使用Sed替換從正則表達式獲得的另一個字符串的字符串

Question

我想用第一行＃之間的任何內容替換字符串（在本例中爲Linux）。

所以我想轉：

#Windows# 
1. Linux Sysadmin, Linux Scripting etc. 
2. Databases - Oracle, mySQL etc. 
3. Security (Firewall, Network, Online Security etc) 
4. Storage in Linux 
5. Productivity (Too many technologies to explore, not much time available) 

到：

1. Windows Sysadmin, Windows Scripting etc. 
2. Databases - Oracle, mySQL etc. 
3. Security (Firewall, Network, Online Security etc) 
4. Storage in Windows 
5. Productivity (Too many technologies to explore, not much time available) 

或

#Solaris# 
1. Linux Sysadmin, Linux Scripting etc. 
2. Databases - Oracle, mySQL etc. 
3. Security (Firewall, Network, Online Security etc) 
4. Storage in Linux 
5. Productivity (Too many technologies to explore, not much time available) 

到：

1. Solaris Sysadmin, Solaris Scripting etc. 
2. Databases - Oracle, mySQL etc. 
3. Security (Firewall, Network, Online Security etc) 
4. Storage in Solaris 
5. Productivity (Too many technologies to explore, not much time available) 

只需使用sed。 我得儘可能： sed的 ' /^＃[A-ZA-Z] */{ ^ h d } /Linux的/ G'

我想我需要做的是運行正則表達式使用保持緩衝區作爲替代模式，我想看起來像s/Linux/g但我找不到一種方法來做到這一點。 任何人有任何想法？

Answer 1

這裏是一個sed的解決方案：

sed ' 
/^#.*#$/{      # if replacement text is matchd 
s/#//g       # remove all pound chars 
h        # copy pattern space to hold space 
d        # delete pattern space and start over 
}        # end if 
:a        # create label "a" 
G        # append hold to pattern 
s/Linux\(.*\)\n\(.*\)/\2\1/ # replace Linux with contents of hold space 
ta        # if above replacement succeeded, go to "a" 
s/\n.*//      # remove replacement text and print pattern space 
' file.txt 

的一行，如果你可以使用;分隔命令：

sed '/^#.*#$/{s/#//g;h;d};:a;G;s/Linux\(.*\)\n\(.*\)/\2\1/;ta;s/\n.*//' file.txt 

Answer 2

你會更好的使用awk。考慮一下這個腳本：

awk '/^#/ {gsub("#", "", $0); txt=$0} {gsub("Linux", txt, $0); print}' file.txt