如何在PostgreSQL中执行与Oracle的" DESCRIBE TABLE"等效的命令(使用psql命令)？

解决方案

psql等效于" DESCRIBE TABLE"是" \ d table"。

有关更多详细信息，请参见PostgreSQL手册的psql部分。

我们可以使用psql斜杠命令执行此操作：

\d myTable describe table

它也适用于其他对象：

\d myView describe view
 \d myIndex describe index
 \d mySequence describe sequence

资料来源：faqs.org

试试看(在psql命令行工具中)：

\d+ tablename

有关更多信息，请参见手册。

除了PostgreSQL方式(\ d'something'或者\ dt'table'或者\ ds'sequence'等)

SQL标准方式，如下所示：

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

许多数据库引擎都支持它。

如果要从查询而不是psql获取它，则可以查询目录架构。这是一个执行此操作的复杂查询：

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，但是确实向我们展示了PostgreSQL系统目录的强大功能和灵活性，应该可以使我们逐步掌握pg_catalog ;-)。确保更改查询中的％s。第一个是模式，第二个是表名。