データ型からフィールド/列を選択できますか?
Postgres 7.4(はい、アップグレードしています)
SELECT *
FROM tbl_name
WHERE tbl_name.column = 'timestamp with time zone'
データ型からフィールド/列を選択できますか?
Postgres 7.4(はい、アップグレードしています)
SELECT *
FROM tbl_name
WHERE tbl_name.column = 'timestamp with time zone'
それにはメタデータが必要です。
select column_name
from information_schema.columns
where table_schema='your_schema'
and table_name='tbl_name'
and data_type='timestamp without time zone'
order by ordinal_position;
ETA:上記のリストと一致する列名を持つテーブルの実際のデータが必要な場合は、それらの列名を取得してコンマ区切りのリストに入れ、からのクエリを解析するユーザー定義関数を設定できます。 tbl_nameを適切に(もちろん、データベースの少し外にあるスクリプト言語で作業している場合、これははるかに簡単です)。
I had to do that recently. To ease things I defined two views :
CREATE VIEW view_table_columns AS
SELECT n.nspname AS schema, cl.relname AS table_name,
a.attname AS column_name, ty.typname AS data_type,
a.attnotnull AS nnull,
a.atthasdef AS hasdef,
descr.description AS descr, cl.oid AS tableoid, a.attnum AS colnum
FROM pg_attribute a
JOIN pg_class cl ON a.attrelid = cl.oid AND cl.relkind = 'r'::"char"
JOIN pg_namespace n ON n.oid = cl.relnamespace
JOIN pg_type ty ON ty.oid = a.atttypid
LEFT JOIN pg_description descr
ON descr.objoid = cl.oid AND descr.objsubid = a.attnum
WHERE a.attnum > 0 AND NOT a.attisdropped
AND n.nspname !~~ 'pg_%'::text
AND n.nspname <> 'information_schema'::name;
COMMENT ON VIEW view_table_columns IS 'Lista all fields of all tables';
CREATE VIEW view_table_columns2 AS
SELECT view_table_columns.*,
( SELECT count(*) AS count
FROM pg_index
WHERE pg_index.indrelid = pg_index.tableoid AND
(view_table_columns.colnum = ANY (pg_index.indkey::smallint[]))) AS indexes
FROM view_table_columns;
COMMENT ON VIEW view_table_columns2 IS 'Adds to view_table_columns info about indexed fields';
That includes, for every field in your tables the following info: