这是我目前使用的，它将列出一个表，它的fkey约束[删除表子句，它将列出当前目录中的所有表]:

SELECT

    current_schema() AS "schema",
    current_catalog AS "database",
    "pg_constraint".conrelid::regclass::text AS "primary_table_name",
    "pg_constraint".confrelid::regclass::text AS "foreign_table_name",

    (
        string_to_array(
            (
                string_to_array(
                    pg_get_constraintdef("pg_constraint".oid),
                    '('
                )
            )[2],
            ')'
        )
    )[1] AS "foreign_column_name",

    "pg_constraint".conindid::regclass::text AS "constraint_name",

    TRIM((
        string_to_array(
            pg_get_constraintdef("pg_constraint".oid),
            '('
        )
    )[1]) AS "constraint_type",

    pg_get_constraintdef("pg_constraint".oid) AS "constraint_definition"

FROM pg_constraint AS "pg_constraint"

JOIN pg_namespace AS "pg_namespace" ON "pg_namespace".oid = "pg_constraint".connamespace

WHERE
    --fkey and pkey constraints
    "pg_constraint".contype IN ( 'f', 'p' )
    AND
    "pg_namespace".nspname = current_schema()
    AND
    "pg_constraint".conrelid::regclass::text IN ('whatever_table_name')

我创建了一个小工具来查询和比较数据库模式: Dump PostgreSQL数据库模式到文本

有关于FK的信息，但ollyc的回复提供了更多的细节。

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;

您可以通过information_schema表来实现这一点。例如:

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 information_schema.constraint_column_usage AS ccu
      ON ccu.constraint_name = tc.constraint_name
      AND ccu.table_schema = tc.table_schema
WHERE tc.constraint_type = 'FOREIGN KEY' AND tc.table_name='mytable';