Выберите счетчик Oracle Query, основываясь на разных условиях

Я новичок с оракулом PL / SQL,
прямо сейчас с помощью оракула sqldeveloper,

мне нужна помощь для моей домашней работы,

как сделать так, чтобы столбец с номерами 2, 3, 4, 5, 6, ... рассчитывал на основе первого значения столбца, DISTINCT OWNER?

Может кто-нибудь мне помочь? или предложить мне какой-нибудь намек?

SELECT DISTINCT OWNER as SCHEMA, 
    (SELECT COUNT(DISTINCT OBJECT_NAME) FROM DBA_OBJECTS where object_type = 'TABLE') "TABLE",
    (SELECT COUNT(DISTINCT OBJECT_NAME) FROM DBA_OBJECTS where object_type = 'VIEW') "VIEW",
    (SELECT COUNT(DISTINCT OBJECT_NAME) FROM DBA_OBJECTS where object_type = 'MATERIALIZED_VIEW') "MATERIALIZED VIEW",
    (SELECT COUNT(DISTINCT OBJECT_NAME) FROM DBA_OBJECTS where object_type = 'INDEX') "INDEX",
    (SELECT COUNT(DISTINCT OBJECT_NAME) FROM DBA_OBJECTS where object_type = 'SEQUENCE') "SEQUENCE",
    (SELECT COUNT(DISTINCT OBJECT_NAME) FROM DBA_OBJECTS where object_type = 'TYPE') "TYPE",
    (SELECT COUNT(DISTINCT OBJECT_NAME) FROM DBA_OBJECTS where object_type = 'SYNONYM') "SYNONYM",
    (SELECT COUNT(DISTINCT OBJECT_NAME) FROM DBA_OBJECTS where object_type = 'DATABASE LINK') "DATABASE LINK",
    (SELECT COUNT(DISTINCT OBJECT_NAME) FROM DBA_OBJECTS where object_type = 'LOB') "LOB",
    (SELECT COUNT(DISTINCT OBJECT_NAME) FROM DBA_OBJECTS where object_type = 'JOB') "JOB",
    (SELECT COUNT(DISTINCT OBJECT_NAME) FROM DBA_OBJECTS where object_type = 'TRIGGER') "TRIGGER",
    (SELECT COUNT(DISTINCT OBJECT_NAME) FROM DBA_OBJECTS where object_type = 'PROCEDURE') "PROCEDURE",
    (SELECT COUNT(DISTINCT OBJECT_NAME) FROM DBA_OBJECTS where object_type = 'FUNCTION') "FUNCTION",
    (SELECT COUNT(DISTINCT OBJECT_NAME) FROM DBA_OBJECTS where object_type = 'PACKAGE') "PACKAGE",
    (SELECT COUNT(DISTINCT OBJECT_NAME) FROM DBA_OBJECTS where object_type = 'PACKAGE BODY') "PACKAGE BODY"
    FROM DBA_OBJECTS
    WHERE OWNER IN ('BERKAS', 'BIAYAPERJALANAN', 'DASHBOARD', 'DIKLAT', 'FLATKKP', 'INVENT', 
    'KKPWEB', 'KKPWEBAUDITOR', 'KONTEN', 'LP2B', 'LPPB', 'MITRAKERJA', 'MOBILE', 'OLTPUSER', 'ONEDEE', 'PEGAWAI', 'PERENCANAAN', 
    'PETA', 'PM', 'PNBP', 'PNBPTRAIN', 'SIMPEG_1703', 'SIMPEG_2702', 'SIPT', 'SKP', 'SURAT', 'SURATBDG', 'TEMATIK', 'TIMKENDALI', 
    'WEBAPIDATA', 'WILAYAH') ORDER BY OWNER;

это неправильно, мне нужно количество n другой столбец на основе первого столбца, что мне делать?

sql oracle

Всего 2 ответа

Вы должны использовать условное агрегирование следующим образом:

