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
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"
其他回答
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。
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
我认为你想要的和@ollyc写的很接近的是:
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';
这将列出所有使用指定表作为外键的表
检查ff帖子的解决方案,不要忘记标记这个,当你认为这是有帮助的
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');
这是我目前使用的,它将列出一个表,它的fkey约束[删除表子句,它将列出当前目录中的所有表]:
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')
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