让 sed 忽略不匹配的行 [英] Have sed ignore non-matching lines

问题描述

如何让 sed 根据某些表达式过滤匹配的行，但忽略不匹配的行，而不是让它们打印?

How can I make sed filter matching lines according to some expression, but ignore non-matching lines, instead of letting them print?

作为一个真实的例子，我想在一组文件上运行 scalac(Scala 编译器)，并从它的 -verbose 输出中读取 .类 文件创建.scalac -verbose 输出一堆消息，但我们只对 [wrote some-class-name.class] 形式的那些感兴趣.我目前正在做的是这个(|& 是 bash 4.0 将 stderr 通过管道传输到下一个程序的方式):

As a real example, I want to run scalac (the Scala compiler) on a set of files, and read from its -verbose output the .class files created. scalac -verbose outputs a bunch of messages, but we're only interested in those of the form [wrote some-class-name.class]. What I'm currently doing is this (|& is bash 4.0's way to pipe stderr to the next program):

$ scalac -verbose some-file.scala ... |& sed 's/^\[wrote \(.*\.class\)\]$/\1/'

这将从我们感兴趣的消息中提取文件名，但也会让所有其他消息不变地通过！当然，我们可以这样做:

This will extract the file names from the messages we're interested in, but will also let all other messages pass through unchanged! Of course we could do instead this:

$ scalac -verbose some-file.scala ... |& grep '^\[wrote .*\.class\]$' |
  sed 's/^\[wrote \(.*\.class\)\]$/\1/'

可行，但看起来很像解决真正的问题，即如何指示 sed 忽略输入中的不匹配行.那么我们该怎么做呢?

which works but looks very much like going around the real problem, which is how to instruct sed to ignore non-matching lines from the input. So how do we do that?

推荐答案

另一种简单的 sed 方法:

Another way with plain sed:

sed -e 's/.../.../;t;d'

s/// 是一个替代，t 不带任何标签有条件地跳过后面的所有命令，d 删除一行.

s/// is a substituion, t without any label conditionally skips all following commands, d deletes line.

不需要 perl 或 grep.

No need for perl or grep.

(根据 Nicholas Riley 的建议编辑)

(edited after Nicholas Riley's suggestion)

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