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
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'
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
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
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
只需替换'您的表名'在下面的查询与您的表名。
简短但贴心的赞,如果对你有用的话。
select * from information_schema.key_column_usage
where constraint_catalog=current_catalog and table_name='your_table_name'
and position_in_unique_constraint notnull;
从最流行的答案改进查询
因为对于postgresql 12+ information_schema是非常慢的
它帮助了我:
SELECT sh.nspname AS table_schema,
tbl.relname AS table_name,
col.attname AS column_name,
referenced_sh.nspname AS foreign_table_schema,
referenced_tbl.relname AS foreign_table_name,
referenced_field.attname AS foreign_column_name
FROM pg_constraint c
INNER JOIN pg_namespace AS sh ON sh.oid = c.connamespace
INNER JOIN (SELECT oid, unnest(conkey) as conkey FROM pg_constraint) con ON c.oid = con.oid
INNER JOIN pg_class tbl ON tbl.oid = c.conrelid
INNER JOIN pg_attribute col ON (col.attrelid = tbl.oid AND col.attnum = con.conkey)
INNER JOIN pg_class referenced_tbl ON c.confrelid = referenced_tbl.oid
INNER JOIN pg_namespace AS referenced_sh ON referenced_sh.oid = referenced_tbl.relnamespace
INNER JOIN (SELECT oid, unnest(confkey) as confkey FROM pg_constraint) conf ON c.oid = conf.oid
INNER JOIN pg_attribute referenced_field ON (referenced_field.attrelid = c.confrelid AND referenced_field.attnum = conf.confkey)
WHERE c.contype = 'f'
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