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# 1  
Old 04-23-2007
Trim

Hello,

I am passing a filename to a script to draw parameters from it. However, I want to use part of the filename as a parameter. The filename is transfer_ccf_3731_10.sh but I only need the 3731_10 part of it.

Is this possible? Any help or suggestions would be appreciated!

Regards,

J.
# 2  
Old 04-23-2007
you can use bash's internal parameter expansion, check the bash man page, or tools like awk/sed.
eg awk
Code:
# echo "transfer_ccf_3731_10.sh" | awk -F "_" '{print $3"_"substr($4,-1,2)}'
3731_10

combination eg:
Code:
# var="transfer_ccf_3731_10.sh"
# echo ${var%.sh}
transfer_ccf_3731_10
# echo ${var%.sh} | awk -F "_" '{print $3"_"$4}'
3731_10

# 3  
Old 04-23-2007
Quote:
Originally Posted by JWilliams
I am passing a filename to a script to draw parameters from it. However, I want to use part of the filename as a parameter. The filename is transfer_ccf_3731_10.sh but I only need the 3731_10 part of it.

Use the shell's parameter expansion:

Code:
filename=transfer_ccf_3731_10.sh
temp=${filename%%[0-9]*}
temp=${filename#"$temp"}
extract=${temp%.*}


Last edited by cfajohnson; 04-23-2007 at 06:44 PM..
# 4  
Old 04-23-2007
Quote:
Originally Posted by JWilliams
Hello,
....The filename is transfer_ccf_3731_10.sh but I only need the 3731_10 part of it.
Code:
# var=transfer_ccf_3731_10.sh
# OIFS="$IFS"
# IFS="[_.]"
# set -- $var
# part="$3_$4"
# IFS="$OIFS"
# echo $part
3731_10

# 5  
Old 04-24-2007
echo transfer_ccf_3731_10| cut -d_ -f3-
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