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Old 01-24-2010, 02:59 PM   #1
Registered: Oct 2004
Location: Atlanta
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Question BASH: if Variable -eq String not working

Here is the code in question

# 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 $SRCDIR/files/aptsources/lenny /etc/apt/sources.list
How ever when I run this script I get the following error
./ line 57: [: too many arguments
Line 57 is the line that reads if [ "$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?

Old 01-24-2010, 03:04 PM   #2
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In the context in which you are using this code, the -eq operator is expecting integers. Just use an = sign instead.
Old 01-24-2010, 03:07 PM   #3
Registered: Oct 2004
Location: Atlanta
Distribution: CentOS, RHEL, HP-UX, OS X
Posts: 567

Original Poster
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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.


bash, if, string, variable

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