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')

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

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"

选择的答案不为我工作，所以张贴我的sql工作。

select 
    con.conname as constraint_name,
    src_schema.nspname as source_schema,
    source.relname as source_table,
    source_col.attname as source_column,
    trg_schema.nspname as target_schema,
    target.relname as target_table,
    target_col.attname as target_column
from 
    pg_constraint con
inner join 
    pg_class source on source.oid = con.conrelid
inner join
    pg_attribute source_col on source_col.attrelid = con.conrelid and source_col.attnum = con.conkey[1] and source_col.attisdropped = false
inner join
    pg_namespace src_schema on src_schema.oid = source.relnamespace
inner join 
    pg_class target on target.oid = con.confrelid
inner join
    pg_attribute target_col on target_col.attrelid = con.confrelid and target_col.attnum = con.confkey[1] and source_col.attisdropped = false    
inner join
    pg_namespace trg_schema on trg_schema.oid = target.relnamespace 

$1 ('my_schema')是模式，$2 ('my_table')是表名:

SELECT ss.conname constraint_name, a.attname column_name, ss.refnamespace fk_table_schema, ss.reflname fk_table_name, af.attname fk_column_name
    FROM  pg_attribute a, pg_attribute af,
        (SELECT r.oid roid, c.conname, rf.relname reflname, information_schema._pg_expandarray(c.conkey) x,
                nrf.nspname refnamespace, rf.oid rfoid, information_schema._pg_expandarray(cf.confkey) xf
            FROM pg_namespace nr, pg_class r, pg_constraint c,
                pg_namespace nrf, pg_class rf, pg_constraint cf
            WHERE nr.oid = r.relnamespace
                AND r.oid = c.conrelid
                AND rf.oid = cf.confrelid
                AND c.conname = cf.conname
                AND nrf.oid = rf.relnamespace
                AND nr.nspname = $1
                AND r.relname = $2) ss
    WHERE ss.roid = a.attrelid AND a.attnum = (ss.x).x AND NOT a.attisdropped
        AND ss.rfoid = af.attrelid AND af.attnum = (ss.xf).x AND NOT af.attisdropped
    ORDER BY ss.conname, a.attname;

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'