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Bashing Your Head in: Local Is a Command

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Bash has a major conceptional drawback. No, I’m not talking about odd syntax, overuse of whitespace delimiters or even the slow performance. Nearly everything in the language is a command. Bash works really well for executing simple Unix commands, parsing the output, and acting on their return codes, but it sometimes returns confusing results when you fail to remember that it’s own builtin commands use this same system for reporting output and results.

For instance, this bit of code is a clean (even if contrived) way to parse a syslog file for all executions of the CRON subsystem on Ubuntu. Quick and readable, runnable at the command line, and just makes sense:

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output=`cat /var/log/syslog | grep CRON | wc -l`
if [ $? -eq 0 ]; then
    echo "Found $output cron log line(s)"
else
    echo "No cron log lines found"
fi

Notice that I am capturing the output of that command into a variable named output. Ideally, you would want that output variable to be local, since output is a common word, and you could be trampling on some other code in a sourced file, or even in your own file. So, let’s do this:

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local output=`cat /var/log/syslog | grep CRON | wc -l`

There! All fixed. Except, it’s not. You probably see the problem, the local builtin is actually a command whose status is evaluated after the execution of your log parsing. Since local output will succeed in pretty much every case, the result of the last command ($?) will always evaluate to 0.

The best practice is always to

  1. Use Local variables
  2. Declare local variables on their own line

So our silly example becomes:

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local output
output=`cat /var/log/syslog | grep CRON | wc -l`
if [ $? -eq 0 ]; then
    echo "Found $output cron log line(s)"
else
    echo "No cron log lines found"
fi

Happy Bashing!

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