Using if elif fi in shell scripts [duplicate]

Answer 1

Use double brackets...

if [[ expression ]]

Answer 2

I have a sample from your code. Try this:

echo "*Select Option:*"
echo "1 - script1"
echo "2 - script2"
echo "3 - script3 "
read option
echo "You have selected" $option"."
if [ $option="1" ]
then
    echo "1"
elif [ $option="2" ]
then
    echo "2"
    exit 0
elif [ $option="3" ]
then
    echo "3"
    exit 0
else
    echo "Please try again from given options only."
fi

This should work. :)

Answer 3

This is working for me,

 # cat checking.sh
 #!/bin/bash
 echo "You have provided the following arguments $arg1 $arg2 $arg3"
 if [ "$arg1" = "$arg2" ] && [ "$arg1" != "$arg3" ]
 then
     echo "Two of the provided args are equal."
     exit 3
 elif [ $arg1 == $arg2 ] && [ $arg1 = $arg3 ]
 then
     echo "All of the specified args are equal"
     exit 0
 else
     echo "All of the specified args are different"
     exit 4
 fi

 # ./checking.sh
 You have provided the following arguments
 All of the specified args are equal

You can add set -x in script to troubleshoot the errors.

Answer 4

Change [ to [[, and ] to ]].

Answer 5

Josh Lee's answer works, but you can use the "&&" operator for better readability like this:

echo "You have provided the following arguments $arg1 $arg2 $arg3"
if [ "$arg1" = "$arg2" ] && [ "$arg1" != "$arg3" ]
then 
    echo "Two of the provided args are equal."
    exit 3
elif [ $arg1 = $arg2 ] && [ $arg1 = $arg3 ]
then
    echo "All of the specified args are equal"
    exit 0
else
    echo "All of the specified args are different"
    exit 4 
fi

Answer 6

sh is interpreting the && as a shell operator. Change it to -a, that’s [’s conjunction operator:

[ "$arg1" = "$arg2" -a "$arg1" != "$arg3" ]

Also, you should always quote the variables, because [ gets confused when you leave off arguments.