[SOLVED] BASH: if Variable -eq String not working

worm5252 · 01-24-2010, 02:59 PM

Here is the code in question

Code:

# Determine which release of Debian is
# being used.
CODENAME=`lsb_release -c`

# Backup /etc/apt/sources.list
mv /etc/apt/sources.list /etc/apt/sources.list.bak

# Install the correct /etc/apt/sources.list
# file.
LENNY='Codename:       lenny'

if [ "$CODENAME" -eq "$LENNY" ] ; then
        cp $SRCDIR/files/aptsources/lenny /etc/apt/sources.list
fi

How ever when I run this script I get the following error

Quote:

./test.sh: line 57: [: too many arguments

Line 57 is the line that reads if [ "$CODENAME" -eq "$LENNY" ] ; then

I just don't get it, I have racked my brain trying to figure out every combination of how I should write this if statement and I can't get it to work. What am I doing wrong and why is it wrong?

Thanks

GrapefruiTgirl · 01-24-2010, 03:04 PM

In the context in which you are using this code, the -eq operator is expecting integers. Just use an = sign instead.

worm5252 · 01-24-2010, 03:07 PM

DOH! It is always something simple. Thanks for pointing that out. Sometimes you just don't see the simplest errors and need a 2nd set of eyes.