使用 BASH 或 awk 或 sed 或其他方法删除文件的前两行
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Deleting the first two lines of a file using BASH or awk or sed or whatever
提问by Amit
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:
我试图通过不将文件打印到另一个文件来删除文件的前两行。我不是在寻找花哨的东西。这是我对 awk 的（失败）尝试：
awk '{ (NR > 2) {print} }' myfile
That throws out the following error:
这会抛出以下错误：
awk: { NR > 2 {print} }
awk: ^ syntax error
Example:
例子：
contents of 'myfile':
“我的文件”的内容：
blah
blahsdfsj
1
2
3
4
What I want the result to be:
我想要的结果是：
1
2
3
4
回答by RobS
Use tail:
使用尾巴：
tail -n+3 file
from the man page:
从手册页：
-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by anubhava
How about:
怎么样：
tail +3 file
OR
或者
awk 'NR>2' file
OR
或者
sed '1,2d' file
回答by Chris J
You're nearly there. Try this instead:
你快到了。试试这个：
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.
awk 是基于规则的，并且规则在它通过时将执行的块之前显示为裸露的（即，没有大括号）。
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:
同样正如 Jaypal 所指出的，在 awk 中，如果您只想打印与规则匹配的行，您甚至可以省略该操作，从而将命令简化为：
awk 'NR > 2' myfile
回答by jaypal singh
awkis based on pattern{action}statements. In your case, the patternis NR>2and the actionyou want to perform is print. This actionis also the default actionof awk.
awk是基于pattern{action}陈述。在您的情况下，patternisNR>2和action您想要执行的是print。这action是同样default action的awk。
So even though
所以即使
awk 'NR>2{print}' filename
awk 'NR>2{print}' filename
would work fine, you can shorten it to
可以正常工作，您可以将其缩短为
awk 'NR>2' filename.
awk 'NR>2' filename.
