0

私はこれらのステートメントに参加しようとしています:

select min(len(Password)) as min, 
       max(len(Password)) as max, 
       avg(len(Password)) as avg 
FROM Customer.dbo.Password 

UNION ALL
select min(len(password2)) as min, 
       max(len(password2)) as max, 
       avg(len(password2)) as avg 
FROM website.dbo.password2

これらに：

SELECT c.name COLLATE SQL_Latin1_General_CP1_CI_AS colName, o.name COLLATE SQL_Latin1_General_CP1_CI_AS tableName, 'website' db FROM website.sys.all_columns as c INNER JOIN website.sys.all_objects o ON c.object_id = o.object_id WHERE c.name like '%password%' AND type = 'U'

UNION
SELECT c.name COLLATE SQL_Latin1_General_CP1_CI_AS colName, o.name COLLATE SQL_Latin1_General_CP1_CI_AS tableName, 'Customer' db FROM Customer.sys.all_columns as c INNER JOIN Customer.sys.all_objects o ON c.object_id = o.object_id WHERE c.name like '%password%' AND type = 'U'

共通の列はありません。最初の 2 つの SELECT ステートメントは、パスワードの長さの最小値、最大値、および平均値の 3 つの列を生成し、次の 2 つの列は、パスワードが保持されている場所の colName、tableName、およびデータベース名を生成します。

純粋に視覚的な理由から、両方の結果セットを 1 つのテーブルに表示して、対応する最小値、最大値、平均値を適切な colName、tableName、および db に取得したいと考えています。

私はこのようにしてみました：

DECLARE @numberTable TABLE (link nvarchar(500), max int, min int, avg int)
INSERT INTO @numberTable

select 'link', min(len(Password)) as min, 
       max(len(Password)) as max, 
       avg(len(Password)) as avg 
FROM Customer.dbo.Password UNION ALL
select 'link', min(len(password2)) as min, 
       max(len(password2)) as max, 
       avg(len(password2)) as avg 
FROM website.dbo.password2

SELECT * FROM @numberTable ORDER BY max

DECLARE @myTable TABLE (link nvarchar(500), colName nvarchar(500), tableName   nvarchar(500), db nvarchar(500))
INSERT INTO @myTable

SELECT 'link', c.name COLLATE SQL_Latin1_General_CP1_CI_AS colName, o.name COLLATE SQL_Latin1_General_CP1_CI_AS tableName, 'website' db FROM website.sys.all_columns as c INNER JOIN website.sys.all_objects o ON c.object_id = o.object_id WHERE c.name like '%password%' AND type = 'U' UNION

SELECT 'link', c.name COLLATE SQL_Latin1_General_CP1_CI_AS colName, o.name COLLATE SQL_Latin1_General_CP1_CI_AS tableName, 'Customer' db FROM Customer.sys.all_columns as c INNER JOIN Customer.sys.all_objects o ON c.object_id = o.object_id WHERE c.name like '%password%' AND type = 'U' 

SELECT * FROM @myTable ORDER BY db

SELECT link
FROM @myTable
LEFT OUTER JOIN @numberTable
ON @myTable.link = @numberTable.link

残念ながら、これはうまくいきませんでした。

Microsoft SQL Server 2008 を使用しています。

私が求めていることが理にかなっていることを願っています。

4

2 に答える 2

1

を使用cross joinして、2つの無関係なテーブルを結合できます。

SELECT  c.name COLLATE SQL_Latin1_General_CP1_CI_AS colName
,       o.name COLLATE SQL_Latin1_General_CP1_CI_AS tableName
,       'website' db 
,       pwd.[min]
,       pwd.[max]
,       pwd.[avg]
FROM    website.sys.all_columns as c 
INNER JOIN 
        website.sys.all_objects o 
ON      c.object_id = o.object_id 
CROSS JOIN
        (
        select min(len(Password)) as min, 
               max(len(Password)) as max, 
               avg(len(Password)) as avg 
        FROM   website.dbo.Password        
        ) as pwd
WHERE   c.name like '%password%' 
        AND type = 'U' 

UNION

SELECT  c.name COLLATE SQL_Latin1_General_CP1_CI_AS colName
,       o.name COLLATE SQL_Latin1_General_CP1_CI_AS tableName
,       'Customer' db 
,       pwd.[min]
,       pwd.[max]
,       pwd.[avg]
FROM    Customer.sys.all_columns as c 
INNER JOIN 
        Customer.sys.all_objects o 
ON      c.object_id = o.object_id 
CROSS JOIN
        (
        select min(len(Password)) as min, 
               max(len(Password)) as max, 
               avg(len(Password)) as avg 
        FROM    Customer.dbo.Password        
        ) as pwd
WHERE   c.name like '%password%' 
        AND type = 'U'
于 2012-10-11T08:45:24.677 に答える
0

彼らが共通して持っている恣意的な何かを作成し、それらに参加します

Select * 
from
(
select 
       1 as RowID,
       min(len(Password)) as min,  
       max(len(Password)) as max,  
       avg(len(Password)) as avg  
FROM Customer.dbo.Password  

UNION ALL 
select 
       2,
       min(len(password2)) as min,  
       max(len(password2)) as max,  
       avg(len(password2)) as avg  
FROM website.dbo.password2 
) t1
inner join
(


SELECT 1 as RowID, c.name COLLATE SQL_Latin1_General_CP1_CI_AS colName, o.name COLLATE SQL_Latin1_General_CP1_CI_AS tableName, 'website' db FROM website.sys.all_columns as c INNER JOIN website.sys.all_objects o ON c.object_id = o.object_id WHERE c.name like '%password%' AND type = 'U' 

UNION 
SELECT 2, c.name COLLATE SQL_Latin1_General_CP1_CI_AS colName, o.name COLLATE SQL_Latin1_General_CP1_CI_AS tableName, 'Customer' db FROM Customer.sys.all_columns as c INNER JOIN Customer.sys.all_objects o ON c.object_id = o.object_id WHERE c.name like '%password%' AND type = 'U' 
) t2
    on t1.rowID = t2.rowID
于 2012-10-11T08:48:18.247 に答える