1

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'
4

2 に答える 2

2

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)。

于 2011-06-29T19:44:53.083 に答える
1

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:

  • schema name
  • table name
  • column name
  • data_type name
  • is nullable?
  • has default value?
  • description (comment)
  • tableoid (handy if you need to get more data from catalog)
  • column number (idem)
  • indexes (in second view - number of indexes that refer to this column)
于 2011-06-29T20:07:09.297 に答える