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
The following `awk` command uses the '-F' option and NR and NF to print the book names after skipping the first book. The '-F' option is used to separate the content of the file base on \t. NR is used to skip the first line, and NF is used to print the first column only.
To delete a line, we'll use the sed “d” command. Note that you have to declare which line to delete. Otherwise, sed will delete all the lines.
sed is a common text processing utility in the Linux command-line. Removing the first line from an input file using the sed command is pretty straightforward. The sed command in the example above isn't hard to understand. The parameter '1d' tells the sed command to apply the 'd' (delete) action on line number '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
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
.
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