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Home/ Questions/Q 1067549
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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T20:10:19+00:00 2026-05-16T20:10:19+00:00

I am trying to use the simple fuinction below. But i get error sayin

  • 0

I am trying to use the simple fuinction below. But i get error sayin unary oprator expected and the output is always one. Can any1 help me correct it.

#!/bin/bash
checkit ()
{
if [ $1 = "none" ]
then
     echo "none"
else
     echo "one"
fi
}
checkit
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  1. Editorial Team
    Editorial Team
    2026-05-16T20:10:20+00:00Added an answer on May 16, 2026 at 8:10 pm

    $1 is an argument to the entire script and not to the function checkit(). So send the same argument to the function too.

    #!/bin/bash
    checkit ()
    {
    if [ $1 = "none" ]
    then
         echo "none"
    else
         echo "one"
    fi
    }
    
    checkit $1
    

    This has to work.

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