CODEWITHSUNDEEP
jsonpostgresqlpostgresql-9.5
Patrick
Patrick

Reputation: 1066

PostgreSQL 9.5: Combine json_agg results into single json object

I am struggling with the following SQL query:

There is a table data_tracks with coordinates describing a trip. Each trip is uniquely identified by a trip_log_id. After reaching the trip's destination, the user needs to participate to a survey. The answers of the survey are stored in a table crowd_sourcing_answers. Each answer belongs to a question, located in table crowd_sourcing_questions.

I have written two SQL queries - one to get all points of a trip as JSON, another one to get all question-answers pairs:

Query for fetching all question-answer pairs of a trip:

SELECT json_agg(answer_single_trip)
FROM (SELECT json_agg(
               json_build_object(
                 'tripId', trip_log_id,
                 'question', qt.question,
                 'answeringOption', qt."answeringOptions",
                 'answer', at.answer
                   )
                 ) as crowdsourcing
      FROM crowd_sourcing_questions as qt
             INNER JOIN crowd_sourcing_answers as at ON at.crowd_sourcing_question_id = qt.id
      GROUP BY trip_log_id) answer_single_trip;

and its output:

[
  {
    "crowdsourcing": [
      {
        "tripId": 92,
        "question": "Gab es auf der Strecke teilweise schlecht befahrbare Streckenabschnitte?",
        "answeringOption": [
          "Ja",
          "Nein"
        ],
        "answer": "2"
      }
    ]
  },
  {
    "crowdsourcing": [
      {
        "tripId": 91,
        "question": "Gab es auf der Strecke teilweise schlecht befahrbare Streckenabschnitte?",
        "answeringOption": [
          "Ja",
          "Nein"
        ],
        "answer": "1"
      }
    ]
  },
  {
    "crowdsourcing": [
      {
        "tripId": 90,
        "question": "Gab es auf der Strecke teilweise schlecht befahrbare Streckenabschnitte?",
        "answeringOption": [
          "Ja",
          "Nein"
        ],
        "answer": "0"
      }
    ]
  }
]

Query for fetching all points belonging to a trip:

SELECT json_agg(
         json_build_object(
           'tripId', trip_log_id,
           'trackId', id,
           'recorded_at', created_at,
           'latitude', latitude,
           'longitude', longitude
             )
           ) as trips
FROM data_tracks
GROUP by trip_log_id;

and its output:

[
  [
    {
      "trip_log_id": 91,
      "recorded_at": "2018-10-05T14:11:44.847",
      "latitude": 52.5242370846803,
      "longitude": 13.3443558528637
    },
    {
      "trip_log_id": 91,
      "recorded_at": "2018-10-05T14:11:44.911",
      "latitude": 52.5242366166393,
      "longitude": 13.3443558656828
    }
  ],
  [
    {
      "trip_log_id": 90,
      "recorded_at": "2018-10-05T13:28:24.452",
      "latitude": 52.5242370846803,
      "longitude": 13.3443558528637
    },
    {
      "trip_log_id": 90,
      "recorded_at": "2018-10-05T13:28:24.489",
      "latitude": 52.5242366166393,
      "longitude": 13.3443558656828
    }
  ]
]

Goal

Now I need to merge these two results such that there is one JSON object for each trip ID, containing the question-answer pairs (key: "crowdsourcing"; array) and the points of the trip (key: "trip"; array). Following an example:

[
  {  // DATA FOR TRIP 1
    "crowdsourcing": [
      {
        "question": "Bitte bewerten Sie die Sicherheit der Radroute!",
        "answeringOption": [
          "Sehr sicher",
          "Eher sicher",
          "Neutral",
          "Eher unsicher",
          "Sehr unsicher"
        ],
        "answer": "2"
      },
      {
        "question": "Würden Sie die gefahrene Route anderen Radfahrenden weiterempfehlen?",
        "answeringOption": [
          "Ja",
          "Nein"
        ],
        "answer": "1"
      }
    ],
    "trip": [
      {
        "recorded_at": "2018-10-11T15:16:33",
        "latitude": 52.506785999999998,
        "longitude": 13.398065000000001
      },
      {
        "recorded_at": "2018-10-11T15:16:32.969",
        "latitude": 52.50647,
        "longitude": 13.397856000000001
      },
      {
        "recorded_at": "2018-10-11T15:16:32.936",
        "latitude": 52.506166,
        "longitude": 13.397593000000001
      }
    ]
  },
  { // DATA FOR TRIP 2
    "crowdsourcing": [
      {
        "question": "Bitte bewerten Sie die Sicherheit der Radroute!",
        "answeringOption": [
          "Sehr sicher",
          "Eher sicher",
          "Neutral",
          "Eher unsicher",
          "Sehr unsicher"
        ],
        "answer": "2"
      }
    ],
    "trip": [
      {
        "recorded_at": "2018-10-11T15:33:33.971999",
        "latitude": 52.506785999999998,
        "longitude": 13.398065000000001
      },
      {
        "recorded_at": "2018-10-11T15:33:33.929",
        "latitude": 52.50647,
        "longitude": 13.397856000000001
      }
    ]
  }
]

Approach

I created a query, please see DB Fiddle. However, it returns duplicate records within the two arrays (question-answers pairs, points of trip). I was thinking it has to do something with the JOIN but all my trials failed.

Upvotes: 3

Views: 4960

Answers (1)

S-Man
S-Man

Reputation: 23676

In your subqueries you have included the trip_log_id into the json part. But if you put them as separate column you'll have the chance to join both parts against it:

demo: db<>fiddle

SELECT
    json_agg(
        json_build_object('crowdsourcing', cs.json_agg, 'trip', t.json_agg)
    )
FROM
(
    SELECT 
        trip_log_id,                          -- 1
        json_agg(
            json_build_object('question', question, 'answeringOption', "answeringOptions", 'answer', answer)
        )
    FROM 
        crowd_sourcing_answers csa
    JOIN crowd_sourcing_questions csq ON csa.crowd_sourcing_question_id = csq.id
    GROUP BY trip_log_id
) cs

JOIN                                         -- 2

(
    SELECT
        trip_log_id,                         -- 1
        json_agg(
           json_build_object('recorded_at', created_at, 'latitude', latitude, 'longitude', longitude)
        )
    FROM data_tracks
    GROUP by trip_log_id 
) t

USING (trip_log_id)                          -- 2
  1. Get out the trip_log_id
  2. using it for joining

Addionally: Please notice that in postgres all column names should be without any capital letters. I would recommend to rename addionalOptions into something like additional_options. With that no additional " chars are needed.

Upvotes: 2

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