检索所有 PK 和 FK

Question

我有一个大数据库，需要从每个表中提取所有主键和外键。

我有 pgAdmin III。

有没有办法自动执行此操作而不手动查看每个表？

Answer 1

您可以pg_get_constraintdef(constraint_oid)在查询中使用该函数，如下所示：

SELECT conrelid::regclass AS table_from
     , conname
     , pg_get_constraintdef(oid)
FROM   pg_constraint
WHERE  contype IN ('f', 'p ')
AND    connamespace = 'public'::regnamespace  -- your schema here
ORDER  BY conrelid::regclass::text, contype DESC;

结果：

 table_from | conname    | pg_get_constraintdef
------------+------------+----------------------
 tbl        | tbl_pkey   | PRIMARY KEY (tbl_id)
 tbl        | tbl_col_fk | FOREIGN KEY (col) REFERENCES tbl2(col) ON UPDATE CASCADE
...

返回给定模式中所有表的所有主键和外键，按表名排序，PK 优先。

关于pg_constraint.

关于对象标识符类型( regclass, regnamespace, ...) 的手册。

Answer 2

基于 Erwin 解决方案：

SELECT conrelid::regclass AS "FK_Table"
      ,CASE WHEN pg_get_constraintdef(c.oid) LIKE 'FOREIGN KEY %' THEN substring(pg_get_constraintdef(c.oid), 14, position(')' in pg_get_constraintdef(c.oid))-14) END AS "FK_Column"
      ,CASE WHEN pg_get_constraintdef(c.oid) LIKE 'FOREIGN KEY %' THEN substring(pg_get_constraintdef(c.oid), position(' REFERENCES ' in pg_get_constraintdef(c.oid))+12, position('(' in substring(pg_get_constraintdef(c.oid), 14))-position(' REFERENCES ' in pg_get_constraintdef(c.oid))+1) END AS "PK_Table"
      ,CASE WHEN pg_get_constraintdef(c.oid) LIKE 'FOREIGN KEY %' THEN substring(pg_get_constraintdef(c.oid), position('(' in substring(pg_get_constraintdef(c.oid), 14))+14, position(')' in substring(pg_get_constraintdef(c.oid), position('(' in substring(pg_get_constraintdef(c.oid), 14))+14))-1) END AS "PK_Column"
FROM   pg_constraint c
JOIN   pg_namespace n ON n.oid = c.connamespace
WHERE  contype IN ('f', 'p ')
AND pg_get_constraintdef(c.oid) LIKE 'FOREIGN KEY %'
ORDER  BY pg_get_constraintdef(c.oid), conrelid::regclass::text, contype DESC;

将返回一个表格：

| FK_Table | FK_Column | PK_Table | PK_Column |

Answer 3

无需解析pg_get_constraintdef()，只需使用表的pg_constraint列来获取其他详细信息（文档）。

这里constraint_type可以是：

• p -主键，
• f -外键，
• u -独特的，
• c -检查约束，
• x -排除，
• ...

基于欧文的回答：

SELECT c.conname                                 AS constraint_name,
   c.contype                                     AS constraint_type,
   sch.nspname                                   AS "self_schema",
   tbl.relname                                   AS "self_table",
   ARRAY_AGG(col.attname ORDER BY u.attposition) AS "self_columns",
   f_sch.nspname                                 AS "foreign_schema",
   f_tbl.relname                                 AS "foreign_table",
   ARRAY_AGG(f_col.attname ORDER BY f_u.attposition) AS "foreign_columns",
   pg_get_constraintdef(c.oid)                   AS definition
FROM pg_constraint c
       LEFT JOIN LATERAL UNNEST(c.conkey) WITH ORDINALITY AS u(attnum, attposition) ON TRUE
       LEFT JOIN LATERAL UNNEST(c.confkey) WITH ORDINALITY AS f_u(attnum, attposition) ON f_u.attposition = u.attposition
       JOIN pg_class tbl ON tbl.oid = c.conrelid
       JOIN pg_namespace sch ON sch.oid = tbl.relnamespace
       LEFT JOIN pg_attribute col ON (col.attrelid = tbl.oid AND col.attnum = u.attnum)
       LEFT JOIN pg_class f_tbl ON f_tbl.oid = c.confrelid
       LEFT JOIN pg_namespace f_sch ON f_sch.oid = f_tbl.relnamespace
       LEFT JOIN pg_attribute f_col ON (f_col.attrelid = f_tbl.oid AND f_col.attnum = f_u.attnum)
GROUP BY constraint_name, constraint_type, "self_schema", "self_table", definition, "foreign_schema", "foreign_table"
ORDER BY "self_schema", "self_table";

结果按schema和排序table。

技术说明：看到这个问题有关with ordinality。