How to display a file, omitting lines that contain a string?
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good effort. note: no need cat. Its useless. many i have come across always do that. Just pass the file to grep. Next, you may have "#" as comment at the back of the line. if you do -v "#" , then you may omit an actual line.
note: no need cat. Its useless. many i have come across always do that. Just pass the file to grep.
Maybe its the Ubuntu permissions system, but the command won't work without cat, even with sudo. Without sudo, permission denied. With sudo, command not found. Maybe its different using a shell owned by root?
Next, you may have "#" as comment at the back of the line. if you do -v "#" , then you may omit an actual line.
Good point. So I could use a caret (^) to specify that grep only selects lines that begin with the string -
cat /boot/grub/menu.lst | grep ^#
The material on regular expressions that I'm looking at also states that you can add an OR operator with | , but I'm not sure how to use this with grep?
I see what you mean now by not needing cat - you give the file name as an argument to grep or awk and it goes to standard output, which by default is printing to the terminal. So to correct my grep usage -