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egemen_ates
Yak Posting Veteran
76 Posts |
Posted - 2013-03-24 : 14:37:03
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I have two server Server A -- Sql server 2008server B MS AccessI define linked server Sql server to Ms Access.using this command.EXEC sp_addlinkedserver @server = N'test4', @provider = N'Microsoft.ACE.OLEDB.12.0', @srvproduct = N'Access', @datasrc = N'\\192.168.1.9\TESTSERVER\ACCESS.mdb', @provstr = N';PWD=ABC'after execute this command SELECT * FROM test4...EMPLOYEESthis commandd execute succesfuly.after create sql server agent job and job step command SELECT * FROM test4...EMPLOYEES and execute job but given error error is :executed as user deneme\egemen cannot initialize the datasource object of oledb provider "Microsoft.ACE.OLEDB.12.0" for linked server test4[sql state 42000] error [7303]test4 returned message:the microsoft access database engine cannont open or write \\192.168.1.9\TESTSERVER\ACCESS.mdb. it is already opened exclusive by another user you need permission view or write datahow can i solve this problem tahnk you |
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James K
Master Smack Fu Yak Hacker
3873 Posts |
Posted - 2013-03-24 : 14:52:37
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Does anyone have the access file opened exclusively?If that is not the problem, check if the login account used to run the SQL Server agent has access to that folder and the mdb file. You can find the login account from SQL Server Configuration manager (in SQL Server Program group) or runwmic service where "name Like '%Agent%'" get Name , StartName |
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egemen_ates
Yak Posting Veteran
76 Posts |
Posted - 2013-03-24 : 15:19:01
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I solve problem i open configuration manager change sql server and sql server agent userThis operation maybe create some problem is it possible ?quote: Originally posted by James K Does anyone have the access file opened exclusively?If that is not the problem, check if the login account used to run the SQL Server agent has access to that folder and the mdb file. You can find the login account from SQL Server Configuration manager (in SQL Server Program group) or runwmic service where "name Like '%Agent%'" get Name , StartName
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