Why is the -z to check if the variable is an string is not working on below script

Shell Scripting
#1

Script is working but is not validating if is string or not, instead it goes to the second if and if i type an string it will tell me (“Invalid month number given. Please enter a valid number - 1 to 12.”) instead of (“No month number given. Please enter a month number as a command line argument.”)

month_number=$1

if [ -z $month_number ]

then

echo “No month number given. Please enter a month number as a command line argument.”

echo “eg: ./print-month-number 5”

exit

fi

if [[ $month_number -lt 1 ]] || [[ $month_number -gt 12 ]]

then

echo “Invalid month number given. Please enter a valid number - 1 to 12.”

exit

fi

if [ $month_number -eq 1 ]

then

echo “January”

elif [ $month_number -eq 2 ]

then

echo “February”

elif [ $month_number -eq 3 ]

then

echo “March”

elif [ $month_number -eq 4 ]

then

echo “April”

elif [ $month_number -eq 5 ]

then

echo “May”

elif [ $month_number -eq 6 ]

then

echo “June”

elif [ $month_number -eq 7 ]

then

echo “July”

elif [ $month_number -eq 8 ]

then

echo “August”

elif [ $month_number -eq 9 ]

then

echo “September”

elif [ $month_number -eq 10 ]

then

echo “October”

elif [ $month_number -eq 11 ]

then

echo “November”

elif [ $month_number -eq 12 ]

then

echo "December"l

fi

#2

Hi,

-z does not test whether the value is a string. It tests whether the value is zero length.

./print-month-number

With nothing after print-month-number then $1 is a zero-length string, and hence so is $month_number by assignment from $1 and thus -z is true and it will print “No month number given. Please enter a month number as a command line argument.” and exit

If you enter a string, e.g.

./print-month-number aaa

then the second if wil be true and you will get “Invalid month number given. Please enter a valid number - 1 to 12.”. The second if checks for the input being a number between 1 and 12

1 Like
#3

From Bash Reference Manual

-z string

True if the length of string is zero.
1 Like
#4

thanks for you time and the response.

#5

Thanks for the info, i was looking for a condition that capture only integer and if string then give you the message that is not an string, but thanks i see -z is not what i thought.

#6

Bash variables are effectively all strings, but if the string is actually an integer number, you can perform numerical operations on it, like -lt, -gt, -eq and shell arithmetic e.g.

x=1
y=2
z=$(($x + $y))
echo $z

3

#7

@Ernesto-Diaz ,

This might help, linux - How do I test if a variable is a number in Bash? - Stack Overflow