Parse - Query users object contained in an array

Question

I'm running a Parse Query on my User class.

I want to retrieve users that are contained in an array of strings (ID).

const usersQuery = new Parse.Query(User).containedIn('objectId', customersArrayId).find({ useMasterKey: true });

Which works, actually. I'm getting a [ParseUser { _objCount: 6, className: '_User', id: 'iYIJ7Zrmms' }], because only 1 user matches.

But well, my issue is that I'm only getting ParseUser { _objCount: 6, className: '_User', id: 'iYIJ7Zrmms' }. This class contains other fields (firstname, lastname, e.g.) that are not returned.

When I performed the same thing, looping on my customersArrayId and performing .get():

const customerId = customersArrayId[index];
promises.push(new Parse.Query(User).get(customerId).then((user) => { 
  return user.toJSON();
}, (error) => console.error(error)));

I'm getting the full object, as expected. But it doesn't seem to be the right way of querying Parse objects from an array of ids.

I can't find anything in the docs about it, any idea why containedIn only returns a part of the queried objects?

enguerranws · Accepted Answer · 2019-12-04 11:18:11Z

0

I actually get the difference:

new Parse.Query(User).get(customerId)

=> returns the Parse Object

new Parse.Query(User).containedIn('objectId', customersArrayId)

=> returns the Parse User, subclass of a Parse Object.

And well, then, this thread was useful: Get user from parse in Javascript

I ended up using :

usersQuery.then(customersResponse => {
    const customers = [];
    for (let index = 0; index < customersResponse.length; index++) {
      const customer = {
        ...customersResponse[index].toJSON(),
        ...customersResponse[index].attributes
      }; 
      ...

Still not sure that the best answer.

EDIT

router.get('/:userId/shop/customers/details', checkUserMatch, (req, res, next) => {


if('shops' in req.jwtData.data) {
    const shopId = req.jwtData.data.shops[0];
    const query = new Parse.Query('OtherStuff');
    const Shop = Parse.Object.extend('Shops');
    query.equalTo('shop', new Shop({id: shopId})).find().then((otherStuffs) => {
      const customersArrayId = otherStuffs.map(otherStuff => otherStuff.toJSON().user.objectId);
      const usersQuery = new Parse.Query('_User').containedIn('objectId', customersArrayId).find({ useMasterKey: true });
      usersQuery.then(customersResponse => {
        const customers = [];
        for (let index = 0; index < customersResponse.length; index++) {
          let customer = {
            ...customersResponse[index].toJSON(),
            ...customersResponse[index].attributes
          }; 
          const customerId = customer.objectId;         
          const stuffQuery = new Parse.Query('Stuff').equalTo('user', new UserModel({objectId: customerId})).find().then((stuff) => {
            return stuff;
          });
          const otherStuffQuery = new Parse.Query('otherStuff').equalTo('user', new UserModel({objectId: customerId})).find().then((otherStuff) => {
            return otherStuff;
          });
          Promise.all([stuffQuery, otherStuffQuery]).then((data) => {
            const stuff = data[0];
            const otherStuff = data[1];
            customer = {
              ...customer,
              stuff,
              otherStuff,
            }
            customers.push(customer);
            if(index === customersResponse.length - 1) { // last customer
              res.json({
                success: true,
                data: customers
              });
            }
          })
        }


      });

    });
  }
});

edited Dec 4, 2019 at 11:18

answered Dec 3, 2019 at 8:53

enguerranws

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4 Comments

Davi Macêdo Over a year ago

Your last code looks good. Can you share the complete version so I can better advise?

enguerranws Over a year ago

I updated my answer, glad someone from B4A is looking at this, since I'm actually using B4A on this project :)

Davi Macêdo Over a year ago

Nice! Your code looks good to me and it should work properly. But you can probably improve it using relational queries. Take a look at matchesQuery and matchesKeyInQuery constraints.

enguerranws Over a year ago

Thanks for the tip, I'll try that and let you know.

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