I'm trying to delete the first two lines of a file by just not printing it to another file. I'm not looking for something fancy. Here's my (failed) attempt at awk:
awk '{ (NR > 2) {print} }' myfile
That throws out the following error:
awk: { NR > 2 {print} }
awk: ^ syntax error
Example:
contents of 'myfile':
blah
blahsdfsj
1
2
3
4
What I want the result to be:
1
2
3
4
Use tail:
tail -n+3 file
from the man page:
-n, --lines=K
output the last K lines, instead of the last 10; or use -n +K
to output lines starting with the Kth
How about:
tail +3 file
OR
awk 'NR>2' file
OR
sed '1,2d' file
You're nearly there. Try this instead:
awk 'NR > 2 { print }' myfile
awk is rule based, and the rule appears bare (i.e., without braces) before the block it woud execute if it passes.
Also as Jaypal has pointed out, in awk if all you want to do is print the line that matches the rules you can even omit the action, thus simplifying the command to:
awk 'NR > 2' myfile
awk is based on pattern{action} statements. In your case, the pattern is NR>2 and the action you want to perform is print. This action is also the default action of awk.
So even though
awk 'NR>2{print}' filename
would work fine, you can shorten it to
awk 'NR>2' filename.
