Question

有没有办法使用SQL列出给定表的所有外键？我知道表名/模式，我可以将其插入。

Answer 1

您可以通过information_schema表执行此操作。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';

Answer 2

psql执行此操作，如果您使用以下命令启动psql：

psql -E

它会准确显示执行的查询。在找到外键的情况下，它是：

SELECT conname,
  pg_catalog.pg_get_constraintdef(r.oid, true) as condef
FROM pg_catalog.pg_constraint r
WHERE r.conrelid = '16485' AND r.contype = 'f' ORDER BY 1

在这种情况下，16485是我正在查看的表的oid - 您可以通过将您的tablename转换为regclass来获取该表：

WHERE r.conrelid = 'mytable'::regclass

如果表名不是唯一的（或search_path中的第一个），则表格对表名进行限定：

WHERE r.conrelid = 'myschema.mytable'::regclass

Answer 3

Ollyc的答案很好，因为它不是Postgres特有的，但是，当外键引用多个列时，它会崩溃。以下查询适用于任意数量的列，但它在很大程度上依赖于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

Answer 4

扩展到ollyc食谱：

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';

Answer 5

在PostgreSQL提示符上发出\d+ tablename，除了显示表列的数据类型外，还会显示索引和外键。

Answer 6

检查你的解决方案的ff帖子，不要忘记在你使用这个有用的

时标记这个

http://errorbank.blogspot.com/2011/03/list-all-foreign-keys-references-for.html

SELECT
  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
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
  o.contype = 'f' AND o.conrelid IN (SELECT oid FROM pg_class c WHERE c.relkind = 'r');

Answer 7

此查询也适用于复合键：

select c.constraint_name
    , x.table_schema as schema_name
    , x.table_name
    , x.column_name
    , y.table_schema as foreign_schema_name
    , y.table_name as foreign_table_name
    , y.column_name as foreign_column_name
from information_schema.referential_constraints c
join information_schema.key_column_usage x
    on x.constraint_name = c.constraint_name
join information_schema.key_column_usage y
    on y.ordinal_position = x.position_in_unique_constraint
    and y.constraint_name = c.unique_constraint_name
order by c.constraint_name, x.ordinal_position

Answer 8

我认为你在寻找和@ollyc写的非常接近的是：

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';

这将列出使用指定表作为外键的所有表

Answer 9

您可以使用PostgreSQL system catalogs。也许你可以查询pg_constraint来询问外键。您也可以使用Information Schema

Answer 10

以下是来自PostgreSQL邮件列表的Andreas Joseph Krogh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;

此解决方案处理引用多个列的外键，并避免重复（其他一些答案无法执行）。我唯一改变的是变量名。

以下示例返回引用employee表的所有permission列：

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

Answer 11

现有的答案都没有以我真正想要的形式给出结果。所以这是我的（庞大的）查询，用于查找有关外键的信息。

一些注意事项：

用于生成from_cols和to_cols的表达式可以在Postgres 9.4及更高版本上使用WITH ORDINALITY大大简化，而不是使用我正在使用的窗口函数hackery。
这些相同的表达式依赖于查询计划程序而不是更改UNNEST的结果的返回顺序。我不认为它会，但我的数据集中没有任何多列外键来测试。添加9.4个细节完全消除了这种可能性。
查询本身需要Postgres 9.0或更高版本（8.x在聚合函数中不允许ORDER BY）
如果您想要一个列数组而不是逗号分隔的字符串，请将STRING_AGG替换为ARRAY_AGG。

-

SELECT
    c.conname AS constraint_name,
    (SELECT n.nspname FROM pg_namespace AS n WHERE n.oid=c.connamespace) AS constraint_schema,

    tf.name AS from_table,
    (
        SELECT STRING_AGG(QUOTE_IDENT(a.attname), ', ' ORDER BY t.seq)
        FROM
            (
                SELECT
                    ROW_NUMBER() OVER (ROWS UNBOUNDED PRECEDING) AS seq,
                    attnum
                FROM
                    UNNEST(c.conkey) AS t(attnum)
            ) AS t
            INNER JOIN pg_attribute AS a ON a.attrelid=c.conrelid AND a.attnum=t.attnum
    ) AS from_cols,

    tt.name AS to_table,
    (
        SELECT STRING_AGG(QUOTE_IDENT(a.attname), ', ' ORDER BY t.seq)
        FROM
            (
                SELECT
                    ROW_NUMBER() OVER (ROWS UNBOUNDED PRECEDING) AS seq,
                    attnum
                FROM
                    UNNEST(c.confkey) AS t(attnum)
            ) AS t
            INNER JOIN pg_attribute AS a ON a.attrelid=c.confrelid AND a.attnum=t.attnum
    ) AS to_cols,

