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Simon DICKENS
1,325 PointsThe message in the final echo command only makes sense if your favorite flavor is the same as mine. Add a conditional around that final echo command that checks if the flavor variable has a value of "cookie dough." (Remember to choose carefully between us
<?php $flavor = "vanilla"; echo "<p>Your favorite flavor of ice cream is "; echo "vanilla"; echo ".</p>"; echo "<p>Randy's favorite flavor is cookie dough, also!</p>"; if ($flavor == "cookie dough") { echo "on"; }; ?>
two hours ive read this question and changed things around..still cant get it!!
HELP PLEASE!
Simon
1 Answer
Angel Cepeda
3,735 PointsYou need to wrap the final echo statement with an 'if' statement. Also, you need to echo your $flavor variable, not "vanilla" on the second echo. The problem with echoing a string (i.e. vanilla) vs. your variable ($flavor), is that when you update your favorite flavor, your program will still echo vanilla. So your code should read like this
<?php $flavor = "vanilla";
echo "Your favorite flavor of ice cream is ";
echo $flavor;
echo ".";
if ($flavor == "cookie dough"){
echo "Randy's favorite flavor is cookie dough, also!";
}
?>