Date: Sat, 18 May 2013 11:09:02 -0400 From: Quartz <quartz@sneakertech.com> To: Devin Teske <dteske@freebsd.org> Cc: Tim Daneliuk <tundra@tundraware.com>, "Teske, Devin" <Devin.Teske@fisglobal.com>, "<freebsd-questions@freebsd.org>" <freebsd-questions@freebsd.org> Subject: Re: check variable content size in sh script Message-ID: <5197998E.6050200@sneakertech.com> In-Reply-To: <13CA24D6AB415D428143D44749F57D7201F4D41F@ltcfiswmsgmb26> References: <5194F65F.6080503@a1poweruser.com> <5194FB0A.9090400@tundraware.com> <13CA24D6AB415D428143D44749F57D7201F4D41F@ltcfiswmsgmb26>
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> However, if the OP wanted to actually truncate $FOO to 51
> characters:
>
> NEWFOO=$( echo "$FOO" | awk -v max=51 '{print substr($0,0,max)}' )
You don't need all that for a simple truncation/substring, you can do it
with a direct assignment:
newfoo=${foo:0:51}
The three params here are "variable", "start position" and "length".
______________________________________
it has a certain smooth-brained appeal
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