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Cursos7 Team
4,745 PointsI have questions to resolve this issue, you can give a tip?
<?php
$flavor = 'vanilla'; echo $flavor;
<?php
$flavor = 'vanilla';
$flavor;
?>
2 Answers
Marileen Mennerich
1,161 PointsWhich step of the challenge are you having troubles with? If it is the 2nd, make sure you do not delete all of the preexisting code. Just replace the plain text flavor with your variable, then it should work out.
Cristian Altin
12,165 PointsProblem is that you deleted the rest of the code. You need to have $flavor = "vanilla"; at line 2 ,then you need to change only line 4 which is the static text of what by default is "your" fav ice cream flavor.
Keep the other lines since the exercise will build on that code.
Btw, don't forget echo.
Cursos7 Team
4,745 PointsChange the value in the flavor variable to cookie dough. Preview the code and make sure the message appears.
<?php
$flavor = "chocolate"; echo "<p>Your favorite flavor of ice cream is "; echo $flavor; echo ".</p>";
if($flavor == "vanilla"){
echo "<p>Randy's favorite flavor is cookie dough, also!</p>"; } ?>
It's wrong. Why?
Cristian Altin
12,165 PointsAlways read the exercise at least twice and start writing when you get all the requirements.
It's telling you to "change the value in the flavor variable" to "cookie dough". Where did you set the $flavor variable? There you need to change it to the new text. ;-)
Oh and be sure to test the variable with the new flavor in the if where vanilla is now outdated, hehe.
Cursos7 Team
4,745 PointsCursos7 Team
4,745 PointsI'm in this question. Modify the command that displays the flavor of ice cream. Instead of it displaying a static piece of text, change it to display the value stored in the 'flavor' variable.