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JWilliams · #1 (**permalink**) 04-23-2007

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.

ghostdog74 · #2 (**permalink**) 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

cfajohnson · #3 (**permalink**) 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%.*}

reborg · #4 (**permalink**) 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

napster_san · #5 (**permalink**) 04-24-2007

echo transfer_ccf_3731_10| cut -d_ -f3-

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