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divan
Posting Yak Master
153 Posts |
 Posted - 2012-10-23 : 07:40:55
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| I have gone into the sql scheduler and have created a step where I need a file copied from one area to another.. DECLARE @SQL VARCHAR(120)--COPY A NEW ONE INSET @SQL = 'xp_cmdshell ' + CHAR(39) + 'COPY C:\Reports\Templates\TRIBUTE_DETAIL_IND.xls C:\Reports\TRIBUTE_DETAIL_IND.xls' + CHAR(39) EXEC (@SQL)Every time I run this I get the following error Executed as user: advocatemd\sqladmin. The process could not be created for step 7 of job 0x56FC8CF8E5D1084B99DCB56D98A901D5 (reason: The system cannot find the file specified). The step failed.the result of the sql command using PRINT SQL xp_cmdshell 'COPY C:\Reports\Templates\TRIBUTE_DETAIL_IND.xls C:\Reports\TRIBUTE_DETAIL_IND.xls'I have run the same sql statement on the same machine in a sql query and it works fine.. So what am I doing wrong.. |
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bandi
Master Smack Fu Yak Hacker
2242 Posts |
Posted - 2012-10-23 : 07:54:32
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| One trail and error method is as follows:Remove that red marked part and then executeDECLARE @SQL VARCHAR(120)--COPY A NEW ONE INSET @SQL = 'xp_cmdshell ' + CHAR(39) + 'COPY C:\Reports\Templates\TRIBUTE_DETAIL_IND.xls C:\Reports\TRIBUTE_DETAIL_IND.xls' + CHAR(39)EXEC (@SQL)--Chandu |
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divan
Posting Yak Master
153 Posts |
Posted - 2012-10-23 : 08:07:24
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| Still getting the same error.... |
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bandi
Master Smack Fu Yak Hacker
2242 Posts |
Posted - 2012-10-23 : 09:29:47
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One more suggestion is remove striked part.. DECLARE @SQL VARCHAR(120)--COPY A NEW ONE INSET @SQL = 'xp_cmdshell ' + CHAR(39) + 'COPY C:\Reports\Templates\TRIBUTE_DETAIL_IND.xls C:\Reports\TRIBUTE_DETAIL_IND.xls' + CHAR(39)EXEC (@SQL)It is working fine for me (I just put my file path)--Chandu |
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Scheduler script