regex - 在 bash 脚本中使用 sed

Question

使用通过 grep 命令获取的行号数组，然后我尝试增加行号并使用 sed 命令检索新行号上的内容，但我假设我的语法有问题(特别是 sed 部分，因为其他一切都有效。)

脚本如下:

#!/bin/bash


#getting array of initial line numbers    

temp=(`egrep -o -n '\<a class\=\"feed\-video\-title title yt\-uix\-contextlink  yt\-uix\-sessionlink  secondary"' index.html |cut -f1 -d:`)

new=( )

#looping through array, increasing the line number, and attempting to add the
#sed result to a new array

for x in ${temp[@]}; do

((x=x+5))

z=sed '"${x}"q;d' index.html

new=( ${new[@]} $z ) 

done

#comparing the two arrays

echo ${temp[@]}
echo ${new[@]}

Answer 1

这可能对你有用:

#!/bin/bash


#getting array of initial line numbers    

temp=(`egrep -o -n '\<a class\=\"feed\-video\-title title yt\-uix\-contextlink  yt\-uix\-sessionlink  secondary"' index.html |cut -f1 -d:`)

new=( )

#looping through array, increasing the line number, and attempting to add the
#sed result to a new array

for x in ${temp[@]}; do

((x=x+5))

z=$(sed ${x}'q;d' index.html) # surrounded sed command by $(...)

new=( "${new[@]}" "$z" ) # quoted variables

done

#comparing the two arrays

echo "${temp[@]}" # quoted variables
echo "${new[@]}"  # quoted variables

你的 sed 命令没问题；它只需要被 $(...) 包围，并删除和重新排列不必要的引号。

顺便说一句

获取模式后五行的行 (GNU sed):

sed '/pattern/,+5!d;//,+4d' file