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awk: isolate a part of a file name

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# 1  
Old 12-16-2010
awk: isolate a part of a file name
hi there,

i have a file named 'x20080613_x20100106.pwr1.gc', i want to isolate the part 'x20080613_x20100106' but by using the following line i isolate the part '.pwr1.gc':

HTML Code:
`awk '$0=substr($0, length($0)-7)' $temp`
how can i reverse that?
thank you!
# 2  
Old 12-17-2010
So many ways to do this:

With awk:
Code:
 echo 'x20080613_x20100106.pwr1.gc' | awk '$0=substr($0, 0, length($0)-8)'

With ksh/bash
Code:
$ F='x20080613_x20100106.pwr1.gc'; echo ${F%%.*}

With sed
Code:
echo 'x20080613_x20100106.pwr1.gc' |  sed 's/........$//'

With sed and extended regular expressions
Code:
echo 'x20080613_x20100106.pwr1.gc' |  sed -r 's/.{8}$//'

# 3  
Old 12-17-2010
An easier way using awk as you requested:
Code:
echo 'x20080613_x20100106.pwr1.gc' | awk -F. '{ print $1 }'

Or to follow the best practice of using the most simple tool for the job, use parameter expansion as mentioned by Chubler_XL:
Quote:
$ F='x20080613_x20100106.pwr1.gc'; echo ${F%%.*}
Another easy way is with cut:
Code:
echo 'x20080613_x20100106.pwr1.gc' | cut -d. -f1

# 4  
Old 12-17-2010
Bug
thank you!
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