTasksPromises.push(task.save());

Tamas Hegedus Over a year ago

Please don't use the explicit promise creation antipattern. If you have to convert one type of promises (mongoose) to another (q), use Promise.resolve(otherPromise) or Q(otherPromise).

miuosh Over a year ago

@Tamas Hegedus : Thanks to point it out. I'm rookie but It isn't excuse to code in bad way :)

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