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

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';

我升级了@ollyc的答案，目前在顶部。 我同意@fionbio，因为key_column_usage和constraint_column_usage在列级上没有相关信息。

如果constraint_column_usage具有像key_column_usage一样的ordinal_position列，则可以将其与该列连接。所以我做了一个ordinal_position到constraint_column_usage如下所示。

我无法确认手动创建的ordinal_position与key_column_usage的顺序完全相同。但我检查了一下，至少在我的箱子里是完全一样的顺序。

SELECT
    tc.table_schema, 
    tc.constraint_name, 
    tc.table_name, 
    kcu.column_name, 
    ccu.table_schema AS foreign_table_schema,
    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
      AND tc.table_schema = kcu.table_schema
    JOIN (select row_number() over (partition by table_schema, table_name, constraint_name order by row_num) ordinal_position,
                 table_schema, table_name, column_name, constraint_name
          from   (select row_number() over (order by 1) row_num, table_schema, table_name, column_name, constraint_name
                  from   information_schema.constraint_column_usage
                 ) t
         ) AS ccu
      ON ccu.constraint_name = tc.constraint_name
      AND ccu.table_schema = tc.table_schema
      AND ccu.ordinal_position = kcu.ordinal_position
WHERE tc.constraint_type = 'FOREIGN KEY' AND tc.table_name = 'mytable'

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;

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"

从最流行的答案改进查询

因为对于postgresql 12+ information_schema是非常慢的

它帮助了我:

SELECT sh.nspname AS table_schema,
  tbl.relname AS table_name,
  col.attname AS column_name,
  referenced_sh.nspname AS foreign_table_schema,
  referenced_tbl.relname AS foreign_table_name,
  referenced_field.attname AS foreign_column_name
FROM pg_constraint c
    INNER JOIN pg_namespace AS sh ON sh.oid = c.connamespace
    INNER JOIN (SELECT oid, unnest(conkey) as conkey FROM pg_constraint) con ON c.oid = con.oid
    INNER JOIN pg_class tbl ON tbl.oid = c.conrelid
    INNER JOIN pg_attribute col ON (col.attrelid = tbl.oid AND col.attnum = con.conkey)
    INNER JOIN pg_class referenced_tbl ON c.confrelid = referenced_tbl.oid
    INNER JOIN pg_namespace AS referenced_sh ON referenced_sh.oid = referenced_tbl.relnamespace
    INNER JOIN (SELECT oid, unnest(confkey) as confkey FROM pg_constraint) conf ON c.oid = conf.oid
    INNER JOIN pg_attribute referenced_field ON (referenced_field.attrelid = c.confrelid AND referenced_field.attnum = conf.confkey)
WHERE c.contype = 'f'