sed难题

__lxmxn__

文件file，内容如下：
开头省略N行……
parameters dstport=80
servername pp.com
listenport 8080
docroot /var/www/htdocs/pp
.....


现在想用sed实现：

找到8080的行，添加内容相同的一行到此行下面，但8080被替换成80，howto？

变成：
parameters dstport=80
servername pp.com
listenport 8080
listenport 80
docroot /var/www/htdocs/pp

Perl_Er

回复 1# __lxmxn__


   

5.5 Append, Insert, and Change
The append (a), insert (i), and change (c) commands provide editing functions that are commonly performed with an interactive editor, such as vi. You may find it strange to use these same commands to "enter" text using a noninteractive editor. The syntax of these commands is unusual for sed because they must be specified over multiple lines. The syntax follows:

append [line-address]a\
text
insert [line-address]i\
text
change [address]c\
text

你的这个就是:

/listenport 8080/a\
listenport 80

lululau

1. 
   
2. 1!{x;/8080/s/8080/80/gp;x};h;
   

复制代码

bbgg1983

1. $ sed '/8080/a\listenport 80' file
   
2. parameters dstport=80
   
3. servername pp.com
   
4. listenport 8080
   
5. listenport 80
   
6. docroot /var/www/htdocs/pp

复制代码

blackold

iori809

[root@commserver: /tmp]#sed 's/8080$/&\nlistenport 80/' test117.txt
dfsdfsad
dfdf
parameters dstport=80
servername pp.com
listenport 8080
listenport 80
docroot /var/www/htdocs/pp

shileiadmin

一楼问的是下面的第四行为什么会是listenport 80
parameters dstport=80
servername pp.com
listenport 8080
listenport 80
docroot /var/www/htdocs/pp

而真正想要的结果是：
parameters dstport=80
servername pp.com
listenport 8080
listenport 8080
docroot /var/www/htdocs/pp

Shell_HAT

最好别把listenport写死
3楼的漂亮

laohuanggua

sed我还没入门
用awk写了一个
awk '/8080/{a=$0;sub("8080","80",a);print $0"\n"a;next}1'
99超人 发表于 2010-06-10 08:13

我的和你类似
cat datafile2
parameters dstport=80
servername pp.com
listenport1 8080
docroot /var/www/htdocs/pp
listenport2 8080
listenport 8081
awk '1;/8080/{gsub(8080,80);print}' datafile2
parameters dstport=80
servername pp.com
listenport1 8080
listenport1 80
docroot /var/www/htdocs/pp
listenport2 8080
listenport2 80
listenport 8081