我可以从数据类型中选择一个字段吗?

时间:2011-06-29 19:08:25

标签: sql postgresql types sqldatatypes columnname

我可以从数据类型中选择字段/列吗?

Postgres 7.4(是的,我们正在升级)

SELECT *
FROM tbl_name
WHERE tbl_name.column = 'timestamp with time zone'

2 个答案:

答案 0 :(得分:2)

这需要元数据。

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:如果您希望表中的实际数据的列名与上面的列表匹配,您可以设置一个用户定义的函数来获取这些列名,将它们放在逗号分隔的列表中,然后解析适当的来自tbl_name的查询(当然,如果您使用的脚本语言略微超出数据库,这会更容易。)

答案 1 :(得分:1)

我最近必须这样做。为了简化问题,我定义了两个视图:

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';

对于表格中的每个字段,包括以下信息:

  • 架构名称
  • 表名称
  • 列名
  • data_type name
  • 可以为空吗?
  • 有默认值吗?
  • 描述(评论)
  • tableoid(如果您需要从目录中获取更多数据,则很方便)
  • 列号(同上)
  • 索引(在第二个视图中 - 引用此列的索引数)