以下是Andreas Joseph Krogh从PostgreSQL邮件列表提供的解决方案:http://www.postgresql.org/message-id/200811072134.44750.andreak@officenet.no

SELECT source_table::regclass, source_attr.attname AS source_column,
    target_table::regclass, target_attr.attname AS target_column
FROM pg_attribute target_attr, pg_attribute source_attr,
  (SELECT source_table, target_table, source_constraints[i] source_constraints, target_constraints[i] AS target_constraints
   FROM
     (SELECT conrelid as source_table, confrelid AS target_table, conkey AS source_constraints, confkey AS target_constraints,
       generate_series(1, array_upper(conkey, 1)) AS i
      FROM pg_constraint
      WHERE contype = 'f'
     ) query1
  ) query2
WHERE target_attr.attnum = target_constraints AND target_attr.attrelid = target_table AND
      source_attr.attnum = source_constraints AND source_attr.attrelid = source_table;

这个解决方案处理引用多个列的外键，并避免重复(其他一些答案无法做到这一点)。我唯一改变的是变量名。

下面是一个示例，返回所有引用权限表的雇员列:

SELECT source_column
FROM foreign_keys
WHERE source_table = 'employee'::regclass AND target_table = 'permission'::regclass;

你可以使用PostgreSQL系统目录。也许您可以查询pg_constraint来请求外键。 您还可以使用信息模式

注意:在读取约束列时，不要忘记列的顺序!

SELECT conname, attname
  FROM pg_catalog.pg_constraint c 
  JOIN pg_catalog.pg_attribute a ON a.attrelid = c.conrelid AND a.attnum = ANY (c.conkey)
 WHERE attrelid = 'schema.table_name'::regclass
 ORDER BY conname, array_position(c.conkey, a.attnum)

 SELECT 
    conrelid::regclass AS table_from,
    conname,
    pg_get_constraintdef(oid) as condef    
    FROM pg_catalog.pg_constraint r

也适用于所有约束条件。例如，使用pysql:

扩展到ollyc配方:

CREATE VIEW foreign_keys_view AS
SELECT
    tc.table_name, kcu.column_name,
    ccu.table_name AS foreign_table_name,
    ccu.column_name AS foreign_column_name
FROM
    information_schema.table_constraints AS tc
    JOIN information_schema.key_column_usage 
        AS kcu ON tc.constraint_name = kcu.constraint_name
    JOIN information_schema.constraint_column_usage 
        AS ccu ON ccu.constraint_name = tc.constraint_name
WHERE constraint_type = 'FOREIGN KEY';

然后:

SELECT * FROM foreign_keys_view WHERE table_name='YourTableNameHere';

最快的验证直接在bash答案完全基于这个答案

IFS='' read -r -d '' sql_code << EOF_SQL_CODE
      SELECT
      o.oid
      , o.conname AS constraint_name
      , (SELECT nspname FROM pg_namespace WHERE oid=m.relnamespace) AS source_schema
      , m.relname AS source_table
      , (SELECT a.attname FROM pg_attribute a
      WHERE a.attrelid = m.oid AND a.attnum = o.conkey[1] AND a.attisdropped = false) AS source_column
      , (SELECT nspname FROM pg_namespace
      WHERE oid=f.relnamespace) AS target_schema
      , f.relname AS target_table
      , (SELECT a.attname FROM pg_attribute a
      WHERE a.attrelid = f.oid AND a.attnum = o.confkey[1] AND a.attisdropped = false) AS target_column
      , ROW_NUMBER () OVER (ORDER BY o.oid) as rowid
      FROM pg_constraint o
      LEFT JOIN pg_class f ON f.oid = o.confrelid
      LEFT JOIN pg_class m ON m.oid = o.conrelid
      WHERE 1=1
      AND o.contype = 'f'
      AND o.conrelid IN (SELECT oid FROM pg_class c WHERE c.relkind = 'r')
EOF_SQL_CODE

psql -d my_db -c "$sql_code"