Question

as we know PK and FKs type must be same, the size CAN be different in Oracle

1. find all the foreign keys,
2. find all the PK them are referenced to
3. compare the length
4. use the PK’s length as standard if FK is short.

anyone got the script?

Answer 1

SELECT cc.table_name,
       CC.COLUMN_NAME,
       TCF.DATA_TYPE || '(' || TCF.DATA_PRECISION || ',' || TCF.DATA_SCALE || ')' DTYPE1,
       rcc.table_name AS r_table_name,
       RCC.COLUMN_NAME AS R_COLUMN_NAME,
       TCP.DATA_TYPE || '(' || TCP.DATA_PRECISION || ',' || TCP.DATA_SCALE || ')' DTYPE2,
       cc.position
FROM   user_constraints c
       JOIN user_cons_columns cc ON c.owner = cc.owner AND c.constraint_name = cc.constraint_name
       JOIN USER_CONS_COLUMNS RCC ON C.OWNER = RCC.OWNER AND C.R_CONSTRAINT_NAME = RCC.CONSTRAINT_NAME AND CC.POSITION = RCC.POSITION
       JOIN USER_TAB_COLS TCF ON TCF.TABLE_NAME = CC.TABLE_NAME AND TCF.COLUMN_NAME = CC.COLUMN_NAME
       JOIN USER_TAB_COLS TCP ON TCP.TABLE_NAME = RCC.TABLE_NAME AND TCP.COLUMN_NAME = RCC.COLUMN_NAME
WHERE TCF.DATA_TYPE || '(' || TCF.DATA_PRECISION || ',' || TCF.DATA_SCALE || ')' != TCP.DATA_TYPE || '(' || TCP.DATA_PRECISION || ',' || TCP.DATA_SCALE || ')'
ORDER BY c.constraint_name, cc.table_name, cc.position;