Adding a new flag to a bash script

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Author	Message
Banker Joined: Sep 29, 2008 Posts: 4 Other Topics	Posted Sep 29, 2008 at 9:47:49 PM Subject: Adding a new flag to a bash script All, Im new to bash and have try to make this work, but am hitting a wall. This script returns the results for all table spaces when the -d flag isnt used and a specific tablespace when the -d flag is used.
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Khabi Joined Apr 21, 2008 Posts: 121 Other Topics	Posted: Sep 30, 2008 1:44:00 AM Subject: Adding a new flag to a bash script Without seeing what the normal output looks like it makes it near impossible to help you out. If it outputs everything to stdout (it looks like it does) you can do what you want by piping the output to 'grep -v' followed by what you want to exclude as a quick and dirty way to get what you want. script.sh -s SID -w 80 -c 90 \| grep -v tablespace
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Banker Joined Sep 29, 2008 Posts: 4 Other Topics	Posted: Sep 30, 2008 3:36:27 AM Subject: Adding a new flag to a bash script Normal out put would return Tablespaces OK. Is there anything i can add to the script so where i can type the following syntax command line: script.sh -s SID -w 80 -c 90 -o tablespace 1 And the output would be: Tablespace2 99% tablespace3 87% Even though tablespace1 would equal 85%. I want to be able to exclude anything designated after the new -o flag from the return. Thanks!
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Banker Joined Sep 29, 2008 Posts: 4 Other Topics	Posted: Sep 30, 2008 10:25:50 PM Subject: Re: Adding a new flag to a bash script BTW the grep -v suggestion doesnt work
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