我有一个脚本，它通过zip文件将数据输入数据库，解压后首先移动数据，然后插入数据库中的数据 CSV ，由解压产生，在表中。现在我想修改这个脚本，只从另一个数据库和表中获取数据，但是我需要能够将这些数据放在 CSV 数据丢失了。

USE [$(DataTools)];
GO

DECLARE @LoadJob  VARCHAR(64) = '208_table';
DECLARE @LoadStep VARCHAR(64) = 'spMoveUnzipLoad';
DECLARE @PreLoad  VARCHAR(128) = CONCAT('etl.spChangeFileEncodingAsMUZLPreLoad ''', @LoadJob, '''');

DELETE FROM [load].LoadParameters WHERE loadJob = @LoadJob;

EXEC [load].setLoadParameter @LoadJob, @LoadStep, 'copyOrMoveFile'     , 'move';
EXEC [load].setLoadParameter @LoadJob, @LoadStep, 'doNotUnzipFile'     , '0';
EXEC [load].setLoadParameter @LoadJob, @LoadStep, 'FileMask'           , 'data.csv.gz';
EXEC [load].setLoadParameter @LoadJob, @LoadStep, 'firstRow'           , '2';
EXEC [load].setLoadParameter @LoadJob, @LoadStep, 'forLoadFolderName'  , '\ForLoad';
EXEC [load].setLoadParameter @LoadJob, @LoadStep, 'formatFile'         , '\208_Customers.fmt';
EXEC [load].setLoadParameter @LoadJob, @LoadStep, 'loadedFolderName'   , '\Loaded';
EXEC [load].setLoadParameter @LoadJob, @LoadStep, 'notLoadedFolderName', '\NotLoaded';
EXEC [load].setLoadParameter @LoadJob, @LoadStep, 'sourceFolderName'   , '$(SourceFileLocation)';
EXEC [load].setLoadParameter @LoadJob, @LoadStep, 'targetDatasetName'  , 'table_Staging';
EXEC [load].setLoadParameter @LoadJob, @LoadStep, 'targetDatabaseName' , '$(DatabaseName)';
EXEC [load].setLoadParameter @LoadJob, @LoadStep, 'truncate'           , '1';
EXEC [load].setLoadParameter @LoadJob, @LoadStep, 'unzipFolderName'    , '\208_Customers\Unzip';
EXEC [load].setLoadParameter @LoadJob, @LoadStep, 'PostLoadSql'        , 'test_PostLoad';
EXEC [load].setLoadParameter @LoadJob, @LoadStep, 'PreLoadSql'         , @PreLoad;
EXEC [load].setLoadParameter @LoadJob, @LoadStep, 'workingFolderName'  , '\208_Customers\Working';
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--                                                     End of Script
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如何更改此脚本以从表中获取数据？它会改变还是需要重新创建？