求助，急

qzhu

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

dbcat

1. 
   
2. sed '
   
3. s/ 0//g
   
4. /\//h
   
5. s/^\/.*//
   
6. /\//!{
   
7. G
   
8. s/\n//
   
9. s/^\/.*//
   
10. s/\(.*\)\/\(.*\)/\2 \1/
    
11. }
    
12. ' file

复制代码

dbcat

修改一下，有点错误：

1. sed '
   
2. /^$/d
   
3. /^\//{
   
4. h
   
5. D
   
6. }
   
7. {
   
8. G
   
9. s/\n//
   
10. s/\([^ ]*\)\(.*\)\(.\)\/\(.*\)/\4 \1 \3/
    
11. }
    
12. ' file
    

复制代码

qzhu

能不能简单解释一下h,D, G 以及最后一行的含义，谢谢

寂寞烈火

http://bbs.chinaunix.net/viewthread.php?tid=336126
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qzhu

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