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私はこのようなEMPLOYEE_AUDITテーブルを持っています

   Employee_id Revision First_name  Last_Name  Gender    Position
   1           1        David       Hem        Male      Developer
   1           2        David       Hem        Male      Manager
   1           3        David       Alter      Male      Manager
   2           1        Agatha      Christie   Female    Business Analyst
   2           2        Agatha      Christie   Female    Group Manager
   2           3        Agatha      Christie   Female    General Manager

そして私はこのような出力を持つSQLクエリが必要です

   Column_name   employee_id Revision Old_Value         new_value
   Position      1           2        Developer         Manager
   Last_name     1           3        Hem               Alter
   Position      2           2        Business Analyst  Group Manager
   Position      2           3        Group Manager     General Manager

oracle 10gを使用しているため、11gの機能は無視してください。

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1 に答える 1

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必ずしも最も美しいSQLである必要はありませんが、これはかなり一般的なCASEロジックを使用して行います。
これは監査テーブルであるため、物事を単純化するためにリビジョンにギャップがないことを前提としています。

SELECT 
 CASE
   WHEN ea1."First_name" <> ea2."First_name" THEN 'First_Name' 
   WHEN ea1."Last_Name"  <> ea2."Last_Name"  THEN 'Last_Name' 
   WHEN ea1."Gender"     <> ea2."Gender"     THEN 'Gender' 
   WHEN ea1."Position"   <> ea2."Position"   THEN 'Position' END "Column_name",
 ea2."Employee_id", ea2."Revision",
 CASE
   WHEN ea1."First_name" <> ea2."First_name" THEN ea1."First_name" 
   WHEN ea1."Last_Name"  <> ea2."Last_Name"  THEN ea1."Last_Name" 
   WHEN ea1."Gender"     <> ea2."Gender"     THEN ea1."Gender" 
   WHEN ea1."Position"   <> ea2."Position"   THEN ea1."Position" END "Old_Value",
 CASE
   WHEN ea1."First_name" <> ea2."First_name" THEN ea2."First_name" 
   WHEN ea1."Last_Name"  <> ea2."Last_Name"  THEN ea2."Last_Name" 
   WHEN ea1."Gender"     <> ea2."Gender"     THEN ea2."Gender" 
   WHEN ea1."Position"   <> ea2."Position"   THEN ea2."Position" END "New_Value"
 FROM EMPLOYEE_AUDIT ea1
 JOIN EMPLOYEE_AUDIT ea2
   ON ea1."Employee_id" = ea2."Employee_id"
  AND ea1."Revision" = ea2."Revision" -1

でテストするSQLFiddle。

編集：あなたが尋ねたので、これはを使用したバージョンLAGであり、大きな改善ではありません。

WITH cte AS (
  SELECT "Employee_id", "Revision", "First_name" fn, "Last_Name" ln,
  "Gender" g, "Position" p, 
  LAG("First_name", 1) OVER 
      (PARTITION BY "Employee_id" ORDER BY "Revision") fn2, 
  LAG("Last_Name", 1)  OVER 
      (PARTITION BY "Employee_id" ORDER BY "Revision") ln2,
  LAG("Gender", 1)  OVER 
      (PARTITION BY "Employee_id" ORDER BY "Revision") g2, 
  LAG("Position",1) OVER 
      (PARTITION BY "Employee_id" ORDER BY "Revision") p2
  FROM EMPLOYEE_AUDIT
)
SELECT 
 CASE
   WHEN fn <> fn2 THEN 'First_Name' WHEN ln <> ln2 THEN 'Last_Name' 
   WHEN g  <> g2  THEN 'Gender' WHEN p  <> p2  THEN 'Position' 
 END "Column_name", "Employee_id", "Revision",
 CASE
   WHEN fn <> fn2 THEN fn2 WHEN ln <> ln2 THEN ln2 
   WHEN g  <> g2  THEN g2  WHEN p  <> p2  THEN p2
 END "Old_Value",
 CASE
   WHEN fn <> fn2 THEN fn  WHEN ln <> ln2 THEN ln 
   WHEN g  <> g2  THEN g   WHEN p  <> p2  THEN p 
 END "New_Value"
FROM cte
WHERE fn2 IS NOT NULL;

別のSQLfiddle。

于 2013-02-26T21:26:35.823 に答える