mjhjmtu

I was testing a Bash script that accepts random arguments and noticed that it would say something like

./Var: line 19: [three: command not found

Here is a minimal working example:

#!/bin/bash
$1 $2 $3

echo "The first argument does $1"
if [$1 >= 2]; then
 echo "$1 has 2 words"
else
 echo "$1 has unknown amount of words"
fi
#^First
echo "The second argument does $2"
if [$2 >= 2]; then
 echo "$2 has 2 words"
else
 echo "$2 has unknown amount of words"
fi
#^Second
echo "The third argument does $3"
if [$3 >= 2]; then
 echo "$3 has 2 words"
fi
#^Third

But it would continue to run the script, is there a way i can have it run without the "command not found" appearing? Or is this just an issue within my code that is making it do this?

The command you want to use is [ (or test), not [whatever, so it should be (for example)

[ "$3" -ge 2 ]

instead of [$3 >= 2], which generates [three: command not found if $3 is three. Additionally:

• quote your variables;


• 
  [ doesn't understand >=, it understands -ge and other options;


• 
  >= is a redirection anyway and it didn't get to the command, regardless if the command was [ or [three. You now have a file named =.

In Bash [ is a builtin. See help [, help test.