On Tue, 2007-08-28 at 10:27 -0400, Stephen Harris wrote:
> On Tue, Aug 28, 2007 at 10:13:00AM -0400, Scott McClanahan wrote:
> > On Tue, 2007-08-28 at 10:08 -0400, Stephen Harris wrote:
> > > > Not a CentOS specific question, although I am running grep on CentOS 4.3
> > > > but how would you grep out a series of lines in a file starting at a
> > > > specific point. For instance, if I have a file named foo and I want to
> > > > grep out the next 5 lines after the first and only instance of the
> > > > string "bar" how could I pull that off? Thanks so much.
> > >
> > > What do you mean by "grep out" ? Do you want to display those lines,
> > > or skip those lines? Do you want to see the "bar" line? Is that included
> > > in the 5 lines?
> > >
> > > Anyway, you probably want to use "sed" here, rather than "grep".
> > >
> > I'd like to skip those lines. I'd like to skip the line with "bar" and
> > the following five lines.
>
> Like this?
>
> $ cat xx
> line 1
> line 2
> line bar
> line after 1
> line after 2
> line after 3
> line after 4
> line after 5
> line after 6
> line after 7
> $ sed '/bar/,+5d' xx
> line 1
> line 2
> line after 6
> line after 7
>
Beautiful man! Hats off. I've never used sed like that but I'll surely
remember that one. Thanks from everybody.
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