反思postgresql 8.3以查找外键

我试图内省postgres 8.3数据库，以检索其详细信息 外键。假设我有以下模式：

CREATE TABLE "a" (
 "id" SERIAL PRIMARY KEY
);

CREATE TABLE "b" (
 "one" integer,
 "two" integer,
 "a_id" integer REFERENCES "a",
 PRIMARY KEY ("one", "two")
);

CREATE TABLE "c" (
 "id" SERIAL PRIMARY KEY,
 "a_id" integer REFERENCES "a",
 "b_one" integer,
 "b_two" integer,
 FOREIGN KEY ("b_one", "b_two") REFERENCES "b"
);

然后我想运行一个查询，该查询产生以下结果：

table | columns        | foreign table | foreign columns
--------------------------------------------------------
 b    | {a_id}         | a             | {id}
 c    | {a_id}         | a             | {id}
 c    | {b_one, b_two} | b             | {one, two}

我最初的努力给了我疑问

SELECT conrelid::regclass as "table",
       conkey as columns, 
       confrelid::regclass as "foreign table", 
       confkey as "foreign columns"
  FROM pg_constraint
 WHERE contype = 'f' ;

 table | columns | foreign table | foreign columns 
-------+---------+---------------+-----------------
 b     | {3}     | a             | {1}
 c     | {2}     | a             | {1}
 c     | {3,4}   | b             | {1,2}

就快到了。但是我将列号转换为列号的努力 名字还没有给我提供想要的结果。谷歌给了我 下面的是同样的，不太正确

SELECT conrelid::regclass as "table",
       a.attname as columns,
       confrelid::regclass as "foreign table",
       af.attname as "foreign columns"
  FROM pg_attribute AS af,
       pg_attribute AS a,
       ( SELECT conrelid,
                confrelid,
                conkey[i] AS conkey,
                confkey[i] as confkey
           FROM ( SELECT conrelid,
                         confrelid, 
                         conkey, 
                         confkey, 
                         generate_series(1, array_upper(conkey, 1)) AS i
                    FROM pg_constraint
     WHERE contype = 'f'
         ) AS ss
       ) AS ss2
 WHERE af.attnum = confkey
   AND af.attrelid = confrelid
   AND a.attnum = conkey
   AND a.attrelid = conrelid ;

 table | columns | foreign table | foreign columns 
-------+---------+---------------+-----------------
 b     | a_id    | a             | id
 c     | a_id    | a             | id
 c     | b_one   | b             | one
 c     | b_two   | b             | two

有人能帮我做最后一步吗

SELECT table, array_agg(columns), foreign_table, array_agg(foreign_columns) FROM (your query here) GROUP BY table, foreign_table;

数组_agg需要PostgreSQL 8.4。对于早期版本，您可以定义自己的（在文档中查找array_accum）。显然，您可以将这个查询合并到您的大查询中，但这应该会给您一个大致的想法

---

数组_agg需要PostgreSQL 8.4。对于早期版本，您可以定义自己的（在文档中查找array_accum）。显然，您可以将这个查询合并到您的大查询中，但这应该会给您一个大致的想法。

充实Peter Eisentrut的答案；对于postgresql 8.3，数组 函数可以定义为

CREATE AGGREGATE array_accum (anyelement)
(
    sfunc = array_append,
    stype = anyarray,
    initcond = '{}'
);

然后得到我想要的答案的完整查询变成

SELECT "table",
       array_accum(columns) AS columns,
       "foreign table",
       array_accum("foreign columns") AS "foreign columns"
  FROM ( SELECT conrelid::regclass AS "table",
                a.attname as columns,
                confrelid::regclass as "foreign table",
                af.attname as "foreign columns"
           FROM pg_attribute AS af,
                pg_attribute AS a,
                ( SELECT conrelid,
                         confrelid,
                         conkey[i] AS conkey,
                         confkey[i] as confkey
                    FROM ( SELECT conrelid,
                                  confrelid, 
                                  conkey, 
                                  confkey, 
                                  generate_series(1, array_upper(conkey, 1)) AS i
                             FROM pg_constraint
              WHERE contype = 'f'
                  ) AS ss
                ) AS ss2
          WHERE af.attnum = confkey
            AND af.attrelid = confrelid
            AND a.attnum = conkey
            AND a.attrelid = conrelid
       ) AS ss3
  GROUP BY "table",
           "foreign table";

请原谅对他的回答的非标准评论方式，我还在学习 如何使用Stackoverflow，并且在第一次创建帐户时

---

实例并没有起到任何作用。

充实了彼得·艾森特的答案；对于postgresql 8.3，数组 函数可以定义为

CREATE AGGREGATE array_accum (anyelement)
(
    sfunc = array_append,
    stype = anyarray,
    initcond = '{}'
);

然后得到我想要的答案的完整查询变成

SELECT "table",
       array_accum(columns) AS columns,
       "foreign table",
       array_accum("foreign columns") AS "foreign columns"
  FROM ( SELECT conrelid::regclass AS "table",
                a.attname as columns,
                confrelid::regclass as "foreign table",
                af.attname as "foreign columns"
           FROM pg_attribute AS af,
                pg_attribute AS a,
                ( SELECT conrelid,
                         confrelid,
                         conkey[i] AS conkey,
                         confkey[i] as confkey
                    FROM ( SELECT conrelid,
                                  confrelid, 
                                  conkey, 
                                  confkey, 
                                  generate_series(1, array_upper(conkey, 1)) AS i
                             FROM pg_constraint
              WHERE contype = 'f'
                  ) AS ss
                ) AS ss2
          WHERE af.attnum = confkey
            AND af.attrelid = confrelid
            AND a.attnum = conkey
            AND a.attrelid = conrelid
       ) AS ss3
  GROUP BY "table",
           "foreign table";

请原谅对他的回答的非标准评论方式，我还在学习 如何使用Stackoverflow，并且在第一次创建帐户时 实例并没有起到任何作用