grep - 之前打印行，不打印匹配 [英] grep - print line before, don't print match

问题描述

如何轻松地在匹配上方打印行并跳过匹配本身?grep -A, -B 和 -o opt 不解决.也许一些 awk 魔法?

How to easily print line above the match and skip the match itself? grep -A, -B and -o opt do not solve it. Maybe some awk magic?

例如:

$ cat foo.txt
bar
foo
baz
foo

$ cat foo.txt | grep foo-SOMETHING
bar
baz

编辑

• 如果第 2 行和第 3 行有foo"，则应打印第 1 行和第 2 行(虽然我在这里不是很严格)

附加功能:考虑示例:

bar
foo
baz
foo
foo

理想情况下应该返回

bar
baz
foo

推荐答案

awk '!/foo/ { line = $0 }
     /foo/ { print line }' foo.txt

第一个子句将每个非 foo 行保存在一个变量中.当行匹配 foo 时，第二个子句打印最近保存的行.

The first clause saves each non-foo line in a variable. The second clause prints the most recent saved line when the line matches foo.

这也有效(并处理连续有两行 foo 的情况):

This also works (and handles the case where you have two foo lines in a row):

awk '/foo/ {print line}
     {line = $0}' foo.txt

使用 grep 你可以:

grep -B 1 foo foo.txt | grep -vE 'foo|^--$'

第二个命令过滤掉 foo 行和匹配块之间打印的分隔符.

The second command filters out the foo lines and the dividers that are printed between the matching blocks.

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