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"

我写了一个解决方案，喜欢和经常使用。代码在http://code.google.com/p/pgutils/。看这些小疙瘩。foreign_keys视图。

不幸的是，输出太冗长，这里不包括。但是，你可以在数据库的公共版本上尝试一下，就像这样:

$ psql -h unison-db.org -U PUBLIC -d unison -c 'select * from pgutils.foreign_keys;

这至少适用于8.3版本。如果需要的话，我预计会在未来几个月内对其进行更新。

莉丝

我认为你想要的和@ollyc写的很接近的是:

SELECT
tc.constraint_name, 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' AND ccu.table_name='YourTableNameHere';

这将列出所有使用指定表作为外键的表

只需替换'您的表名'在下面的查询与您的表名。

简短但贴心的赞，如果对你有用的话。

select * from information_schema.key_column_usage
where constraint_catalog=current_catalog and table_name='your_table_name'
and position_in_unique_constraint notnull;

Ollyc的答案很好，因为它不是特定于postgres的，但是，当外键引用多个列时，它就会崩溃。以下查询适用于任意数量的列，但它严重依赖于Postgres扩展:

select 
    att2.attname as "child_column", 
    cl.relname as "parent_table", 
    att.attname as "parent_column",
    conname
from
   (select 
        unnest(con1.conkey) as "parent", 
        unnest(con1.confkey) as "child", 
        con1.confrelid, 
        con1.conrelid,
        con1.conname
    from 
        pg_class cl
        join pg_namespace ns on cl.relnamespace = ns.oid
        join pg_constraint con1 on con1.conrelid = cl.oid
    where
        cl.relname = 'child_table'
        and ns.nspname = 'child_schema'
        and con1.contype = 'f'
   ) con
   join pg_attribute att on
       att.attrelid = con.confrelid and att.attnum = con.child
   join pg_class cl on
       cl.oid = con.confrelid
   join pg_attribute att2 on
       att2.attrelid = con.conrelid and att2.attnum = con.parent

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;