Parse Cloud Code relational query syntax

Question

I have a cloud function on Parse. When it's called it retrieves a PFObject then adds a relation between that object and the user. This part works fine (Seen towards the end of the function). I'm having trouble getting the query that selects the PFObject to ignore those that the user is already related to

Here is my code:

Parse.Cloud.define("NextMedia", function (request, response) {

    var LikeRequest = Parse.Object.extend("LikeRequest");
    var query = new Parse.Query(LikeRequest);

    query.equalTo("completed", false);
    console.log("user: " + Parse.User.current().id);

    query.notEqualTo("user", Parse.User.current());

    // Also only fetch if never been sent before 
    // HERE SHOULD USE THE BELOW RELATIONSHIP
    var innerQuery = new Parse.Query(Parse.User);
    innerQuery.exists(Parse.User);
    query.matchesQuery("sentTo", innerQuery);

    query.ascending("createdAt");

    query.first({

        success: function (object) {
            // Successfully retrieved the object.
            console.log("Got 1 object: " + object.get('mediaId'));

            // Record that the user has been sent it
            var user = Parse.User.current();
            var relation = object.relation("sentTo"); // RELATION ADDED HERE

            relation.add(user);
            object.save();

            response.success(object);
        },
        error: function (error) {
            console.log("Error: " + error.code + " " + error.message);

            response.error("Error getting next mediaId");
        }
    });
});

I'm sure i'm just not understanding how the relation query syntax works.

On your LikeRequest class do you have a user column? I'm not understanding query.notEqualTo("user", Parse.User.current()); — bmurmistro, Apr 11 '14 at 14:59
In the bottom section where it adds the Relation, that's adds a many to many user column. — Darren, Apr 11 '14 at 18:03
Sorry for the delayed response. Have you tried something like this: var innerQuery = LikeRequest.relation("sendTo").query(); innerQuery.exists(Parse.User); — bmurmistro, Apr 14 '14 at 14:18
I'm away for a few days now, so will try when I get back. Thanks — Darren, Apr 15 '14 at 11:22
This gives error `Object function (){a.apply(this,arguments)} has no method 'relation'` — Darren, Apr 17 '14 at 17:49
Hmm, I just tried with just `query.notEqualTo("sentTo", Parse.User.current());` and it seems to work. Even though none of the syntax is for relations. It must just search the `sentTo` relation for any matches. — Darren, Apr 17 '14 at 18:20
There's a query method on the Relation. Think you want to use that as the inner query: http://www.parse.com/docs/js/symbols/Parse.Relation.html#query — Tom Erik Støwer, Sep 20 '14 at 19:55
Only now realized this is very old question. Did you ever come to a solution? — Tom Erik Støwer, Sep 20 '14 at 20:06
It would be helpful if you show the data type of the involved fields. — eth3lbert, Jan 24 '15 at 18:36
Try to include USER Class - Put `query.include('_User');` before `query.notEqualTo("user", Parse.User.current());` — Vintesh, Jan 26 '15 at 19:02

score 2 · Answer 1 · edited Mar 28 '15 at 07:54

2

This stretch:

// Also only fetch if never been sent before 
// HERE SHOULD USE THE BELOW RELATIONSHIP
var innerQuery = new Parse.Query(Parse.User);
innerQuery.exists(Parse.User);
query.matchesQuery("sentTo", innerQuery);

Could be changed to:

// Also only fetch if never been sent before 
query.notContainedIn("sentTo",[Parse.User.current()])

That works.Parse Query

edited Mar 28 '15 at 07:54

Muath

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answered Feb 10 '15 at 11:39

Oscar Eduardo

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score 1 · Answer 2 · answered Feb 08 '15 at 18:59

what about Query Constraints?

If you want to retrieve objects that do not match any of several values you can use notContainedIn

  // Finds objects from anyone who is neither Jonathan, Dario, nor Shawn
query.notContainedIn("playerName",
                     ["Jonathan Walsh", "Dario Wunsch", "Shawn Simon"]);

score 1 · Answer 3 · answered Feb 09 '15 at 18:10

1

I think Kerem has it partially correct but notContained in is not dynamic enough for your case.

You need to construct the query from the relation and then use doesNotMatchKeyInQuery to exclude those objects from your outer query.

answered Feb 09 '15 at 18:10

Rog

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Parse Cloud Code relational query syntax

3 Answers3