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
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:
其他回答
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
;
使用Key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')
这里'TABLE_NAME_pkey'是外键引用的主键的名称。
现有的答案都没有给出我想要的结果。这是我的(庞大的)查询,用于查找有关外键的信息。
注意事项:
The expressions used to generate from_cols and to_cols could be vastly simplified on Postgres 9.4 and later using WITH ORDINALITY rather than the window-function-using hackery I'm using. Those same expressions are relying on the query planner not altering the returned order of results from UNNEST. I don't think it will, but I don't have any multiple-column foreign keys in my dataset to test with. Adding the 9.4 niceties eliminates this possibility altogether. The query itself requires Postgres 9.0 or later (8.x didn't allow ORDER BY in aggregate functions) Replace STRING_AGG with ARRAY_AGG if you want an array of columns rather than a comma-separated string.
-
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;
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'
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
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