Deleting the first two lines of a file using BASH or awk or sed or whatever

问题

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:

回答1:

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

回答2:

How about:

tail +3 file

awk 'NR>2' file

sed '1,2d' file

回答3:

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

回答4:

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.

来源：https://stackoverflow.com/questions/8857705/deleting-the-first-two-lines-of-a-file-using-bash-or-awk-or-sed-or-whatever

标签

bash