我的团队拥有基于视图构建的视图，因此schema通常会导致灾难和大量的反复试验。
我想要的是一个返回所有依赖对象的查询，这些对象需要在给定某个table和DROP TABLE的情况下以正确的顺序重新创建，以便它们可以自动化并在脚本。我在Redshift WITH dependencies AS (
SELECT DISTINCT
cls1.oid AS tbloid,
nsp1.nspname AS schemaname,
cls1.relname AS name,
nsp2.nspname AS refbyschemaname,
cls2.relname AS refbyname,
cls2.oid AS viewoid
FROM pg_catalog.pg_class cls1
JOIN pg_catalog.pg_depend dep1
ON cls1.relfilenode = dep1.refobjid
JOIN pg_catalog.pg_depend dep2
ON dep1.objid = dep2.objid
JOIN pg_catalog.pg_class cls2
ON dep2.refobjid = cls2.relfilenode
LEFT OUTER JOIN pg_namespace nsp1
ON cls1.relnamespace = nsp1.oid
LEFT OUTER JOIN pg_namespace nsp2
ON cls2.relnamespace = nsp2.oid
WHERE dep2.deptype IN ('i' :: "CHAR", 'n' :: "CHAR")
AND cls2.relkind IN ('v' :: "CHAR", 'r' :: "CHAR")
AND nsp1.nspname NOT IN ('pg_catalog', 'information_schema')
ORDER BY 4, 5
),
joined_to_views AS (
SELECT
d.schemaname,
d.name,
d.refbyschemaname,
d.refbyname,
p.definition
FROM dependencies d
LEFT JOIN pg_views p
ON d.refbyschemaname = p.schemaname AND d.refbyname = p.viewname
)
SELECT *
FROM joined_to_views
文档http://docs.aws.amazon.com/redshift/latest/dg/r_DROP_TABLE.html上使用了依赖项查询的修改版本。
它似乎是返回视图及其依赖项，而不是常规表。我觉得我很接近，我错过了什么？
{{1}}
答案 0 :(得分：10)
这对你有用吗？
SELECT dependent_ns.nspname as dependent_schema
, dependent_view.relname as dependent_view
, source_ns.nspname as source_schema
, source_table.relname as source_table
, pg_attribute.attname as column_name
FROM pg_depend
JOIN pg_rewrite ON pg_depend.objid = pg_rewrite.oid
JOIN pg_class as dependent_view ON pg_rewrite.ev_class = dependent_view.oid
JOIN pg_class as source_table ON pg_depend.refobjid = source_table.oid
JOIN pg_attribute ON pg_depend.refobjid = pg_attribute.attrelid
AND pg_depend.refobjsubid = pg_attribute.attnum
JOIN pg_namespace dependent_ns ON dependent_ns.oid = dependent_view.relnamespace
JOIN pg_namespace source_ns ON source_ns.oid = source_table.relnamespace
WHERE
source_ns.nspname = 'my_schema'
AND source_table.relname = 'my_table'
AND pg_attribute.attnum > 0
AND pg_attribute.attname = 'my_column'
ORDER BY 1,2;
如果我做了一个错误的假设，请发表评论，我将重新调整我的答案。