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compare variables in UNIX shell

I am trying to compare two variables in shell script, but it does not work although the code looks simple.
Basically the idea is to determine when an empty string is entered by a user and display a warning.
This is the code:
#!/bin/sh
echo "Enter value A:"
read VALUEA
echo "$VALUEA"

echo "Enter VALUE B:"
read VALUEB
echo "$VALUEB"

X="5"
echo "$X"

if [ "$X$VALUEA"="$X" ]; then
echo "You did not enter Value A. Please try again $X$VALUEA and $X"
else
echo "not empty"
fi

if [ "$X$VALUEB"="$X" ]; then
echo "You did not enter ending date. Please try again $X$VALUEB and $X"
else
echo "not empty"
fi

exit
---------------------------
The problem is that regardless whether or not I enter a value A or B, it always displays:
You did not enter beginning date. Please try again 5r and 5
You did not enter ending date. Please try again 5 and 5

("r" is a value I entered for the value A. I used that as a debugging display).

Please assist.

pavelmed

ASKER

Please note: in the code below, the displayed message is "You did not enter Value A" and "You did not enter Value B".
I copied and pasted the code, so forgot to change that.
Thanks.

pavelmed

ASKER

I meant, "in the code above" ...

ASKER CERTIFIED SOLUTION

ozo

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Tintin

Much easier to do

if [ -z "$VALUEA" ]
then
      echo "You did not enter VALUE A"
else
     echo "VALUE A is set to $VALUEA"
fi

Open in new window

or if you want to reverse the logic

if [ -n "$VALUEA" ]
then
       echo "VALUE A is set to $VALUEA"
else
     echo "You did not enter VALUE A"
fi

Open in new window

pavelmed

ASKER

Thank you very much ozo!