unix - 如何删除匹配的行和上一行？

Question

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 行和第 2 行。

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。

我试过 How do I delete a matching line, the line above and the one below it, using sed?但无法理解 sed 的用法。

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

testoing
testing

super crap blah

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

Answer 1

你想做一些与 answer given 非常相似的事情

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
}'

并且作为单行

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