SELECT OWNER as SCHEMA, 
    COALESCE(SUM(CASE WHEN object_type = 'TABLE' THEN 1 END), 0) "TABLE",
    COALESCE(SUM(CASE WHEN object_type = 'VIEW' THEN 1 END), 0) "VIEW",
    COALESCE(SUM(CASE WHEN object_type = 'MATERIALIZED_VIEW' THEN 1 END), 0) "MATERIALIZED VIEW",
    COALESCE(SUM(CASE WHEN object_type = 'INDEX' THEN 1 END), 0) "INDEX",
    COALESCE(SUM(CASE WHEN object_type = 'SEQUENCE' THEN 1 END), 0) "SEQUENCE",
    COALESCE(SUM(CASE WHEN object_type = 'TYPE' THEN 1 END), 0) "TYPE",
    COALESCE(SUM(CASE WHEN object_type = 'SYNONYM' THEN 1 END), 0) "SYNONYM",
    COALESCE(SUM(CASE WHEN object_type = 'DATABASE LINK' THEN 1 END), 0) "DATABASE LINK",
    COALESCE(SUM(CASE WHEN object_type = 'LOB' THEN 1 END), 0) "LOB",
    COALESCE(SUM(CASE WHEN object_type = 'JOB' THEN 1 END), 0) "JOB",
    COALESCE(SUM(CASE WHEN object_type = 'TRIGGER' THEN 1 END), 0) "TRIGGER",
    COALESCE(SUM(CASE WHEN object_type = 'PROCEDURE' THEN 1 END), 0) "PROCEDURE",
    COALESCE(SUM(CASE WHEN object_type = 'FUNCTION' THEN 1 END), 0) "FUNCTION",
    COALESCE(SUM(CASE WHEN object_type = 'PACKAGE' THEN 1 END), 0) "PACKAGE",
    COALESCE(SUM(CASE WHEN object_type = 'PACKAGE BODY' THEN 1 END), 0) "PACKAGE BODY"
    FROM DBA_OBJECTS
    WHERE OWNER IN ('BERKAS', 'BIAYAPERJALANAN', 'DASHBOARD', 'DIKLAT', 'FLATKKP', 'INVENT', 
    'KKPWEB', 'KKPWEBAUDITOR', 'KONTEN', 'LP2B', 'LPPB', 'MITRAKERJA', 'MOBILE', 'OLTPUSER', 'ONEDEE', 'PEGAWAI', 'PERENCANAAN', 
    'PETA', 'PM', 'PNBP', 'PNBPTRAIN', 'SIMPEG_1703', 'SIMPEG_2702', 'SIPT', 'SKP', 'SURAT', 'SURATBDG', 'TEMATIK', 'TIMKENDALI', 
    'WEBAPIDATA', 'WILAYAH') 
    GROUP BY OWNER
    ORDER BY OWNER;

Ура !!

Используйте PIVOT :

SELECT *
FROM (
  SELECT owner AS schema,
         object_name,
         object_type
  FROM   DBA_OBJECTS
  WHERE  OWNER IN (
    'BERKAS', 'BIAYAPERJALANAN', 'DASHBOARD', 'DIKLAT', 'FLATKKP', 'INVENT', 
    'KKPWEB', 'KKPWEBAUDITOR', 'KONTEN', 'LP2B', 'LPPB', 'MITRAKERJA', 'MOBILE',
    'OLTPUSER', 'ONEDEE', 'PEGAWAI', 'PERENCANAAN', 'PETA', 'PM', 'PNBP',
    'PNBPTRAIN', 'SIMPEG_1703', 'SIMPEG_2702', 'SIPT', 'SKP', 'SURAT',
    'SURATBDG', 'TEMATIK', 'TIMKENDALI', 'WEBAPIDATA', 'WILAYAH'
  )
)
PIVOT ( COUNT( DISTINCT OBJECT_NAME ) FOR OBJECT_TYPE IN (
  'TABLE' AS "TABLE",
  'VIEW' AS "VIEW",
  'MATERIALIZED_VIEW' AS "MATERIALIZED VIEW",
  'INDEX' AS "INDEX",
  'SEQUENCE' AS "SEQUENCE"
  -- rest of your object types.
) )
ORDER BY schema;