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 -- rgds Stephen