Hi`
I have a difficulties to find the corect patern for the following string value:

"20030815/ABC.SS20_0.dat"


I d like to get word SS20_0,however i still could not get it work.
Do you guys know what is the correct pattern of the regex for my purpose?

i use

echo $myvar | sed -e 's/myfailedpattern//g'



Thanks

Am I missing something ?

$ export myvar="20030815/ABC.SS20_0.dat"
$ echo $myvar
20030815/ABC.SS20_0.dat
$ echo $myvar | sed -e 's/SS20_0//g'
20030815/ABC..dat
$

Cheers

Sathyaram

You want to strip out the first part and the last part and end up with SS20_0?

You need a regex for the first part - I'm lazy, so I'll use: ^.*
A regex for the part you want: SS20_0
And a regex for the last part: \.dat

Hold the part you want in a buffer using \( ... \) which in this instance will be the first buffer \1.

Substitute that whole lot for the string in the buffer...


echo 20030815/ABC.SS20_0.dat | sed 's/^.*\(SS20_0\)\.dat/\1/'

Probably a little clearer if we use a different separator...

echo 20030815/ABC.SS20_0.dat | sed 's#^.*\(SS20_0\)\.dat#\1#'

Hello,

just a try...

echo 20030815/ABC.SS20_0.dat | cut -f2 -d.

it will give u :
SS20_0

Hope this will help,
Pooja

maybe you can use awk to try it.

export myvar="20030815/ABC.SS20_0.dat"

echo $myvar |awk -F\. '{print $2}'

you can get SS20_0