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In my script, I transform data to a more readable format. But I am having an issue. If no arguments are typed, just the script, I want it to come back "Please enter a filename" which I have done. But for some reason, it will not display that when tested. I did it by telling the system if argument is greater than 0, execute, otherwise, echo the message.

*I am not looking for advice on the substitution code. For specific reasons, it cannot be changed, so please do not give advice on that piece. 

for myvar
do

        if [ $# -gt 0 ]
        then

                        if [ -e "$myvar" ]
                        then :
                        else
                                echo "$myvar cannot be found."
                        exit
                        fi

                echo "This is the filname:" $myvar 

                cut -f1 -d, $myvar > social_security
                cut -f2 -d, $myvar > last_name
                cut -f3 -d, $myvar > first_name
                sed 's/^ *//' last_name > last_name_draft
                sed 's/^ *//' first_name > first_name_final
                sed 's/-//' social_security > social_security_draft
                sed 's/-//' social_security_draft > social_security_final
                sed 's/$/:/' last_name_draft > last_name_final
                paste -d\  first_name_final last_name_final social_security_final  > final_draft
                sed '1d' final_draft > final
                cat final

                echo -en '\n'   

        else
                echo "Please enter a filename."
        fi
done
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2
  • you tags are wrong Commented Jul 19, 2015 at 17:04
  • 2
    @mikeserv I would appreciate a little more elaboration. Commented Jul 19, 2015 at 17:23

2 Answers 2

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2

Think about the first line of your script: for myvar. How many times will this for loop be executed if there are no arguments?

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  • Ya but regardless of the loop it should still display the else would it not? Commented Jul 19, 2015 at 17:43
  • The if/else is inside the for loop. So if the for loop is never executed, neither is the if or the else. Commented Jul 19, 2015 at 17:47
  • So the loop needs to be in the if/else Commented Jul 19, 2015 at 17:51
-1

Yes, you should first check if there are any arguments or not and then proceed.

if [ $# -gt 0 ]
then
    for myvar
    do
        if ! [ -e "$myvar" ]
            echo "$myvar cannot be found."
            exit
        fi

        echo "This is the filname:" $myvar 

        cut -f1 -d, $myvar > social_security
        cut -f2 -d, $myvar > last_name
        cut -f3 -d, $myvar > first_name
        sed 's/^ *//' last_name > last_name_draft
        sed 's/^ *//' first_name > first_name_final
        sed 's/-//' social_security > social_security_draft
        sed 's/-//' social_security_draft > social_security_final
        sed 's/$/:/' last_name_draft > last_name_final
        paste -d\  first_name_final last_name_final social_security_final  > final_draft
        sed '1d' final_draft > final
        cat final

        echo -en '\n'   
    done
else 
    echo "Please enter a filename."
fi
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