为什么sed / ^ $ / d只删除空行但/ ^ $ / p会打印所有行？

I'm able to use sed /^$/d <file> to delete all the blank lines in the file, but what if I want to print all the blank lines only? The command sed /^$/p <file> prints all the lines in file.

我可以使用sed / ^ $ / d 删除文件中的所有空白行,但是如果我只想打印所有空白行怎么办?命令sed / ^ $ / p 打印文件中的所有行。

The reason I want to do this is that we use an EDA program (Expedition) that uses regex to run rules on the names of nets. I'm trying to find a way to search for all nets that don't have names assigned. I thought using ^$ would work, but it just ends up finding all nets, which is what /^$/p is doing too. So is there a different way to do this?

我想这样做的原因是我们使用EDA程序(Expedition)使用正则表达式来运行网络名称的规则。我正试图找到一种方法来搜索没有分配名称的所有网络。我认为使用^ $会起作用,但它最终会找到所有网络,这也是/ ^ $ / p正在做的事情。那么有不同的方法吗?

4 个解决方案

#1

Unless otherwise specified sed will print the pattern space when it has finished processing it. If you look carefully at your output you'll notice that you get 2 blank lines for every one in the file. You'll have to use the -n command line switch to stop sed from printing.

除非另有说明,否则sed将在完成处理后打印图案空间。如果仔细查看输出,您会发现文件中的每一行都有2个空行。您必须使用-n命令行开关来停止打印sed。

sed -n /^$/p infile

Should work as you want.

应该按你的意愿工作。

#2

Sed prints every line by default, and so the p flag is useless. To make it useful, you need to give sed the -n switch. Indeed, the following appears to do what you want:

Sed默认打印每一行,因此p标志是无用的。为了使它有用,你需要给sed -n开关。事实上,以下似乎做你想要的:

sed -n /^$/p

#3

You can also use grep as:

你也可以使用grep作为:

grep '^$' infile

#4

think in another way, don't p, but !d

以另一种方式思考,不要p,但是!d

you may try:

你可以尝试:

sed '/^$/!d' yourFile

#1

sed -n /^$/p infile