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
$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;
其他回答
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';
SELECT
conrelid::regclass AS table_from,
conname,
pg_get_constraintdef(oid) as condef
FROM pg_catalog.pg_constraint r
也适用于所有约束条件。例如,使用pysql:
为了扩展Martin的精彩回答,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,以便您可以根据父表中的外键约束查看所有依赖的表/列。
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%'
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'
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"
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