如何使用psql命令在PostgreSQL中执行Oracle的DESCRIBE TABLE?


当前回答

当您的表不是默认模式的一部分时,您应该写:

\d+ schema_name.table_name

否则,您将得到错误消息,即“关系不存在”

其他回答

您可以使用此选项:

SELECT attname 
FROM pg_attribute,pg_class 
WHERE attrelid=pg_class.oid 
AND relname='TableName' 
AND attstattarget <>0; 

如果您想从查询而不是psql获取它,可以查询目录模式。下面是一个复杂的查询,它可以做到这一点:

SELECT  
    f.attnum AS number,  
    f.attname AS name,  
    f.attnum,  
    f.attnotnull AS notnull,  
    pg_catalog.format_type(f.atttypid,f.atttypmod) AS type,  
    CASE  
        WHEN p.contype = 'p' THEN 't'  
        ELSE 'f'  
    END AS primarykey,  
    CASE  
        WHEN p.contype = 'u' THEN 't'  
        ELSE 'f'
    END AS uniquekey,
    CASE
        WHEN p.contype = 'f' THEN g.relname
    END AS foreignkey,
    CASE
        WHEN p.contype = 'f' THEN p.confkey
    END AS foreignkey_fieldnum,
    CASE
        WHEN p.contype = 'f' THEN g.relname
    END AS foreignkey,
    CASE
        WHEN p.contype = 'f' THEN p.conkey
    END AS foreignkey_connnum,
    CASE
        WHEN f.atthasdef = 't' THEN d.adsrc
    END AS default
FROM pg_attribute f  
    JOIN pg_class c ON c.oid = f.attrelid  
    JOIN pg_type t ON t.oid = f.atttypid  
    LEFT JOIN pg_attrdef d ON d.adrelid = c.oid AND d.adnum = f.attnum  
    LEFT JOIN pg_namespace n ON n.oid = c.relnamespace  
    LEFT JOIN pg_constraint p ON p.conrelid = c.oid AND f.attnum = ANY (p.conkey)  
    LEFT JOIN pg_class AS g ON p.confrelid = g.oid  
WHERE c.relkind = 'r'::char  
    AND n.nspname = '%s'  -- Replace with Schema name  
    AND c.relname = '%s'  -- Replace with table name  
    AND f.attnum > 0 ORDER BY number
;

它非常复杂,但它确实向您展示了PostgreSQL系统目录的强大和灵活性,并将帮助您掌握pg_catalog;-)。请确保更改查询中的%s。第一个是Schema,第二个是表名。

除了PostgreSQL方式(\d“something”或\dt“table”或\ds“sequence”等)

SQL标准方式,如下所示:

select column_name, data_type, character_maximum_length, column_default, is_nullable
from INFORMATION_SCHEMA.COLUMNS where table_name = '<name of table>';

它受到许多db引擎的支持。

尝试以下操作(在psql命令行工具中):

\d+ tablename

有关更多信息,请参阅手册。

/dt是列出数据库中所有表的命令。使用/d命令和/d+,我们可以获得表的详细信息。sysntax将如下所示*/d table_name(或)\d+table_name