linux - Shell脚本用另一个字符串替换有空格的字符串

Question

我正在尝试将以下字符串替换为多行，如下所示

setsid /usr/local/bin/Naming_Service ${OPTIONS} &

替换为

setsid /usr/local/bin/Naming_Service ${OPTIONS_13016} &
setsid /usr/local/bin/Naming_Service ${OPTIONS_13018} &

我试过这个命令

sed '0,/setsid \/usr\/var\/run\/Naming_Serivce ${OPTIONS}/s//setsid \/usr\/var\/run\/Naming_Serivce ${OPTIONS_13016}\n\setsid \/usr\/var\/run\/Naming_Serivce ${OPTIONS_13018}\n /' script > new_script

你能帮忙解决一下吗

Answer 1

 sed 's/^\(.*\)\(${OPTIONS}\)\(.*\)$/\1${OPTIONS_13016}\3\n\1${OPTIONS_13018}\3/'  < script > new_script

• (...) - 创建组
• \1\3 - 使用这些组
• \n - 换行符
• .* - 任何字符