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.