Свести один столбец держать других в PostGreSQL

Question

Я создал представление, которое в настоящее время дает данные, как это:

practice_name message_type message_count 
CHC     ALOG_SYNC 1 
CHC     BULKNT 0 
CHC     PIE_SYNC 1 
CHC     PPRV_SYNC 1 
CHC    SYNC_PRACT 3 
CHC     SYNC_PROV 9 
CHC     SYNC_WTXT 3 
CHC     SYNC_XYZ 0 
Midtown    ALOG_SYNC 0 
Midtown     BULKNT 0 
Midtown    PIE_SYNC 0 
Midtown    PPRV_SYNC 0 
Midtown   SYNC_PRACT 3 
Midtown   SYNC_PROV 0 
Midtown   SYNC_WTXT 3 
Midtown    SYNC_XYZ 0 
NextGen MedicalPractice ALOG_SYNC 0 
NextGen MedicalPractice BULKNT 1 
NextGen MedicalPractice PIE_SYNC 0 
NextGen MedicalPractice PPRV_SYNC 0 
NextGen MedicalPractice SYNC_PRACT 3 
NextGen MedicalPractice SYNC_PROV 591 
NextGen MedicalPractice SYNC_WTXT 3 
NextGen MedicalPractice SYNC_XYZ 0 

Мой Вид:

CREATE OR REPLACE VIEW sha.sha_export_queue_view AS 
SELECT q3.practice_name, 
    q3.message_type, 
    q3.share_site_org_key, 
    COALESCE(q2.message_count, '0'::text) AS message_count 
    FROM (SELECT q1.practice_name, 
      mt.message_type, 
      q1.share_site_org_key 
      FROM sha.message_types mt, 
      (SELECT DISTINCT jsonb_array_elements((ai.result_json -> 'Patient Portal Operational Information'::text) -> 'nxmd_export contents by message type'::text) ->> 'Practice Name'::text AS practice_name, 
        ai.share_site_org_key 
        FROM sha.sha_share_site_view ssv 
        LEFT JOIN (SELECT mytable2.assessment_id, 
          mytable2.result_json, 
          mytable2.share_site_org_key, 
          mytable2.rnk 
          FROM (SELECT assessment_info.assessment_id, 
            assessment_info.result_json, 
            assessment_info.share_site_org_key, 
            dense_rank() OVER (PARTITION BY assessment_info.share_site_org_key ORDER BY assessment_info.modified_datetime DESC) AS rnk 
            FROM sha.assessment_info 
            WHERE assessment_info.assessment_id = 8::numeric) mytable2 
          WHERE mytable2.rnk = 1) ai ON ssv.share_site_org_key = ai.share_site_org_key) q1) q3 
    LEFT JOIN (SELECT jsonb_array_elements((ai.result_json -> 'Patient Portal Operational Information'::text) -> 'nxmd_export contents by message type'::text) ->> 'Practice Name'::text AS practice_name, 
      jsonb_array_elements((ai.result_json -> 'Patient Portal Operational Information'::text) -> 'nxmd_export contents by message type'::text) ->> 'Message Type'::text AS message_type, 
      jsonb_array_elements((ai.result_json -> 'Patient Portal Operational Information'::text) -> 'nxmd_export contents by message type'::text) ->> 'Message Count'::text AS message_count 
      FROM sha.sha_share_site_view ssv 
      LEFT JOIN (SELECT mytable2.assessment_id, 
        mytable2.result_json, 
        mytable2.share_site_org_key, 
        mytable2.rnk 
        FROM (SELECT assessment_info.assessment_id, 
          assessment_info.result_json, 
          assessment_info.share_site_org_key, 
          dense_rank() OVER (PARTITION BY assessment_info.share_site_org_key ORDER BY assessment_info.modified_datetime DESC) AS rnk 
          FROM sha.assessment_info 
          WHERE assessment_info.assessment_id = 8::numeric) mytable2 
        WHERE mytable2.rnk = 1) ai ON ssv.share_site_org_key = ai.share_site_org_key) q2 ON q3.message_type::text = q2.message_type AND q3.practice_name = q2.practice_name 
    ORDER BY q3.practice_name; 

Я хочу второй столбец быть сплющенные:

Practice Time Stamp <<message type 1>> <<message type 2>> <<message type 3>> <<message type 4 >> <<message type 5>> <<message type 6>> <<message type 7>> <<message type 8>> 
Practice Name 1 21-12-2016 10:00 23 25 27 29 31 33 35 37 
Practice Name 2 21-12-2016 10:00 24 26 28 30 32 34 36 38 
Practice Name 3 21-12-2016 13:00 25 27 29 31 33 35 37 39 
Practice Name 4 21-12-2016 13:00 26 28 30 32 34 36 38 40 
Practice Name 5 24-12-2016 13:00 27 29 31 33 35 37 39 41 
Practice Name 6 27-12-2016 13:00 28 30 32 34 36 38 40 42 
Practice Name 7 30-12-2016 13:00 29 31 33 35 37 39 41 43 
Practice Name 8 02-01-2017 13:00 30 32 34 36 38 40 42 44 
Practice Name 1 05-01-2017 13:00 31 33 35 37 39 41 43 45 
Practice Name 2 08-01-2017 13:00 32 34 36 38 40 42 44 46 
Practice Name 3 11-01-2017 13:00 33 35 37 39 41 43 45 47 

Есть ли способ, которым я могу это достичь? Извините за небольшую проблему с выравниванием. Значения соответствующего типа сообщения значение

Answer 1

Образца для запроса (идеи в комментариях на вопрос):

SELECT 
    practice_name, 
    sum("ALOG_SYNC") AS "ALOG_SYNC", 
    sum("BULKNT") AS "BULKNT", 
    ... 
FROM (
     SELECT 
      practice_name, 
      CASE WHEN q3.message_type = 'ALOG_SYNC' THEN sum(message_count) END AS "ALOG_SYNC", 
      CASE WHEN q3.message_type = 'BULKNT' THEN sum(message_count) END AS "BULKNT" 
     FROM 
      <your from + where clause> 
    ) AS A 
GROUP BY 1 

Возможно ваш запрос может быть оптимизирован.

Или вы можете использовать функцию перекрестный (https://www.postgresql.org/docs/current/static/tablefunc.html)