如何删除匹配的行和上一行? [英] How do I delete a matching line and the previous one?

问题描述

我需要删除匹配的行和之前的行.例如在下面的文件中，我需要删除第 1 行 &2.

I need delete a matching line and one previous to it. e.g In file below I need to remove lines 1 & 2.

我试过 "grep -v -B 1 "page.of." 1.txt我希望它不会打印匹配的行和上下文.

I tried "grep -v -B 1 "page.of." 1.txt and I expected it to not print the matchning lines and the context.

我尝试了 如何使用 sed 删除匹配的行及其上方和下方的行?但无法理解 sed 的用法.

I tried the How do I delete a matching line, the line above and the one below it, using sed? but could not understand the sed usage.

---1.txt--
**document 1**                         -> 1
**page 1 of 2**                        -> 2

testoing
testing

super crap blah

**document 1**
**page 2 of 2**

推荐答案

你想做一些与 给出的答案

sed -n '
/page . of ./ { #when pattern matches
n #read the next line into the pattern space
x #exchange the pattern and hold space
d #skip the current contents of the pattern space (previous line)
}

x  #for each line, exchange the pattern and hold space
1d #skip the first line
p  #and print the contents of pattern space (previous line)

$ { #on the last line
x #exchange pattern and hold, pattern now contains last line read
p #and print that
}'

并作为一行

sed -n '/page . of ./{n;x;d;};x;1d;p;${x;p;}' 1.txt

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