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”

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。

这应该是解决方案：

SELECT * FROM information_schema.columns
WHERE table_schema = 'your_schema'
   AND table_name   = 'your_table'

描述表的最佳方式，如列、类型、列的修饰符等。

\d+ tablename or \d tablename

除了已经找到的命令行\d+<table_name>之外，还可以使用info_schema.columns使用信息模式查找列数据

SELECT *
FROM info_schema.columns
WHERE table_schema = 'your_schema'
AND table_name   = 'your_table'

您可以使用星号执行\d*搜索模式*，以查找与您感兴趣的搜索模式匹配的表。