shell脚本读取配置文件 有点挑战

cyr1974

一个配置文件如下，想写一个shell script把它读到shell相关变量里，一个难点是，两个部分 common 和 special 有同名的配置项，不怎道怎样去区分？大家有什么好的方法？

[Common]
ClassName = xfer
InstanceName = hua1
LocalHost = 127.0.0.1
Host = 10.70.1.1
#Host = 127.0.0.1
TaskNum =
CheckSize=T
[special]
InstanceName = hua3
Host = 10.70.1.2

wtuter

[W]awk -v RS='[' 'NR==2{print $0}' 3
Common]
ClassName = xfer
InstanceName = hua1
LocalHost = 127.0.0.1
Host = 10.70.1.1
#Host = 127.0.0.1
TaskNum =
CheckSize=T

[W]awk -v RS='[' 'NR==3{print $0}' 3
special]
InstanceName = hua3
Host = 10.70.1.2

[W]cat 3
[Common]
ClassName = xfer
InstanceName = hua1
LocalHost = 127.0.0.1
Host = 10.70.1.1
#Host = 127.0.0.1
TaskNum =
CheckSize=T
[special]
InstanceName = hua3
Host = 10.70.1.2


可以这样分,分完了再在{print $0}里面匹配你要的内容

ywlscpl

以Common为例

1. awk -F "\n" -v RS="[" '$1~/^Common/{for (i=2;i<NF;i++) {sub(
   
2. / = /,"=",$i);print $i}}' file >out
   
3. source out

复制代码

[ 本帖最后由 ywlscpl 于 2009-7-1 10:22 编辑 ]

alexyandy_unix

学习一下，3楼的我已经看明白了，2楼的还没看明白~继续理解

00306

刘德华的那句看得更明白些

sania9

LS让我笑了，刘德华.....

南无小和尚

这个我好像运行的错误