bash syntax

Sushy · 06-04-2007, 05:42 PM

I have simple bash script:

Code:

#!/usr/local/bin/bash
NAME=abc
IP=1.2.3.4
IFACE=eth0
echo "
# $NAME
test_$NAME_rootdir="/directory/$NAME"          # comment1
test_$NAME_hostname="$NAME.example.com"        # comment2
test_$NAME_ip="$IP"                           # comment3
test_$NAME_interface="$IFACE"                 # comment4
" > /test

I know, that something, somewhere is not correct with syntax, cos I have this output:

Code:

# abc
test_=/directory/abc          # comment1
test_=abc.example.com        # comment2
test_=1.2.3.4                           # comment3
test_=eth0                 # comment4

But I need:

Code:

# abc
test_abc_rootdir="/directory/abc"          # comment1
test_abc_hostname="abc.example.com"        # comment2
test_abc_ip="1.2.3.4"                      # comment3
test_abc_interface="eth0"                   # comment4

Enclosing characters in single quotes ' preserves the literal value of each character within the quote, so no $ possible, so this is I don`t need.
Enclosing characters in double quotes " preserves the literal value of all characters within the quotes, with the exception of `$',
so this is what I need, but anyway still some faults in output with _ and ""

ilikejam · 06-04-2007, 06:25 PM

Hi.

Try:

Code:

#!/usr/local/bin/bash
NAME=abc
IP=1.2.3.4
IFACE=eth0
echo "
# $NAME
eval test_${NAME}_rootdir="/directory/$NAME"          # comment1
eval test_${NAME}_hostname="$NAME.example.com"        # comment2
eval test_${NAME}_ip="$IP"                           # comment3
eval test_${NAME}_interface="$IFACE"                 # comment4
" > /test

Also, you might want to escape the '"'s inside the echo.

Dave