redshift - 如何使用listagg与information_schema.columns表

Question

我正在使用redshift并希望创建逗号分隔的列列表。我试图用listagg抓住从信息架构中的列名：

SELECT 
listagg(column_name,',') within group (order by ordinal_position) 
FROM information_schema.columns 
WHERE table_schema = 'my_schema' 
AND  table_name = 'my table'; 

，我发现了以下错误：

[Amazon](500310) Invalid operation: Function (listagg(text,text)) must be applied on at least one user created tables; 

Answer 1

虽然这并没有回答如何在INFORMATION_SCHEMA申请LISTAGG，我可以推荐在pg目录表上使用listagg的替代方法。

试试这个：

SELECT DISTINCT listagg(attname, ',') WITHIN GROUP (ORDER BY a.attsortkeyord) AS "columns" 
FROM pg_attribute a, pg_namespace ns, pg_class c, pg_type t, stv_tbl_perm p, pg_database db 
WHERE t.oid=a.atttypid AND a.attrelid=p.id AND ns.oid = c.relnamespace AND db.oid = p.db_id AND c.oid = a.attrelid 
AND typname NOT IN ('oid','xid','tid','cid') 
AND ns.nspname = 'my_schema' 
AND RTRIM(name) = 'my table'