各テーブルからすべての主キーと外部キーを抽出する必要がある大きなデータベースがあります。
PgAdmin IIIを使用しています。
これを自動的に行い、各テーブルを手動で行わない方法はありますか?
次のようなクエリで関数 pg_get_constraintdef(constraint_oid)
を使用できます。
SELECT conrelid::regclass AS table_from
, conname
, pg_get_constraintdef(oid)
FROM pg_constraint
WHERE contype IN ('f', 'p ')
AND connamespace = 'public'::regnamespace -- your schema here
ORDER BY conrelid::regclass::text, contype DESC;
結果:
table_from | conname | pg_get_constraintdef
------------+------------+----------------------
tbl | tbl_pkey | PRIMARY KEY (tbl_id)
tbl | tbl_col_fk | FOREIGN KEY (col) REFERENCES tbl2(col) ON UPDATE CASCADE
...
指定されたスキーマ内のすべてのテーブルのすべての主キーと外部キーを、PKを最初にtablenameで並べ替えて返します。
オブジェクト識別子のタイプ (regclass
、regnamespace
、...)に関するマニュアル。
Erwinソリューションに基づく:
SELECT conrelid::regclass AS "FK_Table"
,CASE WHEN pg_get_constraintdef(c.oid) LIKE 'FOREIGN KEY %' THEN substring(pg_get_constraintdef(c.oid), 14, position(')' in pg_get_constraintdef(c.oid))-14) END AS "FK_Column"
,CASE WHEN pg_get_constraintdef(c.oid) LIKE 'FOREIGN KEY %' THEN substring(pg_get_constraintdef(c.oid), position(' REFERENCES ' in pg_get_constraintdef(c.oid))+12, position('(' in substring(pg_get_constraintdef(c.oid), 14))-position(' REFERENCES ' in pg_get_constraintdef(c.oid))+1) END AS "PK_Table"
,CASE WHEN pg_get_constraintdef(c.oid) LIKE 'FOREIGN KEY %' THEN substring(pg_get_constraintdef(c.oid), position('(' in substring(pg_get_constraintdef(c.oid), 14))+14, position(')' in substring(pg_get_constraintdef(c.oid), position('(' in substring(pg_get_constraintdef(c.oid), 14))+14))-1) END AS "PK_Column"
FROM pg_constraint c
JOIN pg_namespace n ON n.oid = c.connamespace
WHERE contype IN ('f', 'p ')
AND pg_get_constraintdef(c.oid) LIKE 'FOREIGN KEY %'
ORDER BY pg_get_constraintdef(c.oid), conrelid::regclass::text, contype DESC;
フォームのテーブルを返します:
| FK_Table | FK_Column | PK_Table | PK_Column |
pg_get_constraintdef()
を解析する必要はありません。pg_constraint
取得するテーブルその他の詳細( the docs )。
ここに constraint_type
は次のいずれかです。
アーウィンの答え に基づく:
SELECT c.conname AS constraint_name,
c.contype AS constraint_type,
sch.nspname AS "self_schema",
tbl.relname AS "self_table",
ARRAY_AGG(col.attname ORDER BY u.attposition) AS "self_columns",
f_sch.nspname AS "foreign_schema",
f_tbl.relname AS "foreign_table",
ARRAY_AGG(f_col.attname ORDER BY f_u.attposition) AS "foreign_columns",
pg_get_constraintdef(c.oid) AS definition
FROM pg_constraint c
LEFT JOIN LATERAL UNNEST(c.conkey) WITH ORDINALITY AS u(attnum, attposition) ON TRUE
LEFT JOIN LATERAL UNNEST(c.confkey) WITH ORDINALITY AS f_u(attnum, attposition) ON f_u.attposition = u.attposition
JOIN pg_class tbl ON tbl.oid = c.conrelid
JOIN pg_namespace sch ON sch.oid = tbl.relnamespace
LEFT JOIN pg_attribute col ON (col.attrelid = tbl.oid AND col.attnum = u.attnum)
LEFT JOIN pg_class f_tbl ON f_tbl.oid = c.confrelid
LEFT JOIN pg_namespace f_sch ON f_sch.oid = f_tbl.relnamespace
LEFT JOIN pg_attribute f_col ON (f_col.attrelid = f_tbl.oid AND f_col.attnum = f_u.attnum)
GROUP BY constraint_name, constraint_type, "self_schema", "self_table", definition, "foreign_schema", "foreign_table"
ORDER BY "self_schema", "self_table";
結果はschema
とtable
の順に並べられます。
テクニカルノート:参照 この質問 についてwith ordinality
。
最近、これを、infoスキーマに基づいてCRUDユーティリティを構築するデータアクセスレイヤーに実装する必要がありましたが、最終的にはこれでうまくいきました。
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
"pg_constraint".contype IN ( 'f', 'p' )
AND
"pg_namespace".nspname = current_schema()
AND
"pg_constraint".conrelid::regclass::text IN ('whatever_table_name')