Sql 对于这种情况，最好的数据库建模方法是什么？
对于这种情况，什么是正确的数据建模 在我的例子中，一个应用程序有多个版本，每个版本都有多个计划。因此，每个客户都有一个使用应用程序的合同 基表：Sql 对于这种情况，最好的数据库建模方法是什么？,sql,database,Sql,Database,对于这种情况，什么是正确的数据建模 在我的例子中，一个应用程序有多个版本，每个版本都有多个计划。因此，每个客户都有一个使用应用程序的合同 基表： customers ------------------------------ id int PK INC name string applications ------------------------------ id int PK INC name
customers
------------------------------
id int PK INC
name string
applications
------------------------------
id int PK INC
name string
editions
------------------------------
application_id int PK FK
id string PK (e.g: "STANDARD", "ADVANCED", "PROFESSIONAL")
plans
------------------------------
application_id int PK FK
edition_id string PK FK
id string PK (ex: "MONTHLY", "YEARLY")
contracts
------------------------------
customer_id int FK
application_id int FK
edition_id string FK
plan_id string FK
id int PK INC
editions
------------------------------
application_id int PK FK
id int PK
name string (ex: "STANDARD", "ADVANCED", "PROFESSIONAL")
plans
------------------------------
application_id int PK FK
edition_id int PK FK
id int PK
name string PK (ex: "MONTHLY", "YEARLY")
contracts
------------------------------
customer_id int FK
application_id int FK
edition_id int FK
plan_id int FK
id int PK INC
editions
------------------------------
application_id int FK
id int PK INC
name string (ex: "STANDARD", "ADVANCED", "PROFESSIONAL")
plans
------------------------------
edition_id string FK
id int PK INC
name string (ex: "MONTHLY", "YEARLY")
contracts
------------------------------
customer_id int FK
plan_id int FK
id int PK INC
customers
------------------------------
id int PK INC
name string
applications
------------------------------
id int PK INC
name string
editions
------------------------------
application_id int PK FK
id string PK (e.g: "STANDARD", "ADVANCED", "PROFESSIONAL")
plans
------------------------------
application_id int PK FK
edition_id string PK FK
id string PK (ex: "MONTHLY", "YEARLY")
contracts
------------------------------
customer_id int FK
application_id int FK
edition_id string FK
plan_id string FK
id int PK INC
editions
------------------------------
application_id int PK FK
id int PK
name string (ex: "STANDARD", "ADVANCED", "PROFESSIONAL")
plans
------------------------------
application_id int PK FK
edition_id int PK FK
id int PK
name string PK (ex: "MONTHLY", "YEARLY")
contracts
------------------------------
customer_id int FK
application_id int FK
edition_id int FK
plan_id int FK
id int PK INC
editions
------------------------------
application_id int FK
id int PK INC
name string (ex: "STANDARD", "ADVANCED", "PROFESSIONAL")
plans
------------------------------
edition_id string FK
id int PK INC
name string (ex: "MONTHLY", "YEARLY")
contracts
------------------------------
customer_id int FK
plan_id int FK
id int PK INC
customers
------------------------------
id int PK INC
name string
applications
------------------------------
id int PK INC
name string
editions
------------------------------
application_id int PK FK
id string PK (e.g: "STANDARD", "ADVANCED", "PROFESSIONAL")
plans
------------------------------
application_id int PK FK
edition_id string PK FK
id string PK (ex: "MONTHLY", "YEARLY")
contracts
------------------------------
customer_id int FK
application_id int FK
edition_id string FK
plan_id string FK
id int PK INC
editions
------------------------------
application_id int PK FK
id int PK
name string (ex: "STANDARD", "ADVANCED", "PROFESSIONAL")
plans
------------------------------
application_id int PK FK
edition_id int PK FK
id int PK
name string PK (ex: "MONTHLY", "YEARLY")
contracts
------------------------------
customer_id int FK
application_id int FK
edition_id int FK
plan_id int FK
id int PK INC
editions
------------------------------
application_id int FK
id int PK INC
name string (ex: "STANDARD", "ADVANCED", "PROFESSIONAL")
plans
------------------------------
edition_id string FK
id int PK INC
name string (ex: "MONTHLY", "YEARLY")
contracts
------------------------------
customer_id int FK
plan_id int FK
id int PK INC
customers
------------------------------
id int PK INC
name string
applications
------------------------------
id int PK INC
name string
editions
------------------------------
application_id int PK FK
id string PK (e.g: "STANDARD", "ADVANCED", "PROFESSIONAL")
plans
------------------------------
application_id int PK FK
edition_id string PK FK
id string PK (ex: "MONTHLY", "YEARLY")
contracts
------------------------------
customer_id int FK
application_id int FK
edition_id string FK
plan_id string FK
id int PK INC
editions
------------------------------
application_id int PK FK
id int PK
name string (ex: "STANDARD", "ADVANCED", "PROFESSIONAL")
plans
------------------------------
application_id int PK FK
edition_id int PK FK
id int PK
name string PK (ex: "MONTHLY", "YEARLY")
contracts
------------------------------
customer_id int FK
application_id int FK
edition_id int FK
plan_id int FK
id int PK INC
editions
------------------------------
application_id int FK
id int PK INC
name string (ex: "STANDARD", "ADVANCED", "PROFESSIONAL")
plans
------------------------------
edition_id string FK
id int PK INC
name string (ex: "MONTHLY", "YEARLY")
contracts
------------------------------
customer_id int FK
plan_id int FK
id int PK INC
?
还有另一种方法吗？我认为这些方法都不是“最好的”。它们在体系结构中都有一些奇怪的限制。1和2中都有冗余数据。第三个版本仅限于一份合同。但是……我认为第三个是最接近的。当然，这可能会因为过于基于观点而被关闭。离题-属于。但即使这样，“最佳”也没有什么意义，如果你对你的实际模型和你打算如何使用它没有很好的（真正的、完整的）理解。如果需要，可以将其中一个放回去，这是实际使用的dbms的问题。像这样的问题不属于dba。这是一个数据库设计问题，而不是管理问题。