    CASE confupdtype WHEN 'r' THEN 'restrict' WHEN 'c' THEN 'cascade' WHEN 'n' THEN 'set null' WHEN 'd' THEN 'set default' WHEN 'a' THEN 'no action' ELSE NULL END AS on_update,
    CASE confdeltype WHEN 'r' THEN 'restrict' WHEN 'c' THEN 'cascade' WHEN 'n' THEN 'set null' WHEN 'd' THEN 'set default' WHEN 'a' THEN 'no action' ELSE NULL END AS on_delete,
    CASE confmatchtype::text WHEN 'f' THEN 'full' WHEN 'p' THEN 'partial' WHEN 'u' THEN 'simple' WHEN 's' THEN 'simple' ELSE NULL END AS match_type,  -- In earlier postgres docs, simple was 'u'nspecified, but current versions use 's'imple.  text cast is required.

    pg_catalog.pg_get_constraintdef(c.oid, true) as condef
FROM
    pg_catalog.pg_constraint AS c
    INNER JOIN (
        SELECT pg_class.oid, QUOTE_IDENT(pg_namespace.nspname) || '.' || QUOTE_IDENT(pg_class.relname) AS name
        FROM pg_class INNER JOIN pg_namespace ON pg_class.relnamespace=pg_namespace.oid
    ) AS tf ON tf.oid=c.conrelid
    INNER JOIN (
        SELECT pg_class.oid, QUOTE_IDENT(pg_namespace.nspname) || '.' || QUOTE_IDENT(pg_class.relname) AS name
        FROM pg_class INNER JOIN pg_namespace ON pg_class.relnamespace=pg_namespace.oid
    ) AS tt ON tt.oid=c.confrelid
WHERE c.contype = 'f' ORDER BY 1;

Answer 12

为了扩展Martin的优秀答案，这里有一个查询，它允许您根据父表进行过滤，并显示每个父表的子表的名称，以便您可以根据外部表查看所有相关表/列父表中的键约束。

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

Answer 13

使用密钥所引用的主键的名称并查询information_schema：

select table_name, column_name
from information_schema.key_column_usage
where constraint_name IN (select constraint_name
  from information_schema.referential_constraints 
  where unique_constraint_name = 'TABLE_NAME_pkey')

此处'TABLE_NAME_pkey'是外键引用的主键的名称。

Answer 14

我写了一个喜欢并经常使用的解决方案。代码位于http://code.google.com/p/pgutils/。请参阅pgutils.foreign_keys视图。

不幸的是，输出太过冗长，不能包含在这里。但是，您可以在此处对数据库的公共版本进行尝试，如下所示：

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

这至少适用于8.3。我希望在未来几个月内更新它，如果需要的话。

-Reece

Answer 15

SELECT r.conname
      ,ct.table_name
      ,pg_catalog.pg_get_constraintdef(r.oid, true) as condef
  FROM pg_catalog.pg_constraint r, information_schema.constraint_table_usage ct
 WHERE r.contype = 'f' 
   AND r.conname = ct.constraint_name
 ORDER BY 1

