找出欄位中的關鍵字~

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  • 2008-06-27

摘要:找出欄位中的關鍵字~

 

Declare @key_word varchar(20);
Set @key_word = 'script';
Select 'Select ''欄位'' [COLUMN] , Convert(text,''資料'') [DATA] '
Union All
Select /*DATA_TYPE ,*/
    'Union All' +  char(13) +char(10) +  
    'Select  ''' + TABLE_NAME + '.' + COLUMN_NAME+ ''' ,  [' + COLUMN_NAME + ']  From [' + TABLE_NAME +  
    ']with(nolock) Where ['  +  COLUMN_NAME + '] Like ''%' +@key_word + '%'''  
From INFORMATION_SCHEMA.COLUMNS
Where
    DATA_TYPE In ( 'varchar' , 'text' , 'char' , 'nchar' , 'nvarchar' , 'ntext' )

 


 

/*
CREATE PROC SearchAllTables
(
@SearchStr nvarchar(100)
)
AS
BEGIN
*/

-- Copyright © 2002 Narayana Vyas Kondreddi. All rights reserved.
-- Purpose: To search all columns of all tables for a given search string
-- Written by: Narayana Vyas Kondreddi
-- Site: http://vyaskn.tripod.com
-- Tested on: SQL Server 7.0 and SQL Server 2000
-- Date modified: 28th July 2002 22:50 GMT
DECLARE @SearchStr nvarchar(100)
SET @SearchStr = '<script'

CREATE TABLE #Results (ColumnName nvarchar(370), ColumnValue nvarchar(3630))

SET NOCOUNT ON

DECLARE @TableName nvarchar(256), @ColumnName nvarchar(128), @SearchStr2 nvarchar(110)
SET @TableName = ''
SET @SearchStr2 = QUOTENAME('%' + @SearchStr + '%','''')

WHILE @TableName IS NOT NULL
BEGIN
SET @ColumnName = ''
SET @TableName =  
(
SELECT MIN(QUOTENAME(TABLE_SCHEMA) + '.' + QUOTENAME(TABLE_NAME))
FROM INFORMATION_SCHEMA.TABLES
WHERE TABLE_TYPE = 'BASE TABLE'
AND QUOTENAME(TABLE_SCHEMA) + '.' + QUOTENAME(TABLE_NAME) > @TableName
AND OBJECTPROPERTY(
OBJECT_ID(
QUOTENAME(TABLE_SCHEMA) + '.' + QUOTENAME(TABLE_NAME)
), 'IsMSShipped'
) = 0
)

WHILE (@TableName IS NOT NULL) AND (@ColumnName IS NOT NULL)
BEGIN
SET @ColumnName =
(
SELECT MIN(QUOTENAME(COLUMN_NAME))
FROM INFORMATION_SCHEMA.COLUMNS
WHERE TABLE_SCHEMA = PARSENAME(@TableName, 2)
AND TABLE_NAME = PARSENAME(@TableName, 1)
AND DATA_TYPE IN ('char', 'varchar', 'nchar', 'nvarchar')
AND QUOTENAME(COLUMN_NAME) > @ColumnName
)

IF @ColumnName IS NOT NULL
BEGIN
INSERT INTO #Results
EXEC
(
'SELECT ''' + @TableName + '.' + @ColumnName + ''', LEFT(' + @ColumnName + ', 3630)  
FROM ' + @TableName + ' (NOLOCK) ' +
' WHERE ' + @ColumnName + ' LIKE ' + @SearchStr2
)
END
END  
END

SELECT ColumnName, ColumnValue FROM #Results
--END