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标题: 求助，急 [打印本页]

作者: qzhu 时间: 2006-03-21 09:27
标题: 求助，急
Hi,

I need to parse a file to a new format:

source file format is like this:
/ABC
abc 0 0 0 0 0 1
efg 0 0 0 0 0  3

/XYZ
cde 0 0 0 0 0 1
fgh 0 0 0 0 0  1
uvw 0 0 0 0 0 2

......

need to parse it to a new file like this:
ABC abc  1
ABC efg  3
XYZ cde  1
XYZ fgh 1
XYZ uvw  2
...

basically what need to be done is to get the name after the slash "/" and  combine it with the first and last column of lines immediately after it.
I know a little bit of awk and sed but not enough to solve this problem, any help or hint with a piece of code will be appreciated.

Thanks,
Qing

作者: dbcat 时间: 2006-03-21 09:58

sed '
s/ 0//g
/\//h
s/^\/.*//
/\//!{
G
s/\n//
s/^\/.*//
s/$.*$\/$.*$/\2 \1/
}
' file

复制代码

作者: dbcat 时间: 2006-03-21 12:10

修改一下，有点错误：

sed '
/^$/d
/^\//{
h
D
}
{
G
s/\n//
s/$[^ ]*$$.*$$.$\/$.*$/\4 \1 \3/
}
' file

复制代码

作者: qzhu 时间: 2006-03-22 02:15
能不能简单解释一下h,D, G 以及最后一行的含义，谢谢

作者: 寂寞烈火 时间: 2006-03-22 02:18
http://bbs.chinaunix.net/viewthread.php?tid=336126
BTW:注意论坛发贴要求

作者: qzhu 时间: 2006-03-22 08:46
还是awk简单一点
NF == 1 { name=$1}
NF > 1 {print name, $1, $NF}

然后再把所有的"/"去掉就行了

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