Answer 16

使用information_schema正确解决问题，使用多列密钥，正确连接两个表中不同名称的列并兼容ms sqlsever：

select fks.TABLE_NAME as foreign_key_table_name
, fks.CONSTRAINT_NAME as foreign_key_constraint_name
, kcu_foreign.COLUMN_NAME as foreign_key_column_name
, rc.UNIQUE_CONSTRAINT_NAME as primary_key_constraint_name
, pks.TABLE_NAME as primary_key_table_name
, kcu_primary.COLUMN_NAME as primary_key_column_name
from INFORMATION_SCHEMA.TABLE_CONSTRAINTS fks -- foreign keys
inner join INFORMATION_SCHEMA.KEY_COLUMN_USAGE kcu_foreign -- the columns of the above keys
    on fks.TABLE_CATALOG = kcu_foreign.TABLE_CATALOG
    and fks.TABLE_SCHEMA = kcu_foreign.TABLE_SCHEMA
    and fks.TABLE_NAME = kcu_foreign.TABLE_NAME
    and fks.CONSTRAINT_NAME = kcu_foreign.CONSTRAINT_NAME
inner join INFORMATION_SCHEMA.REFERENTIAL_CONSTRAINTS rc -- referenced constraints
    on rc.CONSTRAINT_CATALOG = fks.CONSTRAINT_CATALOG
    and rc.CONSTRAINT_SCHEMA = fks.CONSTRAINT_SCHEMA
    and rc.CONSTRAINT_NAME = fks.CONSTRAINT_NAME
inner join INFORMATION_SCHEMA.TABLE_CONSTRAINTS pks -- primary keys (referenced by fks)
    on rc.UNIQUE_CONSTRAINT_CATALOG = pks.CONSTRAINT_CATALOG
    and rc.UNIQUE_CONSTRAINT_SCHEMA = pks.CONSTRAINT_SCHEMA
    and rc.UNIQUE_CONSTRAINT_NAME = pks.CONSTRAINT_NAME
inner join INFORMATION_SCHEMA.KEY_COLUMN_USAGE kcu_primary
    on pks.TABLE_CATALOG = kcu_primary.TABLE_CATALOG
    and pks.TABLE_SCHEMA = kcu_primary.TABLE_SCHEMA
    and pks.TABLE_NAME = kcu_primary.TABLE_NAME
    and pks.CONSTRAINT_NAME = kcu_primary.CONSTRAINT_NAME
    and kcu_foreign.ORDINAL_POSITION = kcu_primary.ORDINAL_POSITION -- this joins the columns
where fks.TABLE_SCHEMA = 'dbo' -- replace with schema name
and fks.TABLE_NAME = 'your_table_name' -- replace with table name
and fks.CONSTRAINT_TYPE = 'FOREIGN KEY'
and pks.CONSTRAINT_TYPE = 'PRIMARY KEY'
order by fks.constraint_name, kcu_foreign.ORDINAL_POSITION

注意：information_schema的potgresql和sqlserver实现之间存在一些差异，这使得最佳答案在两个系统上给出不同的结果 - 一个显示外键表的列名，另一个显示主键表的列名。出于这个原因，我决定使用KEY_COLUMN_USAGE视图。

Answer 17

简短但甜美

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

Answer 18

我创建了一个小工具来查询然后比较数据库模式： Dump PostgreSQL db schema to text

有关于FK的信息，但是ollyc响应提供了更多细节。

Answer 19

另一种方式：

WITH foreign_keys AS (
    SELECT
      conname,
      conrelid,
      confrelid,
      unnest(conkey)  AS conkey,
      unnest(confkey) AS confkey
    FROM pg_constraint
    WHERE contype = 'f' -- AND confrelid::regclass = 'your_table'::regclass
)
-- if confrelid, conname pair shows up more than once then it is multicolumn foreign key
SELECT fk.conname as constraint_name,
       fk.confrelid::regclass as referenced_table, af.attname as pkcol,
       fk.conrelid::regclass as referencing_table, a.attname as fkcol
FROM foreign_keys fk
JOIN pg_attribute af ON af.attnum = fk.confkey AND af.attrelid = fk.confrelid
JOIN pg_attribute a ON a.attnum = conkey AND a.attrelid = fk.conrelid
ORDER BY fk.confrelid, fk.conname
;

Answer 20

注意：在阅读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)

Answer 21

这就是我目前正在使用的内容，它将列出一个表及其fkey约束[remove table clause，它将列出当前目录中的所有表格]：

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')

Answer 22

最快基于此answer

的直接验证bash答案的方法

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"

Answer 23

其中$ 1（'my_schema'）是模式，而$ 2（'my_table'）是表的名称：

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;

Answer 24

我升级了@ollyc的答案，该答案目前位于顶部。我同意@fionbio，因为key_column_usage和constraint_column_usage在列一级没有相关信息。

如果constraint_column_usage具有ordinal_positon之类的key_column_usage列，则可以将其与此列合并。因此，我将ordinal_position设置为constraint_column_usage，如下所示。

我无法确认此手动创建的ordinal_position与key_column_usage的顺序完全相同。但是我至少在我的情况下检查过它是否完全相同。

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'

Answer 25

 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。例如。与pysql：

Postgres：SQL列出表外键

25 个答案: