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Answer» Hi,
I have to run a program from a location the following is my batch file code
ECHO off echo Hello this is a test batch file START D:\ORDSG\instances\instance1\bin>ltctctl.bat startall pause
basically I have to run ltctctl.bat program with the syntax as "ltctctl.bat startall" but when I run this batch file it opens the location D:\ORDSG\instances\instance1\bin which contains the file ltctctl.bat instead of running the command "ltctctl.bat startall".
PLEASE SUGGEST
THANKS , Jay Try this:
Code: [Select]echo off echo Hello this is a test batch file START "" D:\ORDSG\instances\instance1\bin\ltctctl.bat startall
Your path statement was incorrect. You had the prompt instead of the patch.
type start /? from a cmd prompt and more info.
Quote from: [email protected] on April 07, 2010, 06:25:06 AM echo off echo Hello this is a test batch file START D:\ORDSG\instances\instance1\bin>ltctctl.bat startall
basically I have to run ltctctl.bat program with the syntax as "ltctctl.bat startall" Please suggest
C:\bin>type ltctctl.bat
Code: [Select]rem C:\bin\ltctctl.bat startall rem the above REQUIRES a command line argument
echo off echo Hello this is a test batch file
rem C:\bin\ltctctl.bat %1
echo ltctctl %1 Output:
C:\bin>ltctctl.bat startall
Hello this is a test batch file ltctctl startall
C:\bin>Hi Thanks to all of you ..
I had actualy made a mistake in the command which Mr. Gregflowers rightly pointed.
Thanks,
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