Uma consulta SQL complicado eu tive de trabalhar fora de dizer-me informações sobre o banco de dados atual.

Esta consulta irá informá-lo
TableName, FieldName, FieldType, FieldSubType, ForeignTableName, ForeignFieldName

selecionar distintas
Rel.rdb $ relation_name TableName,
Rel.rdb $ field_name FieldName,
Fld.rdb $ field_type FieldType,
Fld.rdb $ field_sub_type FieldSubType,
FCon.rdb $ Relation_Name ForeignTableName,
FIseg.rdb $ Field_Name ForeignFieldName
de
RDB $ relation_fields Rel
left join
RDB $ relation_constraints Con
ligado
(Con.rdb $ relation_name = Rel.rdb $ relation_name e
Con.rdb $ constraint_type como 'ESTRANHA%')
left join
RDB $ índices IDX
ligado
IDX.rdb $ index_name = Con.rdb $ index_name
left join
RDB $ index_segments ISEG
ligado
(ISeg.rdb $ index_name = Idx.rdb $ index_name e
ISeg.rdb $ Field_Name = Rel.rdb $ field_name)
left join
RDB Relation_Constraints $ FCon
ligado
FCon.rdb $ index_name = Idx.rdb $ Foreign_Key
left join
RDB $ index_segments FIseg
ligado
(FISeg.rdb $ index_name = Idx.rdb $ Foreign_key e
FISeg.rdb $ Field_Position = ISeg.rdb $ Field_Position),
RDB $ Fld campos,
RDB $ Relation_Fields RFld
onde
Rel.rdb $ relation_name não gosta 'RDB $%' e
Fld.rdb $ field_name = Rel.rdb $ field_source e
RFld.rdb $ Relation_Name = Rel.rdb $ Relation_name e
RFld.rdb $ field_name = Rel.rdb $ field_name
por fim
Rel.rdb $ relation_name,
RFld.rdb $ Field_ID;

Average rating: