awk里怎样做替换

树上开花

i have a problem about awk, thank you for your x!
he have a problem about awk, thank you for your y!
she have a problem about awk, thank you for your z!
我想在每一个有problem的行里面,在行末的x,y,z后面加上help
变成x_help,y_help,z_help,
我用过sub函数不行，大家帮忙一下，最好是用awk,用sed也行。

[ 本帖最后由 树上开花 于 2008-7-3 23:04 编辑 ]

jinl

sed -e '/problem/s/!$/_help!/' urfile

[ 本帖最后由 jinl 于 2008-7-3 23:43 编辑 ]

err7dd

sed '/problem/{s/\(.\)\(.\)$/\1_help\2/}' file

blackold

1. $ awk '/problem/{ if(/[xyz]!$/) sub(/!$/,"_help!"); print} !/problem/' urfile

复制代码

gawk:

1. $ awk '{ print gensub(/(.*problem.*[x-z])!$/,"\\1_help!","g")} ' urfile

复制代码

[ 本帖最后由 blackold 于 2008-7-4 09:00 编辑 ]

edei2004

sed '/\<problem\>/s/!$/_help&/g'  file

edei2004

nawk -F! '/problem/{$1=$1"_help!";print $0}' file