Question

目的是允许插入脚本将数据添加到架构中的所有表，以使其不会在约束中造成任何冲突。我从information_schema.tables中获取表，并从information_schema.table_constraints中获取约束，但不确定如何按外键约束的顺序比较排序表。请帮忙。

以下产生重复的表名：

select a.table_name,b.ordinal_position
from information_schema.tables a left outer join
     information_schema.key_column_usage b
     on a.table_name = b.table_name

Answer 1

您需要一个遍历外键关系的整个依赖关系树的递归查询。

以下查询针对简单的依赖项执行此操作。它不会不处理圆形外键

with recursive fk_tree as (
  -- All tables not referencing anything else
  select t.oid as reloid, 
         t.relname as table_name, 
         s.nspname as schema_name,
         null::text as referenced_table_name,
         null::text as referenced_schema_name,
         1 as level
  from pg_class t
    join pg_namespace s on s.oid = t.relnamespace
  where relkind = 'r'
    and not exists (select *
                    from pg_constraint
                    where contype = 'f'
                      and conrelid = t.oid)
    and s.nspname = 'public' -- limit to one schema 

  union all 

  select ref.oid, 
         ref.relname, 
         rs.nspname,
         p.table_name,
         p.schema_name,
         p.level + 1
  from pg_class ref
    join pg_namespace rs on rs.oid = ref.relnamespace
    join pg_constraint c on c.contype = 'f' and c.conrelid = ref.oid
    join fk_tree p on p.reloid = c.confrelid
), all_tables as (
  -- this picks the highest level for each table
  select schema_name, table_name,
         level, 
         row_number() over (partition by schema_name, table_name order by level desc) as last_table_row
  from fk_tree
)
select schema_name, table_name, level
from all_tables at
where last_table_row = 1
order by level;

对于以下表结构：

create table customer (id integer primary key);
create table product (id integer primary key);
create table manufacturer (id integer primary key);
create table manufactured_by (product_id integer references product, manufacturer_id integer references manufacturer);
create table distributor (id integer primary key);
create table orders (id integer primary key, customer_id integer references customer);
create table order_line (id integer primary key, order_id integer references orders);
create table invoice (id integer, order_id integer references orders);
create table delivery (oder_line_id integer references order_line, distributor_id integer references distributor);

这将返回以下结果：

schema_name  | table_name      | level
-------------+-----------------+------
public       | customer        |     1
public       | distributor     |     1
public       | manufacturer    |     1
public       | product         |     1
public       | manufactured_by |     2
public       | orders          |     2
public       | order_line      |     3
public       | invoice         |     3
public       | delivery        |     4

这意味着您需要先插入customer，然后才能插入orders。具有相同级别的表的插入顺序无关紧要。

使用CTE all_tables的中间步骤仅需要列出每个表一次。否则，表manufactured_by会被列出两次：

schema_name | table_name      | level
------------+-----------------+------
public      | customer        |     1
public      | product         |     1
public      | manufacturer    |     1
public      | distributor     |     1
public      | manufactured_by |     2
public      | manufactured_by |     2
public      | orders          |     2
public      | delivery        |     2
public      | order_line      |     3
public      | invoice         |     3
public      | delivery        |     4

此查询可能需要更多调整（特别是为了防止依赖树中的循环），但它应该为您提供一个开始。

按依赖性顺序对表格进行排序-Postgres

1 个答案: