是否有一种方法使用SQL列出给定表的所有外键?我知道表名/模式,我可以把它插入。
当前回答
在PostgreSQL提示符上发出\d+ tablename,除了显示表列的数据类型外,它还会显示索引和外键。
其他回答
PSQL就是这样做的,如果你用:
psql -E
它将准确地显示执行了哪些查询。在查找外键的情况下,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是我正在寻找的表的oid -你可以通过将你的表名转换为regclass来获得它:
WHERE r.conrelid = 'mytable'::regclass
如果表名不是唯一的(或者是搜索路径中的第一个),则对表名进行模式限定:
WHERE r.conrelid = 'myschema.mytable'::regclass
注意:在读取约束列时,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)
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
我自己的贡献。目标是备份所有外键的定义:
SELECT
'ALTER TABLE ' || tc.table_schema || '.' || tc.table_name || E'\n
ADD FOREIGN KEY (' || kcu.column_name || ')' || E'\n
REFERENCES ' || ccu.table_schema || '.' || ccu.table_name ||
' (' || ccu.column_name || ') ' || E'\n ' ||
CASE WHEN rc.match_option <> 'NONE' THEN E'\n
MATCH ' || rc.match_option ELSE '' END ||
CASE WHEN rc.update_rule <> 'NO ACTION' THEN E'\n
ON UPDATE ' || rc.update_rule || ' ' ELSE '' END ||
CASE WHEN rc.delete_rule <> 'NO ACTION'
THEN 'ON DELETE ' || rc.delete_rule ELSE '' END || ';'
AS add_constraint
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
JOIN information_schema.referential_constraints AS rc
ON tc.constraint_name=rc.constraint_name
WHERE tc.constraint_type = 'FOREIGN KEY'
\t\a\g\a\ta
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'
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