sql - 以编程方式创建用于删除和创建具有依赖顺序的数据库视图的脚本

Question

我只看到这个话题。这与我的问题非常相似。但我不想使用任何第三方来创建脚本。我想以依赖顺序和编程方式创建删除和创建数据库视图的脚本。我怎么能做这样的事？

Answer 1

最后我发现我的答案是这样的，但请记住，这仅适用于具有一个嵌套级别的视图。不多 ！：

SELECT MIN(Row) AS Row, CreateCmd FROM ( 
SELECT Row_Number() OVER (ORDER BY S.Row) As Row, S.CreateCmd FROM ( 
SELECT 0 AS Row, RTRIM(ISNULL(smv.definition, ssmv.definition)) AS CreateCmd 
FROM sys.all_objects AS v 
LEFT OUTER JOIN sys.sql_modules AS smv ON smv.object_id = v.object_id 
LEFT OUTER JOIN sys.system_sql_modules AS ssmv ON ssmv.object_id = v.object_id 
INNER JOIN (SELECT object_id, referenced_major_id FROM sys.sql_dependencies 
GROUP BY object_id, referenced_major_id) AS a ON v.object_id = a.referenced_major_id
WHERE (v.type = 'V' OR v.type = 'P' OR v.type = 'IF' OR v.type = 'TF' OR v.type =    'FN') 
AND SCHEMA_NAME(v.schema_id) = N'dbo' AND is_ms_shipped <> 1 
AND smv.execute_as_principal_id IS NULL AND ssmv.execute_as_principal_id IS NULL 
GROUP BY v.Name, smv.definition, ssmv.definition 
UNION ALL 
SELECT 1 AS Row, RTRIM(ISNULL(smv.definition, ssmv.definition)) AS CreateCmd 
FROM sys.all_objects AS v 
LEFT OUTER JOIN sys.sql_modules AS smv ON smv.object_id = v.object_id 
LEFT OUTER JOIN sys.system_sql_modules AS ssmv ON ssmv.object_id = v.object_id 
INNER JOIN (SELECT object_id, referenced_major_id FROM sys.sql_dependencies 
GROUP BY object_id, referenced_major_id) AS a ON v.object_id = a.object_id
WHERE (v.type = 'V' OR v.type = 'P' OR v.type = 'IF' OR v.type = 'TF' OR v.type = 'FN') 
AND SCHEMA_NAME(v.schema_id) = N'dbo' AND is_ms_shipped <> 1 
AND smv.execute_as_principal_id IS NULL AND ssmv.execute_as_principal_id IS NULL 
GROUP BY v.Name, smv.definition, ssmv.definition 
) S GROUP BY S.Row, CreateCmd 
) D GROUP BY CreateCmd ORDER BY Row