CODEWITHSUNDEEP
javascriptreactjssupabasesupabase-database
vikingsteve
vikingsteve

Reputation: 40428

supabase query for one-to-one foreign key, return single value (not array)

How can I query a one-to-one relationship with foreign key in supabase?

My tables:

  1. games (id, created_at, played_at)
  2. results (game_id, key)

Data: games

("id": 1, "created_at": "2022-03-02T11:01:11+00:00", "played_at": "2022-04-02T12:15:06+00:00")

Data: results

("game_id": 1, "key": "alyktm3cf11bypgg")

My supabase query looks like this:

    const { data, error } = await supabase
        .from('games')
        .select('*, registrations(walletKey), results(key)')
        .order('scheduled_at')
    ;

The result contains an array of registrations, like here:

{
    "id": 1,
    "created_at": "2022-03-02T11:01:11+00:00",
    "played_at": "2022-04-02T12:15:06+00:00",
    "results": [
        {
            "key": "alyktm3cf11bypgg"
        }
    ]
}

But since there will always only be one result, can I modify my query to get this?

{
    "id": 1,
    "created_at": "2022-03-02T11:01:11+00:00",
    "played_at": "2022-04-02T12:15:06+00:00",
    "results": 
    {
        "key": "alyktm3cf11bypgg"
    }

}

Or even this:

{
    "id": 1,
    "created_at": "2022-03-02T11:01:11+00:00",
    "played_at": "2022-04-02T12:15:06+00:00",
    "key": "alyktm3cf11bypgg"
}

Upvotes: 6

Views: 3603

Answers (1)

chipilov
chipilov

Reputation: 569

This can be achieved by using the single() method:

const { data, error } = await supabase
    .from('games')
    .select('*, registrations(walletKey), results(key)')
    .order('scheduled_at')
    .single();

Note, that if the query does NOT return a single row, the method will return an error (i.e. the data field will be undefined or null while the error field will be defined).

Upvotes: 0

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